A hostel has food for 60 days if after 15 days 500 more students joined th hostel and the food lasted for 40days only. How many students were there in hostel

The slope of the line and describe is the negative 0.5 which means in this situation will be negative 0.5 miles per minute.

The slope is the ratio of rising or falling and running. The difference between the ordinate is called rise or fall, and the difference between the abscissa is called run.

Joseph drove from his summer home to his place of work.

To avoid the road construction, Joseph decided to travel the gravel road.

After driving 20 minutes, he was 62 miles away from work, and after driving 40 minutes, he was 52 miles away from work.

This situation is shown in the graph below.

Then the slope of the line and describe what it means in this situation will be

Slope = (52 – 62) / (40 – 20)

Slope = – 10 / 20

Slope = – 0.5 miles per minute

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Answer:

1.a.  skew

1.b.  parallel

1.c.  none of the above

1.d.  perpendicular

1.e.  parallel

1.f.  skew

2.a.  never

2.b.  always

2.c.  always

2.d.  sometimes, sometimes, sometimes

Step-by-step explanation:

For these questions, it is critical to know the definitions of each of the terms.

Problem Breakdown

1.a.  skew  [tex]\overset{\longleftrightarrow}{AB} \text{ and } \overset{\longleftrightarrow}{EF}[/tex]

Point E is not on line AB.  Any three non-linear points form a unique plane (left face), so A, B, and E are coplanar.  Point F is not in plane ABE.  Since line EF and line AB do not intersect and are not coplanar, they are skew.

1.b.  parallel [tex]\overset{\longleftrightarrow}{CD} \text{ and } \overset{\longleftrightarrow}{EH}[/tex]

Line CD is parallel to line AB, and line AB is parallel to line EH.  Parallel lines have a sort of "transitive property of parallelism," so any pair of those three lines is coplanar and parallel to each other.

1.c.  none of the above  [tex]\overset{\longleftrightarrow}{AC} \text{ and } \overset{\longleftrightarrow}{GA}[/tex]

Line AC and line GA share point A, necessarily intersecting at A.  Therefore, line AC and line GA cannot be parallel or skew.

Any three points form a unique plane, so these two lines are coplanar, however, do they form a right angle?  If the figure is cube-shaped, as depicted, then no.

Note that each line segment AC, AG, and CG are all the diagonal of a cube face.  A cube has equal edge lengths, so each of those diagonals would be equal.  Thus, triangle ACG is an equilateral triangle.

All equilateral triangles are equiangluar, and by the triangle sum theorem, the measures of the angles of a planar triangle must sum to 180°, forcing each angle to have a measure of 60° (not a right angle).  So, lines AC and GA are not perpendicular.  (If the shape were a rectangular prism, these lines still aren't perpendicular, but the proof isn't as neat, there is a lot to discuss, and a character limit)

"none of the above"

1.d.  perpendicular  [tex]\overset{\longleftrightarrow}{DG} \text{ and } \overset{\longleftrightarrow}{HG}[/tex]

Line DG and line HG share point G, so they intersect.  Both lines are the edges of a face, so they intersect at a right angle.  Perpendicular

1.e.  parallel  [tex]\overset{\longleftrightarrow}{AC} \text{ and } \overset{\longleftrightarrow}{FH}[/tex]

Line AC is a diagonal across the top face, and line FH is a diagonal across the bottom face.  They are coplanar in a plane that cuts straight through the cube from top to bottom, but diagonally through those faces.

1.f.  skew  [tex]\overset{\longleftrightarrow}{CD} \text{ and } \overset{\longleftrightarrow}{AG}[/tex]

Point A is not on line CD, and any three non-linear points form a unique plane (the top face), so C, D, and A are coplanar.  Point G is not in plane ACD. Since line CD does not intersect and is not coplanar with line AG, by definition, the lines are skew.

2.a.  Two lines on the top of a cube face are never skew

Two lines are on a top face are necessarily coplanar since they are both in the plane of the top face.  By definition, skew lines are not coplanar.  Therefore, these can never be skew.

2.b.  Two parallel lines are always coplanar.

If two lines are parallel, by definition of parallel lines, they are coplanar.

2.c.  Two perpendicular lines are always coplanar

If two lines AB and CD are perpendicular, they form a right angle.  To form a right angle, they must intersect at the right angle's vertex (point P).

Note that A and B are unique points, so either A or B (or both) isn't P; similarly, at least one of C or D isn't P.  Using P, and one point from each line that isn't P, those three points form a unique plane, necessarily containing both lines.  Therefore, they must always be coplanar.

2.d.  A line on the top face of a cube and a line on the right side face of the same cube are sometimes parallel, sometimes skew, and sometimes perpendicular

Consider each of the following cases:

These 4 cases prove it is possible for a pair of top face/right face lines to be parallel, perpendicular, skew, or none of the above.  So, those two lines are neither "always", nor "never", one of those choices.  Therefore, they are each "sometimes" one of them.

Answer:

  A. single triangle. C = 8.74°, A = 1.26°, a = 1.01

Step-by-step explanation:

When two sides and an angle are given, the possibility of two (or zero) triangles exists only when the given angle is opposite the shorter of the two given sides. That is not the case here, so the given measures define one unique triangle.

The law of sines tells us the sides and angles have the relation ...

  sin(A)/a = sin(B)/b = sin(C)/c

We can use this first to find angle C:

  sin(C) = c/b·sin(B)

  C = arcsin(c/b·sin(B)) = arcsin(7/8·sin(170°)) ≈ 8.7394°

  ∠C ≈ 8.74°

Now that we know two angles, we can find the third:

  A = 180° -B -C = 180° -170° -8.74° = 1.26°

  ∠A = 1.26°

Using the law of sines again, we can find the measure of side 'a'.

  a = b·sin(A)/sin(B) = 8·sin(1.26°)/sin(170°) ≈ 1.01346

The measure of segment 'a' is about 1.01 units.

__

Additional comment

If 'a', 'b', and angle A are given, there will be zero triangles if b/a·sin(A) > 1. If b/a·sin(A) < 1 and b > a, there will be two (2) triangles. Otherwise, as here, there will be one unique triangle.

To solve this problem, you will use the distributive property to create an equation that can be rearranged and solved using the quadratic formula.

Use the distributive property to distribute 3x into the term (x + 6):

[tex]3x(x+6)=-10[/tex]

[tex]3x^2+18x=-10[/tex]

To create a quadratic equation, add 10 to both sides of the equation:

[tex]3x^2+18x+10=-10+10[/tex]

[tex]3x^2+18x+10=0[/tex]

The quadratic formula is defined as:

[tex]\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]

The model of a quadratic equation is defined as ax² + bx + c = 0. This can be related to our equation.

Therefore:

Set up the quadratic formula:

[tex]\displaystyle x=\frac{-18 \pm \sqrt{(18)^2 - 4(3)(10)}}{2(3)}[/tex]

Simplify by using BPEMDAS, which is an acronym for the order of operations:

Brackets

Parentheses

Exponents

Multiplication

Division

Addition

Subtraction

Use BPEMDAS:

[tex]\displaystyle x=\frac{-18 \pm \sqrt{324 - 120}}{6}[/tex]

Simplify the radicand:

[tex]\displaystyle x=\frac{-18 \pm \sqrt{204}}{6}[/tex]

Create a factor tree for 204:

204 - 1, 2, 3, 4, 6, 12, 17, 34, 51, 68, 102 and 204.

The largest factor group that creates a perfect square is 4 × 51. Therefore, turn 204 into 4 × 51:

[tex]\sqrt{4\times51}[/tex]

Then, using the Product Property of Square Roots, break this into two radicands:

[tex]\sqrt{4} \times \sqrt{51}[/tex]

Since 4 is a perfect square, it can be evaluated:

[tex]2 \times \sqrt{51}[/tex]

To simplify further for easier reading, remove the multiplication symbol:

[tex]2\sqrt{51}[/tex]

Then, substitute for the quadratic formula:

[tex]\displaystyle x=\frac{-18 \pm 2\sqrt{51}}{6}[/tex]

This gives us a combined root, which we should separate to make things easier on ourselves.

Separate the roots at the plus-minus symbol:

[tex]\displaystyle x=\frac{-18 + 2\sqrt{51}}{6}[/tex]

[tex]\displaystyle x=\frac{-18 - 2\sqrt{51}}{6}[/tex]

Then, simplify the numerator of the roots by factoring 2 out:

[tex]\displaystyle x=\frac{2(-9 + \sqrt{51})}{6}[/tex]

[tex]\displaystyle x=\frac{2(-9 - \sqrt{51})}{6}[/tex]

Then, simplify the fraction by reducing 2/6 to 1/3:

[tex]\boxed{\displaystyle x=\frac{-9 + \sqrt{51}}{3}}[/tex]

[tex]\boxed{\displaystyle x=\frac{-9 - \sqrt{51}}{3}}[/tex]

The final answer to this problem is:

[tex]\displaystyle x=\frac{-9 + \sqrt{51}}{3}[/tex]

[tex]\displaystyle x=\frac{-9 - \sqrt{51}}{3}[/tex]

The length of the infinite zigzag path is 1

Note that the infinite zigzag path is embedded in an isosceles triangle

Using angle π/4, we have that

Tan π/4 = opposite/ adjacent

[tex]tan \frac{3. 412}{4} = \frac{x}{1}[/tex]

[tex]tan 0. 7855 = x[/tex]

[tex]x = 0. 014[/tex]

To find the longer part 'y',  use the Pythagorean theorem

[tex]y^{2} = (0.014)^2 + (1)^2[/tex]

[tex]y^2 = 1.879 * 10^-4 + 1[/tex]
[tex]y^2 = 1. 00[/tex]

[tex]y = \sqrt{1. 000}[/tex]

[tex]y = 1. 000[/tex]

The length of the infinite zigzag path is the same as the length of the longer part of the triangle which equals 1.

Thus, the length of the infinite zigzag path is 1

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Answer:

75%

Step-by-step explanation:

48x=36

x=36/48

x=3/4

x=0.75=75%

Step-by-step explanation:

[tex]t = k(r)( {q}^{2} )[/tex]

[tex]320 = k(5)( {4}^{2} )[/tex]

[tex]320 = 80k[/tex]

[tex]k = 4[/tex]

[tex]196 = 4(1)( {q}^{2} )[/tex]

[tex]49 = {q}^{2} [/tex]

[tex]q = 7[/tex]

According to the information in the table, it can be inferred that the trend of the results is positive but does not have an exact linear fit (option B).

To calculate the average of the results we must add all the results and divide the result by the number of years as shown below.

Additionally, to identify if the trend is positive or negative, we must see if the results increase or decrease year after year. From the above, it is possible to infer that it is a positive trend.

On the other hand, it can be stated that it does not have an exact linear fit because the increase is not constant each year.

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Answer:

Step-by-step explanation:

You are given a system of equations and asked to solve it by the method of addition. That method requires you add a multiple of one equation to the other so that one of the variables is eliminated. In some cases, this is easier if multiples of both equations are added together. The resulting single-variable equation is then solved in the usual way.

__

The given system of equations is ...

We notice the first equation has the variables on opposite sides of the equal sign, and both their coefficients are 1. The second equation has the variables on the same side of the equal sign, and their coefficients are 0.08 and 0.02.

To eliminate a variable by the "addition method," we need to have the variable on the same side of the equal sign with opposite coefficients. Or, we need to have the variable on opposite sides of the equal sign with the same coefficient.

Both of the x-variables are on the left side, so we need opposite coefficients. We can get that by multiplying the first equation by -0.08, or by multiplying the second equation by -12.5. We judge the first of these choices to be easier.

The y-variables are on opposite sides of the equal sign, so we need equal coefficients. We can get that by multiplying the first equation by 0.02, or the second equation by 50.

We choose to multiply the first equation by 0.02, so we can eliminate the y-variable. Here is the result of doing that, then adding the results

  (0.02)(x) +(0.08x +0.02y) = (0.02)(y +1300) +(350)

  0.10x +0.02y = 0.02y +376 . . . . . eliminate parentheses

  0.10x = 376 . . . . . . . . . subtract 0.02y from both sides. y is eliminated

  x = 3760 . . . . . . . . . divide by 0.10

  y = x -1300 = 2460

__

The amount invested at 8% was $3760; the amount invested at 2% was $2460.

_____

Additional comment

We chose to eliminate y for a couple of reasons. x is the amount at the higher rate. We have found that solving for the higher-rate amount usually works best for preventing errors. The other reason is that multiplying by 0.02 results in smaller numbers, which we consider easier to deal with.

Had we multiplied by -0.08 to eliminate x, we would have ...

  -0.08(x) +(0.08x +0.02y) = -0.08(y +1300) +(350)

  0.02y = -0.08y +246

We judge -0.08(1300) +350 harder to calculate mentally, than 0.02(1300) +350.

  0.10y = 246

  y = 2460; x = 2460+1300 = 3760

The alloy of copper will be 600 ounces.

Expression in maths is defined as the collection of the numbers variables and functions by using signs like addition, subtraction, multiplication and division.

Let us represent

Number of ounce of 30% copper = x

Hence, our Equation will be given below:-

x × 30% + 200 × 70% = (x + 200) × 40%

0.3x + 140 = (x + 200)0.4

0.3x + 140 = 0.4x + 80

Collect like terms

0.4x - 0.3x = 140 - 80

0.1x = 60

x = 60/0.1

x = 600 ounces

Therefore the alloy of copper will be 600 ounces.

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Ivanna's score was at the 36th percentile, will be 99.64 and Aldo's score was at the 19th percentile, will be 99.81.

The analysis of mathematical representations is algebra, and the handling of those symbols is logic.

Last year, over 10,000 students took an entrance exam at a certain state university.

Let the maximum score be 100.

Ivanna's score was at the 36th percentile, will be

⇒ [(10,000 – 36) / 10,000] x 100

⇒ 99.64

Aldo's score was at the 19th percentile, will be

⇒ [(10,000 – 19) / 10,000] x 100

⇒ 99.81

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Answer:

The answer will be J

Step-by-step explanation:

Since you see, that the 67,896 will need to be rounded. It rounds up to 70,000 so you can elinmate M & N. 22% can be converted as a decimal to 0.2, also equals to 0.20. So elinmate L. Now you have J and K. You always times with these kinds of percentage.

Therefore the Answer Will Be J: 70,000 timed by 0.20.

Expert-Certifieder of Cousin.

Answer: $80

Step-by-step explanation:

Given:

The laptop consumes 40 watts/hour

The operating cost is $0.20 per kilowatt - hour

So,

The cost for 1 hour of consumption = 40 * 0.20 = $8

The cost for 10 hours of consumption = $8 * 10 hours = $80

The different councils consisting of 5 freshmen and 7 sophomores are possible 9216.

We have given that,

A pool of possible candidates for a student council consists of 14 freshmen and 8 software.

We have to determine the how many different councils consisting of 5 freshmen and 7 sophomores are possible

[tex]_n C_r=\frac{n !}{r ! (n-r) !}_n C_r = number of combinations\\\n = total number of objects in the set\\\r = number of choosing objects from the set[/tex]

The total number of the council is

[tex]_{10} C_5\times _9 C_7[/tex]

=252(36)

=9216

The different councils consisting of 5 freshmen and 7 sophomores are possible 9216.

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The range, standard deviation, and variance of the numbers of medals won by these countries are 28, 8.845, and 78.2353, respectively.

A range is given to a parameter to allow maximum leverage to the parameter. for example, if a vendor wants a rod of diameter 20 cm, then he may give a range of ±1 cm., which means he will accept the rod of 19(20-1) cm to 21(20+1) cm.

The range  of the numbers of medals won by these countries is,

Range =  Max - Min = 29 - 1 = 28

To find the standard deviation we need to know the following details,

Now, the standard deviation of medals won by these countries is,

[tex]\sigma = \sqrt{\dfrac{\sum x^2 - \frac1n (\sum x)^2}{n-1}}\\\\\sigma = \sqrt{\dfrac{\4372 - \frac{234^2}{18}}{18-1}}\\\\\sigma = 8.845[/tex]

The variance of the numbers of medals won by these countries is,

v = σ²

v = 78.2353

Hence, the range, standard deviation, and variance of the numbers of medals won by these countries are 28, 8.845, and 78.2353, respectively.

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The width of a cell with the length (l) of the cell is [tex]\frac{2}{3}[/tex] mm and area (A) of the cell is [tex]\frac{1}{12}[/tex] mm² is [tex]\frac{1}{8}[/tex].

Given that, the length (l) of the cell is [tex]\frac{2}{3}[/tex] mm and area (A) of the cell is [tex]\frac{1}{12}[/tex] mm².

We need to find the width of the cell.

The area of a rectangle is the product of its length and width. So, the area of the rectangle = Length×Width square units.

Now, the area of a cell = [tex]\frac{2}{3}[/tex] ×Width=  [tex]\frac{1}{2}[/tex] mm².

⇒Width=[tex]\frac{\frac{1}{12} }{\frac{2}{3} } ={\frac{1}{12} \times {\frac{3}{2}=\frac{1}{8}[/tex]

Therefore, the width of a cell with the length (l) of the cell is [tex]\frac{2}{3}[/tex] mm and area (A) of the cell is [tex]\frac{1}{12}[/tex] mm² is  [tex]\frac{1}{8}[/tex].

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The cost of goods sold is $10,258 million and the Cash paid to supplier is $10,447 million.

a. Cost of goods sold

Cost of Goods Sold=Beginning Inventories + Purchases – Ending Inventories

Cost of goods sold = $1,997 + $10,392 - $2,131

Cost of goods sold = $10,258

b. Cash paid to supplier

Cash paid=2015 Beginning balance  + Purchases - 2015 Ending balance

Cash paid = $1,181 + $10,392 - $1,126

Cash paid = $10,447

Therefore the cost of goods sold is $10,258 million and the Cash paid to supplier is $10,447 million.

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Answer:

The rate at which the temperature drops.

Step-by-step explanation:

In science, the dependent variable is what you are measuring.
In math, the dependent variable is what you are evaluating.

Same idea, different topic.

Answer:

d=temp drop

t=hours

dependent variable is the temperature drops because the number of hours will tell you how much the temprature drops

this is the equation

total temp drops=t x 5

hope this helps!?

Answer:

Statement 1 is correct

Angles UST and QSR are complementary.

Step-by-step explanation:

Complementary angles are those angles whose addition gives result of 90°

Congruent angles are those angles which are equal to each other. We can say that there values are same.

Here angle RSU is complementary to angle UST.

Therefore there addition is equal to 90°

∠RSU + ∠UST =  90° - (i)

Here angle QSR is congruent to angle RSU.

Therefore their values are equal.

It means ∠QSR =∠RSU -(ii)

From equation (i) and (ii) we get

∠QSR + ∠UST =  90°

So angles UST and QSR are complimentary angles.

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Answer:

Step-by-step explanation:

A

15-(5)/5-10=10/-5=-2 hope it helps!

One line will be formed which will be parallel to the slant edge option third is correct.

It is defined as the curve which is the intersection of cone and plane. There are three major conic sections; parabola, hyperbola, and ellipse (circle is a special type of ellipse).

We have a statement:

Which degenerate conic is formed when a double cone is cut through the top by a plane parallel to the slant edge of the cone?

As we can see in the attached picture if the double cone is cut through the top by a plane parallel to the slant edge of the cone one line will be formed which will be parallel to the slant edge.

Thus, one line will be formed which will be parallel to the slant edge option third is correct.

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Answer:

2a² + 2b²

Step-by-step explanation:

(a - b)² + (a + b)² ← expand both factors using FOIL

= a² - 2ab + b² + a² + 2ab + b² ← collect like terms

= 2a² + 2b²

Step-by-step explanation:

Expand both classic quadratics and combine like terms to find the sum:

(a−b)2+(a+b)2

Answer:

y = 3x - 2

Step-by-step explanation:

The number beside the x is the slope. The number at the end of the equation is the y-intercept.

All the answers are in this form:

y = mx + b

m is the slope.

b is the y-intercept.

With a 3 filled in for slope and -2 filled in for the y-intercept, you get:

y = 3x + -2

is the same as,

y = 3x - 2

Answer:

well the answer is 4 have a good day

Answer:

c. $29,000

Step-by-step explanation:

since 20% of 145,000 = 29,000

to calculate use

(20/100) * 145,000

or

(y/100) * x

y = the precentage

x = the price of the house

therefore your answer is c. $29,000

hope this helps:)

Notice that the difference in the absolute values of consecutive coefficients is constant:

|-7| - 1 = 6

13 - |-7| = 6

|-19| - 13 = 6

and so on. This means the coefficients in the given series

[tex]\displaystyle \sum_{i=1}^\infty c_i a^{i-1} = \sum_{i=1}^\infty |c_i| (-a)^{i-1} = 1 - 7a + 13a^2 - 19a^3 + \cdots[/tex]

occur in arithmetic progression; in particular, we have first value [tex]c_1 = 1[/tex] and for [tex]n>1[/tex], [tex]|c_i|=|c_{i-1}|+6[/tex]. Solving this recurrence, we end up with

[tex]|c_i| = |c_1| + 6(i-1) \implies |c_i| = 6i - 5[/tex]

So, the sum to [tex]n[/tex] terms of this series is

[tex]\displaystyle \sum_{i=1}^n (6i-5) (-a)^{i-1} = 6 \underbrace{\sum_{i=1}^n i (-a)^{i-1}}_{S'} - 5 \underbrace{\sum_{i=1}^n (-a)^{i-1}}_S[/tex]

The second sum [tex]S[/tex] is a standard geometric series, which is easy to compute:

[tex]S = 1 - a + a^2 - a^3 + \cdots + (-a)^{n-1}[/tex]

Multiply both sides by [tex]-a[/tex] :

[tex]-aS = -a + a^2 - a^3 + a^4 - \cdots + (-a)^n[/tex]

Subtract this from [tex]S[/tex] to eliminate the intermediate terms to end up with

[tex]S - (-aS) = 1 - (-a)^n \implies (1-(-a)) S = 1 - (-a)^n \implies S = \dfrac{1 - (-a)^n}{1 + a}[/tex]

The first sum [tex]S'[/tex] can be handled with simple algebraic manipulation.

[tex]S' = \displaystyle \sum_{i=1}^n i (-a)^{i-1}[/tex]

[tex]\displaystyle S' = \sum_{i=0}^{n-1} (i+1) (-a)^i[/tex]

[tex]\displaystyle S' = \sum_{i=0}^{n-1} i (-a)^i + \sum_{i=0}^{n-1} (-a)^i[/tex]

[tex]\displaystyle S' = \sum_{i=1}^{n-1} i (-a)^i + \sum_{i=1}^n (-a)^{i-1}[/tex]

[tex]\displaystyle S' = \sum_{i=1}^n i (-a)^i - n (-a)^n + S[/tex]

[tex]\displaystyle S' = -a \sum_{i=1}^n i (-a)^{i-1} - n (-a)^n + S[/tex]

[tex]\displaystyle S' = -a S' - n (-a)^n + \dfrac{1 - (-a)^n}{1 + a}[/tex]

[tex]\displaystyle (1 + a) S' = \dfrac{1 - (-a)^n - n (1 + a) (-a)^n}{1 + a}[/tex]

[tex]\displaystyle S' = \dfrac{1 - (n+1)(-a)^n + n (-a)^{n+1}}{(1+a)^2}[/tex]

Putting everything together, we have

[tex]\displaystyle \sum_{i=1}^n (6i-5) (-a)^{i-1} = 6 S' - 5 S[/tex]

[tex]\displaystyle \sum_{i=1}^n (6i-5) (-a)^{i-1} = 6 \dfrac{1 - (n+1)(-a)^n + n (-a)^{n+1}}{(1+a)^2} - 5 \dfrac{1 - (-a)^n}{1 + a}[/tex]

[tex]\displaystyle \sum_{i=1}^n (6i-5) (-a)^{i-1} =\boxed{\dfrac{1 - 5a - (6n+1) (-a)^n + (6n-5) (-a)^{n+1}}{(1+a)^2}}[/tex]

Hassan is incorrect because √0.15  is less than 0.4, and the value of  √0.15 is 0.3872 option third is correct.

Iteration is the process of repeating a procedure in order to produce a series of results.

We have:

= √0.15

[tex]= \rm \sqrt{{\dfrac{15}{100}}}[/tex]

[tex]= \rm \sqrt{{\dfrac{3}{20}}}[/tex]

= (1.732)/(4.472)

= 0.3872

√0.15 is less than 0.4.

Thus, Hassan is incorrect because √0.15  is less than 0.4, and the value of  √0.15 is 0.3872 option third is correct.

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Answer:

C

Step-by-step explanation:

Answer:

D

Step-by-step explanation:

We can test each pair making each ratio into its simplest form.

For A 2/3 is already in its simplest form and 9/15=3/5. they are not equivalent.

For B, 5/8 is already in its simplest form and 15/21=5/7. they are not equivalent.

For C, 3/12=1/4, and 6/18=1/3. They are not equivalent.

For D, 4/10=2/5, and 12/30=2/5. They are equivalent.

[tex]\huge\text{Hey there!}[/tex]

[tex]\huge\textbf{Let's see which ones are equivalent}\\\huge\textbf{to each other, shall we?}\\\\\\\huge\textbf{We will convert the given fractions to}\\\huge\textbf{decimals or make the fractions on the}\\\huge\textbf{right to easier to solve or comparing it}\\\huge\textbf{to the one on the left.}[/tex]

[tex]\huge\textsf{Option A.}[/tex]

[tex]\mathsf{\dfrac{2}{3}, \dfrac{9}{15}}[/tex]

[tex]\mathsf{\dfrac{2}{3}}\\\\\mathsf{= 2\div3}\\\\\mathsf{= 0.66\overline{6}7}\\\\\\\mathsf{\dfrac{9}{15}}\\\\\mathsf{= \dfrac{9\div3}{15\div3}}\\\\\mathsf{= \dfrac{3}{5}}\\\\\\\\\mathsf{\dfrac{2}{3} \neq \dfrac{9}{15}}[/tex]

[tex]\huge\textsf{Option B.}[/tex]

[tex]\mathsf{\dfrac{5}{8}, \dfrac{15}{21}}[/tex]

[tex]\mathsf{\dfrac{5}{8}}\\\\\mathsf{= 5\div8}\\\\\mathsf{= 0.625}\\\\\\\mathsf{\dfrac{15}{21}}\\\\\mathsf{= \dfrac{15\div3}{21\div3}}\\\\\mathsf{= \dfrac{5}{7}}\\\\\mathsf{\dfrac{5}{8}\neq \dfrac{15}{21}}[/tex]

[tex]\huge\textsf{Option C.}[/tex]

[tex]\mathsf{\dfrac{3}{12}.\dfrac{6}{18}}[/tex]

[tex]\mathsf{\dfrac{3}{12}}\\\\\mathsf{= 3 \div 12}\\\\\mathsf{= 0.25}\\\\\\\\\mathsf{\dfrac{6}{18}}\\\\\\\mathsf{= \dfrac{6\div3}{18\div3}}\\\\\\\mathsf{= \dfrac{2}{6}}\\\\\\\mathsf{= \dfrac{2\div2}{6\div2}}\\\\\\\mathsf{= \dfrac{1}{3}}\\\\\\\mathsf{\dfrac{3}{12}\neq \dfrac{6}{18}}[/tex]

[tex]\huge\textsf{Option D.}[/tex]

[tex]\mathsf{\dfrac{4}{10}, \dfrac{12}{30}}[/tex]

[tex]\mathsf{\dfrac{4}{10}}\\\\\mathsf{= 4\div10}\\\\\mathsf{= 0.40}\\\\\\\mathsf{\dfrac{6}{18}}\\\\\mathsf{= \dfrac{12\div3}{30\div3}}\\\\\mathsf{= \dfrac{4}{10}}\\\\\mathsf{= \dfrac{4}{10}}\\\\\mathsf{= \dfrac{4\div2}{10\div2}}\\\\\mathsf{= \dfrac{2}{5}}\\\\\mathsf{\dfrac{4}{10} = \dfrac{12}{20}}[/tex]

[tex]\huge\text{Thus, your answer should be: \boxed{\mathsf{Option\ D. \dfrac{4}{10}, \dfrac{12}{30}}}}\huge\checkmark[/tex]

[tex]\huge\text{Good luck on your assignment \& enjoy your day!}[/tex]

~[tex]\frak{Amphitrite1040:)}[/tex]