Which expressions are equivalent to2 ( -6c+3) +4c

Answer:

The answer is 0.45 pounds for an apple and 0.3 pounds for a pear

Step-by-step explanation:

Answer:

Cost of each apple: [tex]0.45[/tex].

Cost of each pear: [tex]0.30[/tex].

Step-by-step explanation:

Let [tex]x[/tex] denote the cost of each apple. Let [tex]y[/tex] denote the cost of each pear.

The question states that the cost of [tex]4[/tex] apples and [tex]3[/tex] pears is [tex]2.70[/tex]. Thus:

[tex]4\, x + 3\, y = 2.70[/tex].

Likewise, since the cost of [tex]2[/tex] apples and [tex]5[/tex] pears is [tex]2.40[/tex]:

[tex]2\, x + 5\, y = 2.40[/tex].

Solve this system of equations for [tex]x[/tex] and [tex]y[/tex] to find the price of each fruit.

[tex]\left\lbrace \begin{aligned} & 4\, x + 3\, y = 2.70 \\ & 2\, x + 5\, y = 2.40\end{aligned}\right.[/tex].

Multiply both sides of the equation [tex]2\, x + 5\, y = 2.40[/tex] by [tex]2[/tex] to match the coefficient of [tex]x[/tex] in the first equation:

[tex]\left\lbrace \begin{aligned} & 4\, x + 3\, y = 2.70 \\ & (2\times 2)\, x + (2 \times 5)\, y = (2 \times 2.40)\end{aligned}\right.[/tex].

[tex]\left\lbrace \begin{aligned} & 4\, x + 3\, y = 2.70 \\ & 4\, x + 10\, y = 4.80 \end{aligned}\right.[/tex].

Substitute the first equation from the new equation to eliminate [tex]x[/tex] and solve for [tex]y[/tex]:

[tex]\begin{aligned}& 4\, x + 10\, y - (4\, x +3\, y) = 4.80 - 2.70\end{aligned}[/tex].

[tex]10\, y - 3\, y = 2.10[/tex].

[tex]7\, y = 2.10[/tex].

[tex]y = 0.30[/tex].

Substitute [tex]y = 0.30[/tex] into an equation from the original system to eliminate [tex]y[/tex] and solve for [tex]x[/tex].

[tex]2\, x + 5\, y = 2.40[/tex].

[tex]2\, x + 5 \times 0.30 = 2.40[/tex].

[tex]2\, x = 0.90[/tex].

[tex]x = 0.45[/tex].

Thus, [tex]x = 0.45[/tex] and [tex]y = 0.30[/tex].

In other words, the price of each apple would be [tex]0.45[/tex]. The price of each pear would be [tex]0.30[/tex].

The solution to a given word problem by writing it as an equation is k = 253/232

The process of expressing word problems in mathematical forms takes a logical and chronological approach while taking the variables and arithmetic operations given into consideration.

From the given information, to express the word problem in mathematical form, we have:

Making k the subject of the equation, we have:

k = 253/232

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Answer:

See below

Step-by-step explanation:

The first one IS decay...as 'x' increases the function decreases

the second and third are NOT ...as 'x' increases the value increases

radioactive decay IS decay (obviously)

the one with e^2x   is not ....it grows exponentially as x grows

and finally the last one is decay because as x gets larger the value of y gets smaller

The value of the angle AED = 99 degree , Angle BEC is 99 degree , angle DEC = 81 degree , angle AEB is 81 degree

The straight line has an angle of 180 degree.

The two lines AC and BD intersects such that

∠AEB:∠BEC = 9 : 11

Let angle BEC be x

Then

Angle AEB is 9x/11

and angle AEC + BEC = 180 degree

(9x/11) + x = 180

9x +11x = 11 * 180

20x = 11 * 180

x = 99 degree

Angle BEC is 99 degree

and angle AEB is 81 degree

As Angle AED is vertically opposite to angle BEC

Therefore angle AED = 99 degree

As Angle DEC is vertically opposite to angle AEB

Therefore angle DEC = 81 degree.

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The solution will be the point of intersection of the two lines. The solution to the system is (1.1, 1.3)

Inequalities are expression not separated by an equal sign. Given the following inequalities

x+2y<4

3x-y >2

They are both written in standard form. The first line will be a dashed line and shaded below the graph while the second line will be a dashed line and shaded above.

The solution will be the point of intersection of the two lines as attached

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Answer:

Graph D is the correct one.

Step-by-step explanation:

Question:

2÷a=(5/2)÷(5/4)

Answer:

a = 1

[tex]\quad \huge \quad \quad \boxed{ \tt \:Answer }[/tex]

[tex]\qquad \tt \rightarrow \:\approx 9.2 \:\: units [/tex]

____________________________________

[tex] \large \tt Solution \: : [/tex]

[tex] \textsf{Using distance formula : - } [/tex]

[tex]\qquad \tt \rightarrow \: \sqrt{(x_2 - x_1) {}^{2} + (y_2 - y_1) {}^{2} } [/tex]

[tex]\qquad \tt \rightarrow \: \sqrt{(2 - 0) {}^{2} + (0 - 9) {}^{2} } [/tex]

[tex]\qquad \tt \rightarrow \: \sqrt{(2) {}^{2} + (- 9) {}^{2} } [/tex]

[tex]\qquad \tt \rightarrow \: \sqrt{ 4 + 81} [/tex]

[tex]\qquad \tt \rightarrow \: \sqrt{85} \: \: units[/tex]

[tex]\qquad \tt \rightarrow \: \approx 9.2 \: \: units[/tex]

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The distance between points A and B is 9.2 units.

We have,

To find the distance between two points (a, b) and (c, d) in a coordinate plane, you can use the distance formula:

Distance = √[(c - a)² + (d - b)²]

In this case, point A is (2, 0) and point B is (0, 9).

Now, plug the values into the distance formula:

Distance = √[(0 - 2)² + (9 - 0)²]

Distance = √[(-2)² + 9²]

Distance = √[4 + 81]

Distance = √85

Now, round the answer to the nearest tenth:

Distance ≈ √85 ≈ 9.2 (rounded to the nearest tenth)

Thus,

The distance between points A and B is 9.2 units.

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Answer: it's D

Step-by-step explanation:

Answer:

B. 1:18

Explanation Down Below

Step-by-step explanation:

Hello!

First, let's find the volume of each cylinder by plugging in the given values.

Volume of a Cylinder: [tex]V = \pi r^2h[/tex]

Since the variables are the same as given in the formula, we can just use the formula as the volume.

[tex]\implies{\boxed{{V = \pi r^2h}}[/tex]

We have to plug in 3r for the radius, and 2h for the height.

[tex]\implies \boxed{ V = 18\pi r^2h}[/tex]

We can see that the Volume of Cylinder B is just 18 times the Volume of Cylinder A, but we can find the same ratio using equations.

The answer is Option B. 1:18.

It's whichever graph is an absolute function and is shifted to the right by 2 and is shifted down by 1.

Since I don't have a picture of the graph this is the best I can do to help.

Answer:

see attached

Step-by-step explanation:

Translations

For a > 0

[tex]f(x+a) \implies f(x) \: \textsf{translated}\:a\:\textsf{units left}[/tex]

[tex]f(x-a) \implies f(x) \: \textsf{translated}\:a\:\textsf{units right}[/tex]

[tex]f(x)+a \implies f(x) \: \textsf{translated}\:a\:\textsf{units up}[/tex]

[tex]f(x)-a \implies f(x) \: \textsf{translated}\:a\:\textsf{units down}[/tex]

Given function:

[tex]r(x) = |x - 2| - 1[/tex]

The parent function of the given function is:  [tex]f(x) = |x|[/tex]

Translated 2 units to the right:  [tex]f(x-2)=|x-2|[/tex]

Translated 1 unit down:  [tex]f(x-2)-1=|x-2|-1[/tex]

Graph of Modulus function

Line y = x where x ≥ 0

Line y = -x where x ≤ 0

Vertex at (0, 0)

Therefore, the graph of the given function is the graph of the modulus function translated 2 units to the right and 1 unit down.  So, its vertex is at (2, -1).

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From the calculations, the height of the tower is 26.4 Km

If we know that the base of the tower is B and he top of the tower is T then to find the  shortest distance between the pilot and the base of the control tower;

Sin 41 = 10.5//BP/

/BP/ = 10.5/Sin 41

/BP/ = 15.9 Km

Now, the height of the control tower is obtained from;

First /AB/ = √(15.9)^2 - (10.5)^2

/AB/ = 11.9 Km

Given that;

Tan 36 = 10.5//BT/

/BT/=  10.5/Tan 36

/BT/=  14.5 Km

Hence the height of the tower is 11.9 Km + 14.5 Km = 26.4 Km

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The length of the three congruent sides is 7 cm and the fourth side length is 12cm.

Given quadrilateral has three congruent sides and perimeter of quadrilateral is 33 cm.

A quadrilateral can be expressed as ABCD and three congruent sides are AB, BC, CD and the fourth side is AD.

Let AD=x

then AB=BC=CD=x-5

This means if all the sides of a quadrilateral are known, we can get its perimeter by adding all its sides.

So, Perimeter = AB + BC + CD + DA.

Substitute values

(x-5)+(x-5)+(x-5)+x=33

4x-15=33

    4x= 48

     x=12

Hence the fourth side length is 12cm and the other sides are 7 cm longer.

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The number of  days would the stack be long enough to reach a star that is about 3 × 10¹³ km away is 64 days

Since from the question, we see that the number of pennies double with each day. It forms a geoemetric progression with

Since the number of pennies after n days equals N = 2ⁿ

Let

So, after n days, the length of the stack of pennies is the geoemetric progression D = 2ⁿ × L

Making n subject of the formula, we have

n = ㏒(D/L)/㏒2

Substituting the values of the variables into the equation, we have

n = ㏒(D/L)/㏒2

n = ㏒(3 × 10¹⁶ m/1.5 × 10⁻³ m)/㏒2

n = ㏒(3/1.5 × 10¹⁹)/㏒2

n = ㏒(2 × 10¹⁹)/㏒2

n = ㏒2 + ㏒10¹⁹/㏒2

n = (19㏒10 + ㏒2)/㏒2

n = (19 + 0.3010)/0.3010

n = 19.3010/0.3010

n = 64.1

n ≅ 64 days

So, the number of  days would the stack be long enough to reach a star that is about 3 × 10¹³ km away is 64 days

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Use the change-of-basis identity,

[tex]\log_x(y) = \dfrac{\ln(y)}{\ln(x)}[/tex]

to write

[tex]xyz = \log_a(bc) \log_b(ac) \log_c(ab) = \dfrac{\ln(bc) \ln(ac) \ln(ab)}{\ln(a) \ln(b) \ln(c)}[/tex]

Use the product-to-sum identity,

[tex]\log_x(yz) = \log_x(y) + \log_x(z)[/tex]

to write

[tex]xyz = \dfrac{(\ln(b) + \ln(c)) (\ln(a) + \ln(c)) (\ln(a) + \ln(b))}{\ln(a) \ln(b) \ln(c)}[/tex]

Redistribute the factors on the left side as

[tex]xyz = \dfrac{\ln(b) + \ln(c)}{\ln(b)} \times \dfrac{\ln(a) + \ln(c)}{\ln(c)} \times \dfrac{\ln(a) + \ln(b)}{\ln(a)}[/tex]

and simplify to

[tex]xyz = \left(1 + \dfrac{\ln(c)}{\ln(b)}\right) \left(1 + \dfrac{\ln(a)}{\ln(c)}\right) \left(1 + \dfrac{\ln(b)}{\ln(a)}\right)[/tex]

Now expand the right side:

[tex]xyz = 1 + \dfrac{\ln(c)}{\ln(b)} + \dfrac{\ln(a)}{\ln(c)} + \dfrac{\ln(b)}{\ln(a)} \\\\ ~~~~~~~~~~~~+ \dfrac{\ln(c)\ln(a)}{\ln(b)\ln(c)} + \dfrac{\ln(c)\ln(b)}{\ln(b)\ln(a)} + \dfrac{\ln(a)\ln(b)}{\ln(c)\ln(a)} \\\\ ~~~~~~~~~~~~ + \dfrac{\ln(c)\ln(a)\ln(b)}{\ln(b)\ln(c)\ln(a)}[/tex]

Simplify and rewrite using the logarithm properties mentioned earlier.

[tex]xyz = 1 + \dfrac{\ln(c)}{\ln(b)} + \dfrac{\ln(a)}{\ln(c)} + \dfrac{\ln(b)}{\ln(a)} + \dfrac{\ln(a)}{\ln(b)} + \dfrac{\ln(c)}{\ln(a)} + \dfrac{\ln(b)}{\ln(c)} + 1[/tex]

[tex]xyz = 2 + \dfrac{\ln(c)+\ln(a)}{\ln(b)} + \dfrac{\ln(a)+\ln(b)}{\ln(c)} + \dfrac{\ln(b)+\ln(c)}{\ln(a)}[/tex]

[tex]xyz = 2 + \dfrac{\ln(ac)}{\ln(b)} + \dfrac{\ln(ab)}{\ln(c)} + \dfrac{\ln(bc)}{\ln(a)}[/tex]

[tex]xyz = 2 + \log_b(ac) + \log_c(ab) + \log_a(bc)[/tex]

[tex]\implies \boxed{xyz = x + y + z + 2}[/tex]

(C)

Answer:

C: 1/2

Step-by-step explanation:

[tex]\quad \huge \quad \quad \boxed{ \tt \:Answer }[/tex]

[tex]\qquad \tt \rightarrow \: C. \cfrac{1}{2}[/tex]

____________________________________

[tex] \large \tt Solution \: : [/tex]

[tex] \textsf{Slope of line - } [/tex]

[tex]\qquad \tt \rightarrow \: m = \cfrac{rise}{run} [/tex]

[tex]\qquad \tt \rightarrow \: m = \cfrac{0 - ( - 4)}{8 - 0} [/tex]

[tex]\qquad \tt \rightarrow \: m = \cfrac{ 4}{8 } [/tex]

[tex]\qquad \tt \rightarrow \: m = \cfrac{ 1}{2} [/tex]

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Answer:

(3,-4)

Step-by-step explanation:

[tex]\quad \huge \quad \quad \boxed{ \tt \:Answer }[/tex]

[tex]\qquad \tt \rightarrow \: \cfrac{ y{ }^{2} }{(p - y) {}^{y} }[/tex]

____________________________________

[tex] \large \tt Solution \: : [/tex]

[tex]\qquad \tt \rightarrow \: \cfrac{ {p}^{2} }{(p - y) {}^{y} } + \cfrac{ - 2p}{(p - y) {}^{y - 1} } + \cfrac{1}{(p - y) {}^{y - 2} } [/tex]

[tex]\qquad \tt \rightarrow \: \cfrac{ {p}^{2} + ( - 2p)(p - y) + 1(p - y) { }^{2} }{(p - y) {}^{y} } [/tex]

[tex]\qquad \tt \rightarrow \: \cfrac{ {p}^{2}- 2p {}^{2} + 2py + p {}^{2} + y{ }^{2} - 2py}{(p - y) {}^{y} } [/tex]

[tex]\qquad \tt \rightarrow \: \cfrac{ {p}^{2} + {p}^{2} - 2p {}^{2} + 2py - 2py + y{ }^{2} }{(p - y) {}^{y} }[/tex]

[tex]\qquad \tt \rightarrow \: \cfrac{ y{ }^{2} }{(p - y) {}^{y} }[/tex]

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Answer:

1

Step-by-step explanation:

[tex] log_{3}(5) \times log_{25}(9) = \frac{ log(5) }{ log(3) } \times \frac{ log(9) }{ log(25) } = \frac{ log(5) }{ log(3) } \times \frac{2 log(3) }{ 2log(5) } = \frac{2}{2} = 1[/tex]

Answer:

5/7 or D

Step-by-step explanation:

i just know

Answer:

In order to make the function continuous, the first and second pairs of lines should have common points.

1. Intersection of the first two lines:

This determines the value of a:

2. Intersection of the second two lines:

This determines the value of b:

See the graph attached

A fraction is a way to describe a part of a whole. The value of x is 6.

A fraction is a way to describe a part of a whole. such as the fraction ¼ can be described as 0.25.

The value of x can be written as,

[tex]\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{x}}} = \dfrac{7}{13}\\\\\\\dfrac{1}{1+\dfrac{1}{\dfrac{x+1}{x}}} = \dfrac{7}{13}\\\\\\\dfrac{1}{1+\dfrac{x}{x+1}} = \dfrac{7}{13}\\\\\\\dfrac{1}{\dfrac{x+1+x}{x+1}}= \dfrac{7}{13}\\\\\\\dfrac{x+1}{x+1+x}= \dfrac{7}{13}\\\\[/tex]

13x + 13 = 7x + 7 + 7x

13x + 13 = 14x + 7

x = 6

Hence, the value of x is 6.

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Answer:

[tex]y=-8x+5[/tex]

Slope-intercept form:

[tex]\implies y=mx+b[/tex] [tex](\text{m=slope};\text{b=y-intercept})[/tex]

Using the slope-intercept form, let's identify the other slopes:

Option A:

[tex]y=-8x+5[/tex]  

[tex]\text{slope}=-8[/tex]

Option B:

[tex]y-9=-2(x+1)[/tex] (This equation is in point-slope form, the slope it too the left, making it -9)

[tex]\text{slope}=-9[/tex]

Option C:

[tex]y=7x-3[/tex]

[tex]\text{slope}=7[/tex]

Option D:

[tex]y+2=6(x+10)[/tex]

[tex]\text{slope}=2[/tex]  (Using point-slope terms, we can find the slope.)

⇒ A 'steep'est slope is closest to almost having a vertical slope or pitch, or  relatively high gradient, as a hill, an ascent, stairs, etc.

[tex]\text{This could be a line }[/tex] ⇒ ([tex]\text{positive or negative}[/tex])

Hence, the linear function with the steepest slope is Option A: [tex]y=-8x+5[/tex].

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Answer:

-163

Step-by-step explanation:

-2,-25,-163

-2×6=-12 -12-13=-25

-25×6=-150 -150-13=-163

Answer:

The third term is = -163

Step-by-step explanation:

The first term is = -2

The second term = -2 ×6 = -12 Then - 13 = - 25

The second term = - 25

The third term = -25 × 6 = - 150 Then -13 = -163

The third term is -163

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Answer:

auto correlation is the answer to the question

Answer:

60

Step-by-step explanation:

10(2 + 2) + 20

=10(4)+20

=40+20

=60

Using the Fundamental Counting Theorem, it is found that the number of outcomes possible in this scenario is given by:

D. 240.

It is a theorem that states that if there are n things, each with [tex]n_1, n_2, \cdots, n_n[/tex] ways to be done, each thing independent of the other, the number of ways they can be done is:

[tex]N = n_1 \times n_2 \times \cdots \times n_n[/tex]

The number of options for each selection are given as follows:

[tex]n_1 = 6, n_2 = 4, n_3 = 5, n_4 = 2[/tex]

Hence the number of outcomes is given by:

N = 6 x 4 x 5 x 2 = 240.

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Answer:

The "power rule" tells us that to raise a power to a power, just multiply the exponents. Here you see that 52 raised to the 3rd power is equal to 56.

Step-by-step explanation:

Answer:

(- 2, - 2 )

Step-by-step explanation:

the solution to the system is at the point of intersection of the 2 lines.

the 2 lines intersect at (- 2, - 2) , then

solution to the system is (2, - 2 )

Answer:

888

Step-by-step explanation:

let the number be n then divide by 8 , that is [tex]\frac{n}{8}[/tex]

now add 8 to this

[tex]\frac{n}{8}[/tex] + 8 and finally multiply this by 8 and equate to 952

8([tex]\frac{n}{8}[/tex] + 8) = 952 ( divide both sides by 8 )

[tex]\frac{n}{8}[/tex] + 8 = 119 ( subtract 8 from both sides )

[tex]\frac{n}{8}[/tex] = 111 ( multiply both sides by 8 to clear the fraction )

n = 888

the number chosen was 888