Tìm số dư trong phép chia 3^{2020} chia cho 13

Rewriting the equation of the given line in slope-intercept form,

[tex]2x-3y=13\\\\2x-13=3y\\\\y=\frac{2}{3}x-\frac{13}{3}[/tex]

Thus, the slope of the given line is 2/3. Since perpendicular lines have slopes that are negative reciprocals of each other, we want to find the line with a slope of -3/2.

Substituting into point-slope form,

[tex]y-5=-\frac{3}{2}(x+6)\\\\y-5=-\frac{3}{2}x-9\\\\\boxed{y=-\frac{3}{2}x-4}[/tex]

Answer:

1. x - 1

2. [tex]\sqrt{x^2-1}[/tex]

Step-by-step explanation:

[tex]f(g(x)) = \sqrt{x+1}^2 - 2\\ = x + 1 - 2\\= x - 1[/tex]

[tex]g(f(x)) =\sqrt{x^2-2+1}\\ = \sqrt{x^2-1}[/tex]

Answer:

1/4

Step-by-step explanation:

Since y/x = 3/12 = 1/4, the answer is 1/4.

Answer:

the value of x is between -4 and 4

Answer:

-3 and -7

Step-by-step explanation:

Factors of 21: 1, 3, 7, 21

3 and 7 add up to 10, but what about -10? Well, -3 * -7 is also 21, and -3 + -7 gives you -10.

Brainliest, please :)

The best approximation for the median is 3.6.

An illustration of a numerical distribution with continuous results is a density curve. A density curve is, in other words, the graph of a continuous distribution. This implies that density curves can represent continuous quantities like time and weight rather than discrete events like rolling a die (which would be discrete). As seen by the bell-shaped "normal distribution," density curves either lie above or on a horizontal line (one of the most common density curves).

For the given density curve, the area under the density curve is

A=(1/2)×5×(1/2)=5/4

now, for the median, the area on the left and the area on the right of the median.

Let the median be x, then

The area criteria we get

(1/2)x(x/10)=(5/4)/2

⇒x²=5*20/4=25/2

So, we get x=5/([tex]\sqrt2[/tex])=3.5 ≈ 3.6

Hence the required answer is option 3.6

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Answer:

Step-by-step explanation:

A function is defined so that for each input, there is no more than one output.

Answer:

Tim is 8 years old.

Explanation:

8 is the cube number of 2, as 2 × 2 × 2 = 8. Additionally, 8 is double 4, which is the square number of 2, and half of 16, which is the square number of 4.

Tim who goes to school is 8 years old.

Good numerical ability describes your ability to think and conclude logical thoughts in your daily life also in your working role how you handle data in form of numbers, graphs, statements etc.

According to the given question Tim goes to school.

His age is a cube number let that number be x therefore age of Tim would be x³.

His age can also be defined as double of 1 square number which is 2x² and half of a different square of a numbers.

Let us equate x³ = 2x² and by hit and trial we get to know that x = 2 satisfies this condition.

So, Age of Tim could be 8 years,Now lets check our third statement half a different square number 16 is square of 4 so half of 16 is also 8.

Thus Tim is 8 years old.

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Answer: d

Step-by-step explanation:

divide 300 by 50 to get 6 then multiply 2.5 by 6 to get 15 therefore the answer is d 15ml

Answer:

96

Step-by-step explanation:

The formula for average = [tex]\frac{Sum \hspace{0.15cm} of\hspace{0.15cm} Items}{No.\hspace{0.15cm} of \hspace{0.15cm} Items}[/tex]

After taking 3 tests, the average is 88.

∴ 88 = Sum of Items / 3

Sum of Items = 88 x 3

                      = 264

This means Lenny's total score after 3 tests is 264.

We now have to calculate how many marks needs to be added to 264 so that after four tests, the average is 90.

Now the number of items (tests) = 4, and average = 90.

Let the marks required in 4th test = [tex]x[/tex]

∴ [tex]\frac{264 + x}{4} = 90[/tex]

264 + [tex]x[/tex] = 360

[tex]x[/tex] = 360 - 264

  = 96

Answer:

y=3x+8

Step-by-step explanation:

First of all, remember what the equation of a line is:

y = mx+b

Where:

m is the slope, and

b is the y-intercept

you find the slope by using this formula:

then we figure out it is 3

then graph to find the y-intercept which is 8

Answer:

y = 3x + 8

Step-by-step explanation:

Below is the image for the formula of the slope.

1. Subtract y values:
17 - 5 - 12
2. Subtract x values:
3 - (-1) = 4
3. Divide y/x
12 / 4 = 3
Therefore, the slope is 3.

Next, you need to find the y-intercept. This is our equation so far:
y = 3x + b
We can substitute any of the points given to us from the problem. I will use the first point: (-1, 5).

When substituting, you get:
5 = 3 ( -1 ) + b
5 = -3 + b
b = 8

Therefore, the equation is y = 3x + 8

Answer:

f=- 6, x=-1

f = - 6, x=-3

Step-by-step explanation:

Solve for the first variable in one of the equations, then substitute the result into the other equation.

Answer:

f = 4 [tex]\frac{5}{37}[/tex]

Step-by-step explanation:

divide 1000 on both sides.

f = 4.135 repeating (the 135 is repeating.)

Put the .135135135... over 999 because it is repeating. F becomes 4[tex]\frac{135}{999}[/tex]

Simplify. f = [tex]4\frac{5}{37}[/tex]

The packet should be dropped  3636.55 horizontal meters before the target in order to hit the target

Trajectory is the path followed by a moving object when it is falling under gravitational force.

x = 90 t

y = -4.9 t²2 + 4500

t ≥ 0

y = 0

-4.9t² + 4500 = 0

4.9t²2 = 4500

[tex]\rm t = \sqrt {\dfrac{4500}{4.9}}[/tex]

t = 30.30 seconds

substituting the value to determine the horizontal distance

x = 120 * 30.30

x = 3636.55 meters

Therefore the packet should be dropped  3636.55 horizontal meters before the target in order to hit the target

The complete question is

a drone traveling horizontally at 120m s over a flat ground at an elevation of 4500 meters must drop an emergency package on a target on the ground the trajectory of the package is given by x 120t y 4.9t 2 4500 where the origin is the point on the ground directly beneath the drone at the moment of release. How many horizontal meters before the target should the package be released in order to hit the target? Round to the nearest meter.

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A function assigns the values. The value of f(-3) and f(3) + f(-3) is 8 and -8, respectively.

A function assigns the value of each element of one set to the other specific element of another set.

Given the function f(x)=-x²-4x+5, therefore, the value of f(x) when the value of x is -3 is,

f(-3) = -(-3)² - 4(-3) + 5

f(-3) = -9 + 12 + 5

f(-3) = 8

Now, the value of f(x) when the value of x is 3 is,

f(3) = -(3)²-4(3)+5

f(3) = - 9 - 12 + 5

f(3) = -16

further, the value of f(3) + f(-3) is,

f(3) + f(-3) = -16 + 8

f(3) + f(-3) = -8

Hence, the value of f(-3) and f(3) + f(-3) is 8 and -8, respectively.

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Answer:

21 grandchildren

Explanation:

2/7ths + 5/7th = all

5/7ths = 15   <- divide both sides by 3 to get 1/7th

1/7th = 3 (multiply both sides by 7 to get "1")

  7/7ths    = 21

so, eva has 21 grandchildren

Answer:

21

Step-by-step explanation:

If 2/7th of Eva's grandchildren live in one location, we know that 5/7th (the remaining portion of Eva's grandchildren) must live in the other location

Wales + Australia = all of the kids (100% = 1/1  (which is equal to 7/7) )

Wales + Australia = 7/7ths

{x is the total number of kids}

Australia = 5/7ths

15 = [tex]\frac{5}{7}x[/tex]   {multiply both sides by 7 to get rid of fraction}

105 = 5x   {divide both sides to find 1x}

21 = x

check:

2/7th of 21 = 6

5/7th of 21 = 15  {which we know}

6 + 15 = 21

21 = 21

(TRUE)

So, Eva has 21 grandchildren

hope this helps!!

Answer:

The number in the green box should be 3.

Step-by-step explanation:

[tex]c + 4 {c}^{2} - 2 + 2c - {c}^{2} + 5 = 3 {c}^{2} + 3c + 3[/tex]

If the geometric series has first term [tex]a[/tex] and common ratio [tex]r[/tex], then its [tex]N[/tex]-th partial sum is

[tex]\displaystyle S_N = \sum_{n=1}^N ar^{n-1} = a + ar + ar^2 + \cdots + ar^{N-1}[/tex]

Multiply both sides by [tex]r[/tex], then subtract [tex]rS_N[/tex] from [tex]S_N[/tex] to eliminate all the middle terms and solve for [tex]S_N[/tex] :

[tex]rS_N = ar + ar^2 + ar^3 + \cdots + ar^N[/tex]

[tex]\implies (1 - r) S_N = a - ar^N[/tex]

[tex]\implies S_N = \dfrac{a(1-r^N)}{1-r}[/tex]

The [tex]N[/tex]-th partial sum for the series of reciprocal terms (denoted by [tex]S'_N[/tex]) can be computed similarly:

[tex]\displaystyle S'_N = \sum_{n=1}^N \frac1{ar^{N-1}} = \frac1a + \frac1{ar} + \frac1{ar^2} + \cdots + \frac1{ar^{N-1}}[/tex]

[tex]\dfrac{S'_N}r = \dfrac1{ar} + \dfrac1{ar^2} + \dfrac1{ar^3} + \cdots + \dfrac1{ar^N}[/tex]

[tex]\implies \left(1 - \dfrac1r\right) S'_N = \dfrac1a - \dfrac1{ar^N}[/tex]

[tex]\implies S'_N = \dfrac{1 - \frac1{r^N}}{a\left(1 - \frac1r\right)} = \dfrac{r^N - 1}{a(r^N - r^{N-1})} = \dfrac{1 - r^N}{a r^{N-1} (1 - r)}[/tex]

We're given that [tex]a=1[/tex], and the sum of the first [tex]n[/tex] terms of the series is

[tex]S_n = \dfrac{1-r^n}{1-r} = 364[/tex]

and the sum of their reciprocals is

[tex]S'_n = \dfrac{1 - r^n}{r^{n-1}(1 - r)} = \dfrac{364}{243}[/tex]

By substitution,

[tex]\dfrac{1 - r^n}{r^{n-1}(1-r)} = \dfrac{364}{r^{n-1}} = \dfrac{364}{243} \implies r^{n-1} = 243[/tex]

Manipulating the [tex]S_n[/tex] equation gives

[tex]\dfrac{1 - r^n}{1-r} = 364 \implies r (364 - r^{n-1}) = 363[/tex]

so that substituting again yields

[tex]r (364 - 243) = 363 \implies 121r = 363 \implies \boxed{r=3}[/tex]

and it follows that

[tex]r^{n-1} = 243 \implies 3^{n-1} = 3^5 \implies n-1 = 5 \implies \boxed{n=6}[/tex]

Answer:

54.6 ÷ 8.4 = 6.5 hours

6 hours, 30 minutes:)

Step-by-step explanation:

Answer:

[tex]\textsf{Zeros}: \quad x=\dfrac {\sqrt{3}}{2}, \:\:x=-\dfrac {4\sqrt{3}}{3}[/tex]

Step-by-step explanation:

Rewrite the given polynomial in the form ax² + bx + c:

[tex]f(x)=2 \sqrt{3}x^2+5x-4 \sqrt{3}[/tex]

To find the zeros, set the function to zero and solve for x using the quadratic formula.

[tex]\implies 2 \sqrt{3}x^2+5x-4 \sqrt{3}=0[/tex]

Quadratic formula:

[tex]x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0[/tex]

Therefore,

Substituting the values into the quadratic formula:

[tex]\implies x=\dfrac{-5 \pm \sqrt{5^2-4(2\sqrt{3})(-4\sqrt{3})} }{2(2\sqrt{3})}[/tex]

[tex]\implies x=\dfrac {-5 \pm \sqrt {121}}{4\sqrt{3}}[/tex]

[tex]\implies x=\dfrac {-5 \pm 11}{4\sqrt{3}}[/tex]

[tex]\implies x=\dfrac {6}{4\sqrt{3}}, \:\:x=\dfrac {-16}{4\sqrt{3}}[/tex]

[tex]\implies x=\dfrac {3}{2\sqrt{3}}, \:\:x=-\dfrac {4}{\sqrt{3}}[/tex]

[tex]\implies x=\dfrac {\sqrt{3}}{2}, \:\:x=-\dfrac {4\sqrt{3}}{3}[/tex]

The sum of the roots of a polynomial is -b/a:

[tex]\implies -\dfrac{b}{a}=-\dfrac{5}{2 \sqrt{3}}=-\dfrac{5\sqrt{3}}{6}[/tex]

The sum of the found roots is:

[tex]\implies \left(\dfrac {\sqrt{3}}{2}\right)+\left(-\dfrac {4\sqrt{3}}{3}\right)=-\dfrac{5\sqrt{3}}{6}[/tex]

Hence proving the sum of the roots is -b/a

The product of the roots of a polynomial is:  c/a

[tex]\implies \dfrac{c}{a}=\dfrac{-4\sqrt{3}}{2\sqrt{3}}=-2[/tex]

The product of the found roots is:

[tex]\implies \left(\dfrac {\sqrt{3}}{2}\right)\left(-\dfrac {4\sqrt{3}}{3}\right)=-\dfrac{12}{6}=-2[/tex]

Hence proving the product of the roots is c/a

Therefore, the relationship between the roots and the coefficients is verified.

Answer:

Step-by-step explanation:

Rewrite the given polynomial in the form ax² + bx + c:

To find the zeros, set the function to zero and solve for x using the quadratic formula.

Quadratic formula:

Therefore,

a = 2√3

b = 5

c = - 4√3

Substituting the values into the quadratic formula:

The sum of the roots of a polynomial is -b/a:

The sum of the found roots is:

Hence proving the sum of the roots is -b/a

The product of the roots of a polynomial is: c/a

The product of the found roots is:

Hence proving the product of the roots is c/a

Therefore, the relationship between the roots and the coefficients is verified.

The transformation that can be used to show that ABCDE and its image are congruent is a 90° counterclockwise.

It should be noted that when we rotate a point through 90° counterclockwise, the mapping will be: (x, y) to (-y, x).

In this situation, it van be deduced that the x and y coordinates swapped positions. This illustrates a 90° counterclockwise rotation about the origin..

In this situation, since the rotation is a rigid motion, the two shapes are congruent.

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(9x³+5)/(2x - 3)

split up (9x³+5)

9x³/2x-3 + 5/2x-3

attached solution.

Answer:

y=−6x−36

Step-by-step explanation:

ez math

The formula that should be entered in A3 is = A1 * B1

The question implies that:

A3 = A1 times B1

In mathematics, the term "times" means *

So, we have:

A3 = A1 * B1

Remove the variable A3

= A1 * B1

Hence, the formula that should be entered in A3 is = A1 * B1

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lol lol lol lol lol lol lol lol lol lol lol lol lol lol lol lol lol lol lol lol lol lol lol lol

There will be four ways a circle and a parabola can intersect because the solution of the quartic equation will be 4.

It is described as a set of points, where each point is at the same distance from a fixed point (called the center of a circle)

We have a circle and a parabola:

As we know the standard form of a circle:

[tex]\rm (x-h)^2 +(y-k)^2=r^2[/tex]

The standard form of the parabola:

[tex]\rm y = a(x-h)^2+k[/tex]  

If we plug the value of y from the parabola equation in the circle equation, we get a quartic equation(4th order equation)

The solution of the quartic equation will be 4.

Thus, there will be four ways a circle and a parabola can intersect because the solution of the quartic equation will be 4.

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The part that represents the solution to the inequality will be Part B shaded below first line and above second line.

From the information given, the equation of the first line will be:

y - 3 = (-3 - 3/2 + 4)(x + 4)

y - 3 = -1(x + 4)

y + x = -4 + 3

x + y = -1

The equation of the second line will be:

y + 3 = -1(x + 4)

y = x + 4 - 3

y = x + 1

This is plotted on the graph attached.

From the systems of equations, the statement that is correct about the two systems of equations is that They will have the same solution because the first equation of System B is obtained by adding the first equation of System A to 3 times the second equation of System A.

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Answer:

added in the picture

Step-by-step explanation:

added in the picture

The range of the function on the graph is C. all the real numbers greater than or equal to –3 .

We can see by the graph maximum value of y of function is -3 and the graph continues to extend upward .

So, the range contain all the real numbers greater than or equal to –3 .

Therefore, the range contain all the real numbers greater than or equal to –3 .

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