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2 years ago

Write the equation of the line that passes through the points (4,-6) and (-2,3)

Mathematics

1 answer:

yarga [219]2 years ago

7 0

Here is your answer and an explanation

Answer: y = -2/3x

Explanation:

This can be determined by calculating the gradient of the straight line, using:

m=ΔyΔx

=−6−34−(−2)

=−96

=−32

Then we use the slope-point form of the straight line:

y−y1=m(x−x1)

to give:

y−3=−32(x−(−2))

∴y−3=−32(x+2)

∴y−3=−32x−3

∴y=−32x

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Moe painted the lid of a barrel. The circumference of the lid is 32 pi( pie) inches. Which measurement is closest to the area of

JulijaS [17]

Answer:

Area of lid = 804.35 square inches.

Step-by-step explanation:

Given the following data;

Circumference of lid = 32π inches

To find the area of the lid;

First of all, we would find the radius of the lid (circle).

Circumference of circle = 2πr

Substituting into the formula, we have;

32π = 2πr

Radius, r = 32π/2π

Radius, r = 16 inches.

Now, we can solve for the area of the circle using the formula;

Area of circle = πr²

Area of lid = 3.142*16²

Area of lid = 3.142 * 256

Area of lid = 804.35 square inches.

In terms of pie, area of lid = 256π square inches.

Therefore, the measurement which is closest to the area of the lid in square inches is 804.35 or 256π.

6 0

3 years ago

What is the distance between the points (-6, 7) and (-1, 1) ? Round to the nearest whole number

pogonyaev

Answer:

7.8 rounded to the nearest whole number would be 8.

Step-by-step explanation:

-6 would be x1

7 would be y1

-1 would be x2

1 would be y2

Take these numbers and plug them into the distance formula:

$\sqrt{(x2-x1)^{2} + (y2-y1)^2}$

$\sqrt{(-1 - -6)^2 + (1-7)^2} \\\sqrt{(5)^2 + (6)^2\\\} \\\sqrt{25 + 36} \\\sqrt{61} = 7.81$

8 0

3 years ago

Helppp meee !!!!!!!!!!!!!

Vitek1552 [10]

Answer:

Step-by-step explanation:

Since the denominator of each of those rational exponents is a 4, that means that the radical is a 4th root. The numerator of each exponent serves as the power on the given base. For example,

$2^{\frac{3}{5}}$ can be rewritten as

$\sqrt[5]{2^3}$

The little number that sits outside the radical, resting in the bend, is called the index. Our index is 4 (same as saying the 4th root). Put everything under the 4th root and let the numerator be the powers on each base:

$\sqrt[4]{6^1*b^3*c^1}$ which is written simpler as: