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3 years ago

A. If you know the diameter of a sphere, you can find its radius.

1 answer:

7 0

A. If you know the diameter, you can find the radius of a sphere by dividing the diameter in half. If you know the circumference, you can find the radius by dividing the circumference by 2 times pi

B. π × r2 × h

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1. What units cannot be used to measure weight? A. ton O B. inch O C. pound D. ounce

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Answer:

B. Inches

Step-by-step explanation:

Because it measures the length of something, Not the weight.

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3 years ago

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Find the sum. io + 10 5

cestrela7 [59]

Answer:

The answer is -0.6

Step-by-step explanation:

I hope this helps. Sorry if I am wrong.

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3 years ago

The perimeter of a rectangle is 34 units. it's width is 6.5 units. Write an equation to determine the length of the rectangle

gayaneshka [121]

The equation would be 34x2+6.5x2

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Find the perimeter of the window to the nearest hundredth.

Natalka [10]

7.71 feet is your answer.

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2 years ago

2 Here are two equations:

MatroZZZ [7]

a. (3,4) is only the solution to Equation 1.

(4, 2.5) is the solution to both equations

(5,5) is the solution to Equation 2

(3,2) is not the solution to any equation.

b. No, it is not possible to have more than one (x,y) pair that is solution to both equations

Step-by-step explanation:

a. Decide whether each neither of the equations,

i (3,4)

ii. (4,2.5)

ill. (5,5)

iv. (3,2)

To decide whether each point is solution to equations or not we will put the point in the equations

Equations are:

Equation 1: 6x + 4y = 34

Equation 2: 5x – 2y = 15

i (3,4)

Putting in Equation 1:

$6(3) + 4(4) = 34\\18+16=34\\34=34\\$

Putting in Equation 2:

$5(3) - 2(4) = 15\\15-8 = 15\\7\neq 15$

ii. (4,2.5)

Putting in Equation 1:

$6(4) + 4(2.5) = 34\\24+10=34\\34=34\\$

Putting in Equation 2:

$5(4) - 2(2.5) = 15\\20-5 = 15\\15=15$

ill. (5,5)

$6(5) + 4(5) = 34\\30+20=34\\50\neq 34$

Putting in Equation 2:

$5(5) - 2(5) = 15\\25-10 = 15\\15=15$

iv. (3,2)

$6(3) + 4(2) = 34\\18+8=34\\26\neq 34$

Putting in Equation 2:

$5(3) - 2(2) = 15\\15-4 = 15\\11\neq 15$

Hence,

(3,4) is only the solution to Equation 1.

(4, 2.5) is the solution to both equations

(5,5) is the solution to Equation 2

(3,2) is not the solution to any equation.

b. Is it possible to have more than one (x, y) pair that is a solution to both

equations?

The simultaneous linear equations' solution is the point on which the lines intersect. Two lines can intersect only on one point. So a linear system cannot have more than one point as a solution

So,

a. (3,4) is only the solution to Equation 1.

(4, 2.5) is the solution to both equations

(5,5) is the solution to Equation 2

(3,2) is not the solution to any equation.

b. No, it is not possible to have more than one (x,y) pair that is solution to both equations

Keywords: Linear equations, Ordered pairs

Learn more about linear equations at:

#LearnwithBrainly

8 0

3 years ago