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3 years ago

What is the answer to this problem

Mathematics

1 answer:

madam [21]3 years ago

4 0

Please take another photo of this problem sheet and share it, showing the problem statement right side up. Next, please do as much as you can on the first part of this problem. Then I'd gladly comment on your work. If you can't do that, then at least ask questions the answers to which might help you get started.

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The area of the square is 25 inches what is the measure on one side of the square??

Sindrei [870]

Answer:

Step-by-step explanation:

area of a square= side x side

$\sqrt{25}$ = 5

3 0

3 years ago

Sue needs to drive 372 miles to visit her aunt. She stopped at a gas station one-third of the way in to her trip. How many miles

IRINA_888 [86]

Answer:

Sue has traveled 124 miles.

Step-by-step explanation:

Here, the distance covered by Sue is a fraction of the total distance to be covered.

The total distance to her aunt's = 372 miles

but the gas station is at one-third of the distance to be covered.

Thus,

Miles traveled by Sue = $\frac{1}{3}$ x 372

= 124

Miles traveled by Sue = 124 miles.

Therefore Sue has traveled 124 miles.

6 0

3 years ago

Question 5 (1 point) What is the mean of this set of data: 78, 79, 80, 80, 89, 89, 92, 93, 94, 95, 95, 95 Round your answer to t

Len [333]

88.3

add all the numbers, you get 1,059 and divide that by how many numbers there are which is 12 and you get 88.3

7 0

3 years ago

Sharon drops a rubber ball from a height of 48 feet. Every bounce sends the

Reil [10]

Answer:

1.75 feet

Step-by-step explanation:

48 ÷ 2 = 14

14 ÷ 2 = 7

7 ÷ 2 = 3.5

3.5 ÷ 2 = 1.75

3 0

3 years ago

Write a coordinate proof for the following statement: Any triangle ABC formed so that vertex C is on the perpendicular bisector

AnnyKZ [126]

Answer:

Answer is contained in explanation.

Step-by-step explanation:

Description of visual:

I started with the first picture. This is a picture of triangle ABC.

Now I'm going to draw a line segment from vertex C such that it is a perpendicular bisector of AB.

Proof:

CM is a perpendicular bisectors of AB is a given.

From this we can concluded by definition of perpendicular angles that angle AMC and angle BMC are right angles.

Since angles AMC and BMC are right angles, then they are congruent to each other.

By the definition of bisector and since CM bisects AB, then AM is congruent to MB.

By the reflexive property, we have that CM is congruent to CM.

We can conclude the two triangles, triangle CMA and CMB, are congruent by SAS Postulate.

Since triangles CMA and CMB are congruent, we can conclude that their corresponding parts are congruent.

Since their corresponding parts are congruent, then we now know that side CA and side CB are congruent.

Since two sides of the triangle ABC are congruent to each other, namely side CA and side CB, then the triangle ABC is an isosceles triangle.

Setup for coordinate geometry proof:

M is the midpoint of AB since CM is a bisector of AB.

Since M is the midpoint of AB, then M is located at the coordinates $(\frac{0+b}{2},\frac{0+0}{2})=(\frac{b}{2},0)$ .

We found this point such that the length AM is equal to the length MB.

That is, the distance between A and M is the same as the distance between M and B.

Let's check.

$AM=\sqrt{(\frac{b}{2}-0)^2+(0-0)^2}$

$AM=\sqrt{(\frac{b}{2})^2+0}$

$AM=\sqrt{\frac{b^2}{4}}$

$AM=\frac{\sqrt{b^2}}{\sqrt{4}}$

$AM=\frac{b}{2}$

$MB=\sqrt{(b-\frac{b}{2})^2+(0-0)^2}$

$MB=\sqrt{(\frac{b}{2})^2+0}$

$MB=\sqrt{\frac{b^2}{4}}$

$MB=\frac{\sqrt{b^2}}{\sqrt{4}}$

$MB=\frac{b}{2}$

We have confirmed that AM=MB.

(Based on the picture, we could have taken a slightly easier route to calculate the distance between M and A, then the distance between B and M. They are both a horizontal distance. So $MB=b-\frac{b}{2}=\frac{b}{2}$ where as $AM=\frac{b}{2}-0=\frac{b}{2}$ .)

Now we also want to assume that the line segment CM is perpendicular to AB. I have drawn the base of the triangle on the x-axis so a vertical line would be perpendicular to it. Also this would make point $C=(c,d)=(\frac{b}{2},d)$ . The $y$ -coordinate is $d$ because we don't know how high above the $x$ -axis the point $C$ is.