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3 years ago

6

Clark Kent walks 3 miles west and then 4 miles north. How far is it for Superman to fly in a straight line back to his starting

point? (It makes a right triangle.)
Question 16 options:

3 miles

5 miles

7 miles

9 miles

2 answers:

KonstantinChe [14]3 years ago

7 0

It would be 5 miles if the triangle was a right angle

kozerog [31]3 years ago

5 0

You can use the pythagorean theorem a² + b² = c², where a and b are perpendicular sides of a triangle and c is the hypotenuse.

For this problem, a = 3, b = 4
c = √(3² + 4²) = √5² = 5

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Circle O has a circumference of 36π cm.

ehidna [41]

Answer:

r = 18 cm

Step-by-step explanation:

The formula for the circumference of a circle is 2πr.

Hence,

2πr = 36π

r = 18 cm

3 0

2 years ago

What’s the answer?!!!!

mel-nik [20]

Answer: YES

Step-by-step explanation:

<u>Concept</u>

In a right triangle, the square of both legs is equal to the length of hypotenuse squared. This is the Pythagorean theorem: a²+b²=c². For more graphical explanation, please refer to the attachment below.

<u>Given</u>

Leg a = 36 in

Leg b = 48 in

Hypotenuse c = 60 in

<u>Solve</u>

a² + b² = c²

(36)² + (48)² = 3600

(60)² = 3600

3600 = 3600

Therefore, the answer is <u>YES</u>

Hope this helps!! :)

Please let me know if you have any questions

6 0

3 years ago

How to find X in a parallelogram

myrzilka [38]

Answer:

X=7

Step-by-step explanation:

15-8

4 0

2 years ago

PLZ HELPPPP ME ASAP!!!!

zepelin [54]

answer: m=1

work:

3/2 + 2m = 7/2

3/2 (-3/2) + 2m = 7/2 (-3/2) subtract 3/2 from both sides

2m = 2

2m/2 = 2/2 divide 2 on both sides

m = 1

5 0

3 years ago

The coordinate plane shows the floor plan for a swimming pool. What is the area of the pool’s border?

Ivanshal [37]

we know that

The area of a rectangle is equal to

$A=L*W$

where

L is the length side of the rectangle

W is the width side of the rectangle

see the attached figure with letters to better understand the problem

The area of the pool's border is equal to the area of rectangle ABCD minus the area of rectangle EFGH

The formula to calculate the distance between two points is equal to

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

Step 1

<u>Find the distance AB</u>

Let

$A(0,12)\ B(8,18)$

substitute the values in the formula

$d=\sqrt{(18-12)^{2}+(8-0)^{2}}$

$d=\sqrt{(6)^{2}+(8)^{2}}$

$d=\sqrt{100}$

$dAB=10\ m$

Step 2

<u>Find the distance BC</u>

Let

$B(8,18)\ C(17,6)$

substitute the values in the formula

$d=\sqrt{(6-18)^{2}+(17-8)^{2}}$

$d=\sqrt{(-12)^{2}+(9)^{2}}$

$d=\sqrt{225}$

$dBC=15\ m$

Step 3

<u>Find the distance EF</u>

Let

$E(3,12)\ F(7,15)$

substitute the values in the formula

$d=\sqrt{(15-12)^{2}+(7-3)^{2}}$

$d=\sqrt{(3)^{2}+(4)^{2}}$

$d=\sqrt{25}$

$d.EF=5\ m$

Step 4

<u>Find the distance FG</u>

Let

$F(7,15)\ G(13,7)$

substitute the values in the formula

$d=\sqrt{(7-15)^{2}+(13-7)^{2}}$

$d=\sqrt{(-8)^{2}+(6)^{2}}$

$d=\sqrt{100}$

$dFG=10\ m$

Step 5

<u>Find the area of the pool's border</u>

The area of the pool's border is equal to the area of rectangle ABCD minus the area of rectangle EFGH

<u>Area of the rectangle ABCD</u>

$A1=AB*BC=10\ m*15\ m=150\ m^{2}$

<u>Area of the rectangle EFGH</u>

$A2=EF*FG=5\ m*10\ m=50\ m^{2}$

<u>Area of the pool's border</u>

$A=A1-A2=150\ m^{2}-50\ m^{2}=100\ m^{2}$

therefore

<u>the answer is</u>

$100\ m^{2}$

8 0

3 years ago

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