3 years ago

Help?

1 answer:

4 0

This is just good 'ol Pythagorean Theorem.

a^2 + b^2 = c^2

The diagonal will be c because c is always across from the right angle of a right triangle.

3.6^2 + 2.4^2 = c^2
12.96 + 5.76 = c^2
18.72 = c^2

Now take the square root of 18.72 to find c. You get 4.32666...

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5.3, CD r- 8.8, and AD - 14.3, find AB to the nearest tenth.<br>If BC

pychu [463]

5.3/8.8=x/14.3
AB is x,
x= 2.43

Hope I helped.

7 0

3 years ago

How do I solve this?

blagie [28]

Answer:

We'll solve that in parts.

The upper part is a right triangle that is 4 by (2 + 2 +6)

Area = (4 * 10) /2 = 20

The bottom of that figure is made of 3 parts.

A) a 2 by 2 right triangle. area equals (2*2)/2 = 2

B) a 2 by 2 square whose area equals 2*2 = 4

C) a 6 by 2 right triangle whose area equals (6*2) / 2 = 6

Total Area = 20 + 2 + 4 + 6 = 32

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Angle 0 lies in the second quadrant and sin 0 = 3/5 cos= coy =

Elza [17]

Consider the right triangle ABC with legs AB=4, AC=3 and hypotenuse BC=5. Angle B has
$\sin B=\frac{opposite}{hypotenuse} = \frac{3}{5}$ and
$\cos B = \frac{adjacent}{hypotenuse} = \frac{4}{5}$ .

Since O lies in second quadrant $\cos O\ \textless \ 0$ and
$\cos O= -\cos B =- \frac{4}{5}$ .
Answer: $\cos O=- \frac{4}{5}$ .

3 0

3 years ago

HELP ASAP!!!!!!!!!!

chubhunter [2.5K]

Answer:

7.5

Step-by-step explanation:

$\sqrt{0\\.3}$ = 0.5477225575