All categories

2 years ago

I really need help Really

Mathematics

1 answer:

DIA [1.3K]2 years ago

6 0

Answer:

920cm^2

Step-by-step explanation:

look at the attachment below hope it helps pls mark brainliest :)

You might be interested in

What figure has one endpoint and continues forever in one direction? A. Angle B. Line C. Ray D. Line Segment

quester [9]

The answer to your question is Ray

7 0

2 years ago

Read 2 more answers

Por la seguridad de su personal y sus clientes una agencia bancaria se instalara una camara de video en un soporte de pared de m

yulyashka [42]

Answer:

16.24°

Step-by-step explanation:

Are you trying to find the angle of depression from the camera to the cashier?

Assume that the camera is 2.24 m high and the cashier is 7.69 m away.

1. Angle of elevation from cashier

tanθ = 2.24/7.69 = 0.2913

θ = arctan(0.2913) = 16.24°

2. Angle of depression from camera

∠ of depression = ∠ of elevation = 16.24°

6 0

2 years ago

1c) What is 25% of 40? 8.5

grin007 [14]

25% of 40

=25/100×40

=0.25×40

=10 answer.

8 0

2 years ago

AHHH HELPPP ME PLEASEE

Mamont248 [21]

Answer:

the third one is hard to see i think its 77 so the answer is 54.75

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

Verify: cos(2A)=(cotA-tanA)/cscAsecA

kolbaska11 [484]

Answer:

see explanation

Step-by-step explanation:

Using the trigonometric identities

cot A = $\frac{cosA}{sinA}$ , tanA = $\frac{sinA}{cosA}$ , cscA = $\frac{1}{sinA}$ , secA = $\frac{1}{cosA}$

Consider the right side

$\frac{cotA-tanA}{cscAsecA}$

= $\frac{\frac{cosA}{sinA}-\frac{sinA}{cosA} }{\frac{1}{sinA}.\frac{1}{cosA} }$

= $\frac{\frac{cos^2A-sin^2A}{sinAcosA} }{\frac{1}{sinAcosA} }$

= $\frac{cos^2A-sin^2A}{sinAcosA}$ × sinAcosA ( cancel sinAcosA )

= cos²A - sin²A

= cos2A

= left side ⇒ verified

5 0

2 years ago