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3 years ago

How to evaluate s-r/-7 when r=-3 and s=4?

Mathematics

1 answer:

jasenka [17]3 years ago

3 0

Plug in r and s into your equation
You should get -1 for your final answer

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Multiply.<br> 7.8 x 100<br> 7.83 x 10,000

luda_lava [24]

7.8 x 100 = 780
7.83 x 10,000 = 78,300

6 0

3 years ago

Read 2 more answers

A guy wire is used to support flagpoles in windy areas. These wires are anchored in the ground to provide constant stability. Yo

r-ruslan [8.4K]

Answer: a) Length of wire to replace old one=9ft

b) Length of wire=3.09ft

Leftover wire=0.91ft

Step-by-step explanation: from the right angle triangle, Tan&=20/15

&=53.13°

Length of wire,x=cos 31.13=x/15

15cos31.13

X=9

b) Tan&=10/6

Tan^-110/6

&=59.05

Length= 6cos59.04

X=3.09

4-3.09=.91= leftover wire

8 0

3 years ago

Which of the following trigonometric ratios are correct?

notka56 [123]

9514 1404 393

Answer:

D, F

Step-by-step explanation:

The mnemonic SOH CAH TOA can help you with this. It reminds you ...

Sin = Opposite/Hypotenus

sin(40°) = x/b . . . . matches D

Cos = Adjacent/Hypotenuse

<em> cos(40°) = a/b . . . . no match</em>

Tan = Opposite/Adjacent

tan(40°) = x/a . . . . matches F

8 0

3 years ago

Two packages weigh 3.92 pounds and 2.8 pounds. How many times heavier Is the first package than the second?

avanturin [10]

3.92/2.8=1.4

Unless it’s reverse.

2.8/3.92=0.714285714285 and so on.

3 0

3 years ago

Find the integration of (1-cos2x)/(1+cos2x)

slega [8]

Given:

The expression is:

$\dfrac{1-\cos 2x}{1+\cos 2x}$

To find:

The integration of the given expression.

Solution:

We need to find the integration of $\dfrac{1-\cos 2x}{1+\cos 2x}$ .

Let us consider,

$I=\int \dfrac{1-\cos 2x}{1+\cos 2x}dx$

$I=\int \dfrac{2\sin^2x}{2\cos^2x}dx$ $[\because 1+\cos 2x=2\cos^2x,1-\cos 2x=2\sin^2x]$

$I=\int \dfrac{\sin^2x}{\cos^2x}dx$

$I=\int \tan^2xdx$ $\left[\because \tan \theta =\dfrac{\sin \theta}{\cos \theta}\right]$

It can be written as:

$I=\int (\sec^2x-1)dx$ $[\because 1+\tan^2 \theta =\sec^2 \theta]$

$I=\int \sec^2xdx-\int 1dx$

$I=\tan x-x+C$

Therefore, the integration of $\dfrac{1-\cos 2x}{1+\cos 2x}$ is $I=\tan x-x+C$ .

8 0

3 years ago