All categories

3 years ago

Please help me with the questions in the image

Mathematics

1 answer:

elena-s [515]3 years ago

5 0

A) the image on the lower left is correct. Because the equation is set up as “F=“, F goes on the y-axis. The y-intercept is positive 32.

B) the slope is 9/5 because the equation is in the form of y=mc+b, where m is the slope. F increases 9/5 degrees for each increase of 1 C. The F-intercept is 32 and it represents the temperature 32 F corresponding to 0 C.

You might be interested in

Solve the inequality 13×-4 < 12x-1

Murrr4er [49]

Answer:

x <3

Step-by-step explanation:

13x-4 < 12x-1

Subtract 12x from each side

13x-12x-4 < 12x-12x-1

x-4 <-1

Add 4 to each side

x-4+4<-1+4

x <3

7 0

3 years ago

Find the inverse function of f(x)=-4x-10

Dvinal [7]

$f(x)=-4x-10\\
y=-4x-10\\
-4x=y+10\\
x=-\frac{1}{4}y-\frac{5}{2}\\
f^{-1}(x)=-\frac{1}{4}x-\frac{5}{2}$

6 0

3 years ago

HELP QUICK!!!

sattari [20]

Answer:

Step-by-step explanation:

let the length of shortest side be x

and length of second side be 2x

and third side be x+17

given , perimeter = 173 feet

therefore , x+2x+x+17 =173

4x =173-17

4x =156

x = 156/4

x = 39

therefore ,

length of :

shortest side = 39 feet

second side =78 feet

third side =56 feet

hope it helps.. please mark me as the brainliest...

8 0

3 years ago

Rewrite 2tan3x in terms of tanx

blsea [12.9K]

$2\tan3x=2\tan(2x+x)=\dfrac{2\tan2x+2\tan x}{1-\tan2x\tan x}$

Use the same identity to expand $\tan2x$ .

$\tan2x=\tan(x+x)=\dfrac{\tan x+\tan x}{1-\tan x\tan x}=\dfrac{2\tan x}{1-\tan^2x}$

$\implies2\tan3x=\dfrac{2\dfrac{2\tan x}{1-\tan^2x}+2\tan x}{1-\dfrac{2\tan x}{1-\tan^2x}\tan x}$
$2\tan3x=\dfrac{2\tan x\left(\dfrac2{1-\tan^2x}+1\right)}{1-\dfrac{2\tan^2x}{1-\tan^2x}}$
$2\tan3x=\dfrac{2\tan x\left(2+1-\tan^2x\right)}{1-\tan^2x-2\tan^2x}$
$2\tan3x=\dfrac{2\tan x(3-\tan^2x)}{1-3\tan^2x}$

4 0

3 years ago

Which expression is equivalent to

TiliK225 [7]

$x^2+3x+3+\dfrac{2x^2+13x+15}{x+5}=x^2+3x+3+\dfrac{2x^2+10x+3x+15}{x+5}=\\\\\\=
x^2+3x+3+\dfrac{2x(x+5)+3(x+5)}{x+5}=\\\\\\=x^2+3x+3+\dfrac{(2x+3)(x+5)}{x+5}=x^2+3x+3+2x+3=\\\\\\=\boxed{x^2+5x+6}$

Answer B.

3 0

4 years ago