Ask question

Ask question

All categories

Brilliant_brown [7]

3 years ago

15

G(x) = -3x + 1 solve for x (which is 22)

2 answers:

olga2289 [7]3 years ago

6 0

Answer:

-65

Step-by-step explanation:

wolverine [178]3 years ago

5 0

Answer:

-65

Step-by-step explanation:

-3(22) + 1

-65

You might be interested in

The class of math is mapped on a coordinate grid with the origin being at the center point of the hall. Mary’s seat is located a

nignag [31]

To find the distance b/t 2 points you need to use the distance formula. that is

distance = square root of (x2-x1)^2 + (y2-y1)^2

to fill it in, it is

(it doesnt matter which point you start with, as long as its consistent)

x1 = 9
x2 = 5
y1 = -8
y2 = -10

square root of (5-9)^2 + (-10-8)^2

lets do algebra!

square root of (-4)^2 + (-18)^2

square root of 16 + 324

square root of 340

Distance = 18.4391 units.

I hope this helps!

7 0

3 years ago

Which expressions are less that 610

lesantik [10]

Answer:

uhhh you didnt give any answer choices

Step-by-step explanation:

4 0

3 years ago

Evaluate the expression x= -2

Juli2301 [7.4K]

Answer:

x equals negative two

Step-by-step explanation:

that's how you would say it

7 0

3 years ago

What is the measure of angle 3?

Marrrta [24]

Answer:

can you insert a picture or other information to answer this pls

Step-by-step explanation:

6 0

4 years ago

Derive these identities using the addition or subtraction formulas for sine or cosine: sinacosb=(sin(a+b)+sin(a-b))/2

Sergeu [11.5K]

Answer:

The work is in the explanation.

Step-by-step explanation:

The sine addition identity is:

$\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$ .

The sine difference identity is:

$\sin(a-b)=\sin(a)\cos(b)-\cos(a)\sin(a)$ .

The cosine addition identity is:

$\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$ .

The cosine difference identity is:

$\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)$ .

We need to find a way to put some or all of these together to get:

$\sin(a)\cos(b)=\frac{\sin(a+b)+\sin(a-b)}{2}$ .

So I do notice on the right hand side the $\sin(a+b)$ and the $\sin(a-b)$ .

Let's start there then.

There is a plus sign in between them so let's add those together:

$\sin(a+b)+\sin(a-b)$

$=[\sin(a+b)]+[\sin(a-b)]$

$=[\sin(a)\cos(b)+\cos(a)\sin(b)]+[\sin(a)\cos(b)-\cos(a)\sin(b)]$

There are two pairs of like terms. I will gather them together so you can see it more clearly:

$=[\sin(a)\cos(b)+\sin(a)\cos(b)]+[\cos(a)\sin(b)-\cos(a)\sin(b)]$

$=2\sin(a)\cos(b)+0$

$=2\sin(a)\cos(b)$

So this implies:

$\sin(a+b)+\sin(a-b)=2\sin(a)\cos(b)$

Divide both sides by 2:

$\frac{\sin(a+b)+\sin(a-b)}{2}=\sin(a)\cos(b)$

By the symmetric property we can write:

$\sin(a)\cos(b)=\frac{\sin(a+b)+\sin(a-b)}{2}$

3 0

3 years ago