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kkurt [141]
3 years ago
14

Answer asap, please needed by 2:00 pm

Mathematics
1 answer:
myrzilka [38]3 years ago
7 0

Answer:

x < -3

Step-by-step explanation:

The domain in which the function is increasing or getting bigger is from x is negative infinity to  is -3

x < -3

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I'll GIVE BRAINLIST TO THE FRIST PERSON!!!!!!!!!!!!!
hram777 [196]

Answer: the answer should be c

Step-by-step explanation: -3x - 13 is easily found by the perimeter of time (y as a representative)

5 0
4 years ago
Read 2 more answers
15. Write the slope-intercept form of the line described in the following:Parallel to 4x + 5y=20and passing through (12,4)
Amiraneli [1.4K]

Problem

Write the slope-intercept form of the line described in the following:

Parallel to 4x + 5y=20

and passing through (12,4)

Solution>

For this case we need to have the same slope, and if we write the equation given we see:

5y = 20 -4x

y = 4 -4/5 x

then the slope m = -4/5

and we also know a point given x= 12, y= 4 and we can do the following:

4 = -4/5 (12) +b

4 = -48/5 + b

And if we solve for the intercept we got:

b= 4 +48/5= -28/5

And our equation would be given by:

y = -4/5 x -28/5

3 0
1 year ago
For altitudes up to 36,000 feet, the relationship between ground temperature and atmospheric
Anon25 [30]

Answer:

a=16400 feet

Step-by-step explanation:

t = -0.0035 a +g

-17.40 = -0.0035 a+40

-17.40-40=-0.0035a+40-40

-57.40=-0.0035a

a = -57.40/-0.0035

a=16400 feet

7 0
3 years ago
Find the sun and express it in simplest form.(-3s-4c+1)+(-3s+3c)
DiKsa [7]

Answer:

-6s-c+1

Step-by-step explanation:

(-3s-4c+1)+(-3s+3c)

We have been given the above expression. To find the sum, we simply collect the like terms and combine them;

(-3s-4c+1)+(-3s+3c) = -3s + -3s -4c + 3c + 1

-3s + -3s -4c + 3c + 1 = -3s - 3s + 3c - 4c + 1

-3s - 3s + 3c - 4c + 1 = -6s - c + 1

Therefore;

(-3s-4c+1)+(-3s+3c) = -6s-c+1

5 0
3 years ago
Suppose z equals f (x comma y ), where x (u comma v )space equals space 2 u plus space v squared, y (u comma v )space equals spa
barxatty [35]

z=f(x(u,v),y(u,v)),\begin{cases}x(u,v)=2u+v^2\\y(u,v)=3u-v\end{cases}

We're given that f_x(6,1)=3 and f_y(6,1)=-1, and want to find \frac{\partial z}{\partial v}(1,2).

By the chain rule, we have

\dfrac{\partial z}{\partial v}=\dfrac{\partial z}{\partial x}\dfrac{\partial x}{\partial v}+\dfrac{\partial z}{\partial y}\dfrac{\partial y}{\partial v}

and

\dfrac{\partial x}{\partial v}=2v

\dfrac{\partial y}{\partial v}=-1

Then

\dfrac{\partial z}{\partial v}(1,2)=\dfrac{\partial z}{\partial x}(6,1)\dfrac{\partial x}{\partial v}(1,2)+\dfrac{\partial z}{\partial y}(6,1)\dfrac{\partial y}{\partial v}(1,2)

(because the point (x,y)=(6,1) corresponds to (u,v)=(1,2))

\implies\dfrac{\partial z}{\partial v}(1,2)=3\cdot2\cdot2+(-1)\cdot(-1)=\boxed{13}

4 0
3 years ago
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