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Liula [17]
3 years ago
8

This is 1 day late- help. ASAP

Mathematics
2 answers:
White raven [17]3 years ago
7 0

Answer:

the third option

Step-by-step explanation:

do you want an explanation?

Sliva [168]3 years ago
3 0
The answer to this question is x=5
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HELP HELP HELP I NEED HELPPP can anyone answer this the question is on the picture it would help a lot please!!!
forsale [732]
It’s -1/2 (negative 1 over 2)
4 0
3 years ago
I am really confused about this question for math.. any help would be really appreciated:
Elina [12.6K]

y= (\frac 1 4 )^x

A reflection about the x axis, about y=0, is the mapping (x',y')=(x,-y) so

y'= -y = - (\frac 1 4)^{x'}

A dilation of 2 is the mapping (x'',y'')=(2x', 2y')

So

x'=x''/2, y'=y''/2

y''/2= - (\frac 1 4)^{x''/2}

y'' =  - 2((\frac 1 4)^{1/2})^{x''}

y''= - 2(\frac 1 2)^{x''}

We can rewrite that without the primes and combine the powers of 2.

y =  - 2^{1-x}

Let's graph these and see if we're close,

Plot y= (1/4)^x, y= - (1/4)^{x}, y = - 2^{1-x}

6 0
3 years ago
Solve: -2/5 (1/4 x -4)
Nimfa-mama [501]

Answer:

2/5 is tje correct answer for your question

8 0
3 years ago
Read 2 more answers
"The sum of a number and 4."
kenny6666 [7]

Answer:

n + 4

Step-by-step explanation:

It´s like a normal sum, but since the number is unknown just put n.

7 0
3 years ago
Read 2 more answers
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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