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ki77a [65]
3 years ago
9

Help please? A system of linear equations is graphed. Which ordered pair is the best estimate for the solution to the system? A.

(0,7) B. (-4, 2 1/2) C. (0,-2) D. (-4 1/2, 2 1/2)

Mathematics
1 answer:
kherson [118]3 years ago
7 0

Answer:

B. (-4, 2½)

Step-by-step explanation:

The lines intercept at the point -4 on the x-axis and 2½ on the y-axis

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Simplify. 7√32−6√72 Enter your answer, in simplest radical form, in the box.
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Answer-

-8\sqrt2

Step-by-step explanation:

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What is a way that you can write 83.041 in notation form?
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4 years ago
For the expression: 4(2x-3) simplified I know it is 8x-12.
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Let z = ln(x 2 + y), x = ret . and y = ter . Use the Chain Rule to compute ∂z ∂r and ∂z ∂t at the point where (r, t) = (1, 2).\
Natali [406]

By the chain rule,

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

where u\in\{r,t\}.

We have component partial derivatives

\dfrac{\partial z}{\partial x}=\dfrac{2x}{x^2+y}=\dfrac{2re^t}{r^2e^{2t}+te^r}

\dfrac{\partial z}{\partial y}=\dfrac1{x^2+y}=\dfrac1{r^2e^{2t}+te^r}

\dfrac{\partial x}{\partial r}=e^t

\dfrac{\partial x}{\partial t}=re^t

\dfrac{\partial y}{\partial r}=te^r

\dfrac{\partial y}{\partial t}=e^r

Putting the appropriate pieces together and setting (r,t)=(1,2), we get

\dfrac{\partial z}{\partial r}(1,2)=\dfrac{2e^3+2}{e^3+2}

\dfrac{\partial z}{\partial t}(1,2)=\dfrac{2e^3+1}{e^3+2}

3 0
3 years ago
Log 24 base 5 - log 6 base 5
Jet001 [13]
To subtract logs, you need to use the log rule, that when you subtract logs, you divide the two arguments and keep the base if it is the same. So divide 24 by 6 and you get 4. So the answer to your question is Log 4 base 5. Hope this helps. 
4 0
4 years ago
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