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lina2011 [118]
3 years ago
5

Solve for the variable: -11t + 23 = 56

Mathematics
1 answer:
Annette [7]3 years ago
3 0
Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation : 

                    <span> 3*x+23-(56)=0 </span>
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Please help me on this math question
skelet666 [1.2K]

Answer:

x = 6 , y = 3

Step-by-step explanation:

x− 2 y = 0 ⇒ x = 2 y  

Substitute x=2y into 4x-3y=15, we get : 4 ( 2 y ) − 3 y = 15

8 y− 3 y = 15  5 y = 15 ⇒ y = 3  x = 2 y = 2 × 3 = 6

8 0
3 years ago
Multiply 154.18 and 5 together. Round to the nearest hundredth when necessary.
koban [17]

Answer:

that will be 725.10 im not sure

6 0
2 years ago
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Find the derivative
zalisa [80]

First use the chain rule; take y=\dfrac{x+5}{x^2+3}. Then

\dfrac{\mathrm df}{\mathrm dx}=\dfrac{\mathrm df}{\mathrm dy}\cdot\dfrac{\mathrm dy}{\mathrm dx}

By the power rule,

f(x)=y^2\implies\dfrac{\mathrm df}{\mathrm dy}=2y=\dfrac{2(x+5)}{x^2+3}

By the quotient rule,

y=\dfrac{x+5}{x^2+3}\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{(x^2+3)\frac{\mathrm d(x+5)}{\mathrm dx}-(x+5)\frac{\mathrm d(x^2+3)}{\mathrm dx}}{(x^2+3)^2}

\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{(x^2+3)-(x+5)(2x)}{(x^2+3)^2}

\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{3-10x-x^2}{(x^2+3)^2}

So

\dfrac{\mathrm df}{\mathrm dx}=\dfrac{2(x+5)}{x^2+3}\cdot\dfrac{3-10x-x^2}{(x^2+3)^2}

\implies\dfrac{\mathrm df}{\mathrm dx}=\dfrac{2(x+5)(3-10x-x^2)}{(x^2+3)^3}

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snow_lady [41]
The missing one on the top row is 28. The one missing on the bottom is 4
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ololo11 [35]

Answer:

-1

Step-by-step explanation:

It is negative 1 because the signs for the -3 are the same so you keep the -4, change the - to a +, then flip the negative 3.

Hope this helps!

7 0
2 years ago
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