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GenaCL600 [577]
2 years ago
8

What is 4.5a+7+3.5a+2

Mathematics
1 answer:
Nitella [24]2 years ago
3 0

Answer:

8a+9

Step-by-step explanation:

You have to add the like terms.

4.5a+7+3.5a+2

4.5a+3.5a=8a

7+2= 9

8a+9

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Find the derivative of the following. please show the steps when you answer :1) f(x) = 8xe^x2) y= 5xe^x^43) f(x)= x^8+5/x4) f(t)
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Answer:

Since,

\frac{d}{dx}x^n = nx^{n-1}

\frac{d}{dx}(f(x).g(x)) = f(x).\frac{d}{dx}(g(x)) + g(x).\frac{d}{dx}(f(x))

\frac{d}{dx}(\frac{f(x)}{g(x)})=\frac{g(x).f'(x) - f(x) g'(x)}{(g(x))^2}

1) y = 8x e^x

Differentiating with respect to x,

\frac{dy}{dx}=8( x \times e^x + e^x) = 8(xe^x + e^x) = 8e^x(x+1)

2) y = 5x e^{x^4}

Differentiating w. r. t x,

\frac{dy}{dx}=5(x\times 4x^3 e^{x^4}+e^{x^4})=5e^{x^4}(4x^4+1)

3) y = x^8 + \frac{5}{x^4}

Differentiating w. r. t. x,

\frac{dy}{dx}=8x^7 - \frac{5}{x^5}\times 4 = 8x^7 - \frac{20}{x^5}=\frac{8x^{12}-20}{x^5}

4) f(t) = te^{11}-6t^5

Differentiating w. r. t. t,

f'(t) = e^{11} - 30t^4

5) g(p) = p\ln(2p+3)

Differentiating w. r. t. p,

g'(p) = p\frac{1}{2p+3}(2) + \ln(2p+3) = \frac{2p}{2p+3}+\ln(2p+3)

6) z = (te^{6t}+e^{5t})^7

Differentiating w. r. t. t,

\frac{dz}{dt}=7(te^{6t}+e^{5t})^6 ( 6te^{6t}+e^{6t} + 5e^{5t})

7) w =\frac{2y + y^2}{7+y}

Differentiating w. r. t. y,

\frac{dw}{dy} = \frac{(7+y)(2+2y)-(2y+y^2)}{(7+y)^2} = \frac{14 + 2y + 14y +2y^2 - 2y - y^2}{(7+y)^2}=\frac{14+14y+y^2}{(7+y)^2}

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Juan has blood type A, Lilly has blood type AB, and Aaron has blood type O. Is this relation a function? Explain.
nexus9112 [7]

Answer: Yes, this relation is function.

Step-by-step explanation:

A relation is called function if each input has unique output.

It is given that, Juan has blood type A, Lilly has blood type AB, and Aaron has blood type O.

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Lilly is related to blood type AB.

Aaron is related to blood type O.

Here, each input has unique output because here each person has unique blood group.

Therefore, this relation is a function.

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3 years ago
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