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Lapatulllka [165]
3 years ago
15

Which of the following square root of -80

Mathematics
1 answer:
Marta_Voda [28]3 years ago
8 0

You can factor -80 as

-80 = (-1)\cdot 16 \cdot 5

So, we have

\sqrt{-80} = \sqrt{(-1)\cdot 16 \cdot 5}

The square root of a product is the product of the square roots:

\sqrt{-80} = \sqrt{(-1)}\sqrt{16}\sqrt{5}

Since i^2=-1 and 4^2=16, we have

\sqrt{-80} = 4i\sqrt{5}

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What is the y-intercept of the function f(x)=4 – 5x?<br> A:-5<br> B:-4<br> C:4<br> D:-5
Bogdan [553]

Answer: option C is the correct answer

Step-by-step explanation:

The y-intercept of the function is the point on the y or vertical axis at which the straight line cuts through it. The value on the y axis is read off as the y-intercept .

The equation of a straight line is modelled using the slope - intercept equation of

y = mx + c

Where m is the slope( change in value of y/ change in value of x.

c = the y-intercept

Looking at the function given,

f(x)=4 – 5x? Where f(x) is the same as y in our slope intercept equation.

Rearranging the function, it becomes

f(x)= – 5x +4

The intercept is 4

Option C is the correct answer.

7 0
3 years ago
Find the measure of each angle if c||d show work
weeeeeb [17]

Answer:

Part a) <1=72°

Part b) <2=108°

Part c) <3=72°

Part d) <4=108°

Step-by-step explanation:

step 1

Find the measure of angle 1

we know that

<1+108°=180° -----> by supplementary angles

so

<1=180°-108°=72°

step 2

Find the measure angle 2

we know that

<2=108° -----> by corresponding angles

step 3

Find the measure angle 3

we know that

<3=<1-----> by corresponding angles

so

<3=72°

step 4

Find the measure angle 4

we know that

<4=108° -----> by alternate exterior angles

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3 years ago
Find an explicit solution to the Bernoulli equation. y'-1/3 y = 1/3 xe^xln(x)y^-2
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y'-\dfrac13y=\dfrac13xe^x\ln x\,y^{-2}

Divide both sides by \dfrac13y^{-2}(x):

3y^2y'-y^3=xe^x\ln x

Substitute v(x)=y(x)^3, so that v'(x)=3y(x)^2y'(x).

v'-v=xe^x\ln x

Multiply both sides by e^{-x}:

e^{-x}v'-e^{-x}v=x\ln x

The left side can be condensed into the derivative of a product.

(e^{-x}v)'=x\ln x

Integrate both sides to get

e^{-x}v=\dfrac12x^2\ln x-\dfrac14x^2+C

Solve for v(x):

v=\dfrac12x^2e^x\ln x-\dfrac14x^2e^x+Ce^x

Solve for y(x):

y^3=\dfrac12x^2e^x\ln x-\dfrac14x^2e^x+Ce^x

\implies\boxed{y(x)=\sqrt[3]{\dfrac14x^2e^x(2\ln x-1)+Ce^x}}

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