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3 years ago

Use the product rule to simplify sq root 147x^2

Mathematics

1 answer:

klio [65]3 years ago

3 0

Answer:7x√3

Step-by-step explanation:

√147x^2

√49 (3) x^2

√7^2 ⋅ 3x^2

√7^2x^2 ⋅ 3

√(7x)^2 ⋅ 3

7x√3

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Find an equation of the line that passes through the given points. (Let x be the independent variable and y be the dependent var

lara [203]

Answer:

y=5x-6

Step-by-step explanation:

Hello

let´s see , we have P1(3,9) and P2(4.14)

first let's find its slope

$m=\frac{y2-y1}{x2-x1} = \frac{14-9}{4-3}$

$m=\frac{5}{1} \\m=5$

the slope is 5.

$y-y_{0} = m(x-x_{0} )\\\\using P1\\\\ y-9 = 5(x-3)\\\\y=5x-15+9\\\\y=5x-6$

where x is the independent variable and y is the dependent variable

Have a great day.

8 0

3 years ago

PLEASE HELP I NEED TO PASS MATH!

Delicious77 [7]

D is the answer because 2+5 = 7 and ya

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3 years ago

Which of the following is a solution of x^2 + 2x + 4?

Kaylis [27]

Answer:

$\displaystyle x_1=-1+\sqrt{3}i$

$\displaystyle x_2=-1-\sqrt{3}i$

Step-by-step explanation:

<u>Second-Degree Equation</u>

The second-degree equation or quadratic equation has the general form

$ax^2+bx+c=0$

where a is non-zero.

There are many methods to solve the equation, one of the most-used is by using the solver formula:

$\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

The equation of the question has the values: a=1, b=2, c=4, thus the values of x are

$\displaystyle x=\frac{-2\pm \sqrt{2^2-4\cdot 1\cdot 4}}{2\cdot 1}$

$\displaystyle x=\frac{-2\pm \sqrt{-12}}{2}$

Since the square root has a negative argument, both solutions for x are imaginary or complex. Simplifying the radical

$\displaystyle x=\frac{-2\pm 2\sqrt{-3}}{2}=-1\pm\sqrt{3}i$

The solutions are

$\displaystyle x_1=-1+\sqrt{3}i$

$\displaystyle x_2=-1-\sqrt{3}i$

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3 years ago

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vova2212 [387]

Answer:

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Step-by-step explanation:

::'cx

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Arte-miy333 [17]

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Step-by-step explanation:

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