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3 years ago

PLS HELP ME ON THIS QUESTION I WILL MARK YOU AS BRAINLIEST IF YOU KNOW THE ANSWER PLS GIVE ME A STEP BY STEP EXPLANATION!!

Mathematics

1 answer:

Vanyuwa [196]3 years ago

7 0

Answer:

I'm not so sure but i guess it's option D y=5x-3

the question said that the number of box of nails are purchased depending on how many houses they planned to build. so for knowing the boxes of nails they need to purchase they need to subtract the boxes which were left and the boxes we need now

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Hey How are u wanna join my class on remind

Nostrana [21]

Step-by-step explanation:

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5 0

3 years ago

Read 2 more answers

Differentiate with respect to x and simplify your answer. Show all the appropriate steps? 1.e^-2xlog(ln x)^3 2.e^-2x(log(ln x))^

serious [3.7K]

(1) I assume "log" on its own refers to the base-10 logarithm.

$\left(e^{-2x}\log(\ln x)^3\right)'=\left(e^{-2x}\right)'\log(\ln x)^3+e^{-2x}\left(\log(\ln x)^3\right)'$

$=-2e^{-2x}\log(\ln x)^3+\dfrac{e^{-2x}}{\ln10(\ln x)^3}\left((\ln x)^3\right)'$

$=-2e^{-2x}\log(\ln x)^3+\dfrac{3e^{-2x}(\ln x)^2}{\ln10(\ln x)^3}\left(\ln x\right)'$

$=-2e^{-2x}\log(\ln x)^3+\dfrac{3e^{-2x}(\ln x)^2}{\ln10\,x(\ln x)^3}$

$=-2e^{-2x}\log(\ln x)^3+\dfrac{3e^{-2x}}{\ln10\,x\ln x}$

Note that writing $\log(\ln x)^3=3\log(\ln x)$ is one way to avoid using the power rule.

(2)

$\left(e^{-2x}(\log(\ln x))^3\right)'=(e^{-2x})'(\log(\ln x))^3+e^{-2x}\left(\log(\ln x))^3\right)'$

$=-2e^{-2x}(\log(\ln x))^3+3e^{-2x}(\log(\ln x))^2(\log(\ln x))'$

$=-2e^{-2x}(\log(\ln x))^3+3e^{-2x}(\log(\ln x))^2\dfrac{(\ln x)'}{\ln10\,\ln x}$

$=-2e^{-2x}(\log(\ln x))^3+\dfrac{3e^{-2x}(\log(\ln x))^2}{\ln10\,x\ln x}$

(3)

$\left(\sin(xe^x)^3\right)'=\left(\sin(x^3e^{3x})\right)'=\cos(x^3e^{3x}(x^3e^{3x})'$

$=\cos(x^3e^{3x})((x^3)'e^{3x}+x^3(e^{3x})')$

$=\cos(x^3e^{3x})(3x^2e^{3x}+3x^3e^{3x})$

$=3x^2e^{3x}(1+x)\cos(x^3e^{3x})$

(4)

$\left(\sin^3(xe^x)\right)'=3\sin^2(xe^x)\left(\sin(xe^x)\right)'$

$=3\sin^2(xe^x)\cos(xe^x)(xe^x)'$

$=3\sin^2(xe^x)\cos(xe^x)(x'e^x+x(e^x)')$

$=3\sin^2(xe^x)\cos(xe^x)(e^x+xe^x)$

$=3e^x(1+x)\sin^2(xe^x)\cos(xe^x)$

(5) Use implicit differentiation here.

$(\ln(xy))'=(e^{2y})'$

$\dfrac{(xy)'}{xy}=2e^{2y}y'$

$\dfrac{x'y+xy'}{xy}=2e^{2y}y'$

$y+xy'=2xye^{2y}y'$

$y=(2xye^{2y}-x)y'$

$y'=\dfrac y{2xye^{2y}-x}$

8 0

3 years ago

What is the determinant of 15 18 154

IgorC [24]

Answer:

The determinant is 15.

Step-by-step explanation:

You need to calculate the determinant of the given matrix.

1. Subtract column 3 multiplied by 3 from column 1 (C1=C1−(3)C3):

$\left[\begin{array}{ccc}-25&-23&9\\0&3&1\\-5&5&3\end{array}\right]$

2. Subtract column 3 multiplied by 3 from column 2 (C2=C2−(3)C3):

$\left[\begin{array}{ccc}-25&-23&9\\0&0&1\\-5&-4&3\end{array}\right]$

3. Expand along the row 2: (See attached picture).

We get that the answer is 15. The determinant is 15.

6 0

4 years ago

A flower pots space and six size they're all the same length. Each side measures X -6 units. The bases perimeter is 78 units. Wh

irakobra [83]

Answer:

x = 19

Step-by-step explanation:

The shape of the flower base has six sides of equal lengths. This means that the flower base is a regular hexagon. A regular hexagon is a closed polygon with six equal sides and six equal angles.

Each side of the base measures x - 6, and the perimeter of the base is 78 units.

Length of each side = x - 6

The total length of the base = perimeter of the base.

The total length of the base is gotten by summing all the sides of the base. Since the base has 6 sides, therefore the total length of the base = 6(x - 6). Hence:

Perimeter of base = 6(x - 6)

78 = 6x - 36

6x = 78 + 36

6x = 114

x = 114 / 6

x = 19

Length of each side of base = x - 6 = 19 - 6 = 13 units

4 0

3 years ago

How to factor a2b2-1 or 144y2-49 can someome help me out on this please

Sonja [21]

A^2b^2-1

The factor of that is (ab-1)(ab+1)

144y^2-49

The factor of that is (12y-7)(12y+7)

5 0

4 years ago