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3 years ago

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Sphere with a volume of 1767.1 . what is the radius?

1 answer:

Llana [10]3 years ago

3 0

$\bf \textit{volume of a sphere}\\\\ V=\cfrac{4\pi r^3}{3}~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ V=1767.1 \end{cases}\implies 1767.1=\cfrac{4\pi r^3}{3}\implies 5301.3=4\pi r^3 \\\\\\ \cfrac{5301.3}{4\pi }=r^3\implies \sqrt[3]{\cfrac{5301.3}{4\pi }}=r\implies 7.4999\approx r$

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Right triangles abc and dbc with right angle c are given below. If cos(a)=15,ab=12 and cd=2, find the length of bd.

dmitriy555 [2]

see the attached figure to better understand the problem

we have that

$cos(A)=\frac{1}{5} \\ AB=12\ units\\ CD=2\ units$

Step 1

<u>Find the value of AC</u>

we know that

in the right triangle ABC

$cos (A)=(AC/AB)\\AC=AB*cos(A)$

substitute the values in the formula

$AC=12*(1/5)\\ AC=2.4\ units$

Step 2

<u>Find the value of BC</u>

we know that

in the right triangle ABC

Applying the Pythagorean Theorem

$AB^{2} =AC^{2}+BC^{2}\\ BC^{2}=AB^{2} -AC^{2}$

substitute the values

$BC^{2}=12^{2} -2.4^{2}\\BC^{2}= 138.24\\ BC=11.76\ units$

Step 3

<u>Find the value of BD</u>

we know that

in the right triangle BCD

Applying the Pythagorean Theorem

$BD^{2} =DC^{2}+BC^{2}$

substitute the values

$BD^{2} =2^{2}+11.76^{2}$

$BD=11.93\ units$

therefore

<u>the answer is</u>

the length of BD is 11.93 units

8 0

3 years ago

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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

Triangle BAC was rotated 90° clockwise and dilated at a scale factor of 2 from the origin to create triangle XYZ. Based on these

ANEK [815]

Answer:

The correct option is;

∠A ≅ ∠X

Step-by-step explanation:

The given coordinates of the points of triangle ACB are;

A(-4, 4), C(-1, 3), B(-4, 0)

The given coordinates of the points of triangle XYZ are;

X(0, 8), Y(8, 8), Z(6, 2), therefore, we have

The length. l. of segment is given by the following formula;

$l = \sqrt{\left (y_{2}-y_{1} \right )^{2}+\left (x_{2}-x_{1} \right )^{2}}$

For the length of the segment AC; (x₁, y₁) = (-4, 4), (x₂, y₂) = (-1, 3), l = √(10)

For the length of the segment AB; (x₁, y₁) = (-4, 4), (x₂, y₂) = (-4, 0), l = 4

For the length of the segment BC; (x₁, y₁) = (-4, 0), (x₂, y₂) = (-1, 3), l = 3·√2

For the length of the segment XY; (x₁, y₁) = (0, 8), (x₂, y₂) = (8, 8), l = 8

For the length of the segment XZ; (x₁, y₁) = (0, 8), (x₂, y₂) = (6, 2), l = 6·√2

For the length of the segment ZY; (x₁, y₁) = (6, 2), (x₂, y₂) = (8, 8), l = 2·√(10

Therefore;

XY ~ AB, XZ ~ BC, ZY ~ AC

Which gives;

∠A ≅ ∠X, ∠B ≅ ∠Y, ∠C ≅ ∠Z

8 0

3 years ago

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Which of the following always passes through a vertex of a triangle . choose all that apply.

Helga [31]

Just finished the test and the write answers are :

1) A,D,E

2)A,B

3)B,E

4) D

5) A

6) C

7) A

8) B

9) part A is C

Part B is A

Score 100%

7 0

3 years ago

The first three steps of completing the square to solve the quadratic equation x^2 + 8x +5 = 0, are shown below.

evablogger [386]

Answer:

see explanation

Step-by-step explanation:

step 4 take the square root of both sides

$\sqrt{(x+4)^2}$ = ± $\sqrt{11}$

x + 4 = ± $\sqrt{11}$

step 5 subtract 4 from both sides

x = - 4 ± $\sqrt{11}$

step 6 solve for x

x = - 4 - $\sqrt{11}$ or x = - 4 + $\sqrt{11}$

8 0

3 years ago

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