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3 years ago

How do I do this?Look at the directions in the photo

Mathematics

1 answer:

lora16 [44]3 years ago

7 0

Answer:

$Area\ of\ material\ required\ for\ the\ first\ box=384\ inches^2\\Area\ of\ material\ required\ for\ the\ second\ box=486\ inches^2\\Area\ of\ material\ required\ for\ the\ first\ box=600\ inches^2\\Total\ Area\ of\ material\ required=1470\ inches^2$

Step-by-step explanation:

$We\ are\ given:\\Diameter\ of\ the\ first\ volleyball=8\ inches \\Diameter\ of\ the\ second\ volleyball=9\ inches\\Diameter\ of\ the\ third\ volleyball= 10\ inches.\\Hence,\\We\ know\ that,\\If\ the\ side\ of\ the\ cube\ box\ is\ s, it's\ Total\ Surface\ Area\ =No.\ of\\ faces\ in\ a\ regular\ polyhedron\ *Area\ of\ each\ face\ of\ the\ polyhedron=6*s^2=6s^2\\Hence,\\Lets\ apply\ this\ equation\ in\ finding\ the\ area\ of\ material\ required\ for\ the\\ three\ cases.\\$

$As\ the\ volleyball\ should\ wholly\ fit\ into\ the\ box,\ the\ diameter\ of\ the\\ volleyballs\ would\ be\ the\ side\ of\ the\ cube\ box.\\Hence,\\For\ the\ first\ volleyball,\\Diameter\ of\ the\ first\ volleyball=8\ inches\\Hence,\\Side\ of\ the\ cubical\ box\ for\ the\ first\ volleyball=8\ inches.\\Hence,\\The\ Total\ Surface\ Area\ of\ the\ first\ box=6s^2=6*8*8=384\ inches^2$

$For\ the\ second\ volleyball,\\Diameter\ of\ the\ second\ volleyball=9\ inches\\Hence,\\Side\ of\ the\ cubical\ box\ for\ the\ second\ volleyball=9\ inches.\\Hence,\\The\ Total\ Surface\ Area\ of\ the\ second\ box=6s^2=6*9*9=486\ inches^2$

$For\ the\ third\ volleyball,\\Diameter\ of\ the\ third\ volleyball=8\ inches\\Hence,\\Side\ of\ the\ cubical\ box\ for\ the\ third\ volleyball=10\ inches.\\Hence,\\The\ Total\ Surface\ Area\ of\ the\ third\ box=6s^2=6*10*10=600\ inches^2$

$Hence,\\If\ you\ are\ asked\ the\ Total\ Area\ to\ make\ all\ the\ boxes,\\ you\ just\ add\ them\ together.\\Hence,\\Total\ Area\ of\ Material\ required\ to\ make\ the\ three\ boxes=384+486+600=1470\ inches^2$

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babunello [35]

Answer:

x = 49.84

Step-by-step explanation:

We are given an equation of unknown x and we have to solve the equation for x.

$2\ln(e\ln5x)-2\ln15 =0$

⇒ $ln(e\ln5x) = \ln15$

⇒ $\ln e + \ln \ln 5x = \ln 15$ {Since we know that ln AB =ln A + ln B}

⇒ $\ln \ln 5x = \ln 15 -\ln e$

⇒ $\ln \ln 5x=\ln \frac{15}{e}$ {Since we know that ln A/B = ln A - ln B}

⇒ $\ln 5x = \frac{15}{e}$

⇒ $5x = e^{\frac{15}{e} }$

⇒ $x = \frac{1}{5} e^{\frac{15}{e} }$ {Converting logarithm to exponent form}

⇒ x = 49.84 (Approximate) (Answer)

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4 years ago

Determine the domain and range of the quadratic function. (Enter your answers using interval notation.) f(x) = −4(x + 1)2 − 4

zavuch27 [327]

Answer:

Domain -∞ < x < +∞

Range f(x) ≤ - 4

Step-by-step explanation:

The given quadratic function is f(x) = - 4 (x + 1)² - 4

Now, we have to find the domain and range of the quadratic function.

Now, it is clear from the given function that f(x) will have real value for all real value of x.

Therefore, the domain of the function is -∞ < x < +∞ (Answer)

Now, (x + 1)² is always ≥ 0

i.e. (x + 1)² ≥ 0

⇒ - 4 (x + 1)² ≤ 0 {Since the inequality sign changes for multiplying both sides of the equation by a negative term}

⇒ - 4 (x + 1)² - 4 ≤ - 4

⇒ f(x) ≤ - 4

Therefore, this is the required range of the function f(x). (Answer)

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4 years ago

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mixas84 [53]

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3 years ago

Read 2 more answers

Whats the parallel equation of y=1/3x-2 goes through point (0,6)

loris [4]

You know that two lines are parallel if they have the same slope, so the new line will also have slope

$m=\frac{1}{3}$

Because the equation of the given line is written in the form

$\begin{gathered} y=mx+b \\ \text{Where} \\ m\colon\text{slope of the line} \\ b\colon\text{ y-intercept} \end{gathered}$

Then, you can use point slope equation, which is

$\begin{gathered} y-y_1=m(x_{}-x_1) \\ \text{Where} \\ (x_1,y_1)\colon\text{ point through which the line passes.} \\ m\colon\text{slope of the line} \end{gathered}$

So,

$\begin{gathered} (x_1,y_1)=(0,6) \\ \text{ And you have} \\ y-y_1=m(x_{}-x_1) \\ y-6=\frac{1}{3}(x-0) \\ y-6=\frac{1}{3}x-0 \\ y-6=\frac{1}{3}x \\ \text{ Add 6 to both sides of the equation} \\ y-6+6=\frac{1}{3}x+6 \\ y=\frac{1}{3}x+6 \end{gathered}$

Therefore, the equation of the line that is parallel to y=1/3x-2 and goes through point (0,6) is

$y=\frac{1}{3}x+6$

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1 year ago

elijah has a flower box in the shape of a rectangular prism with interior dimensions that are 15 inches in length 8 inches in he

xxTIMURxx [149]

Answer:

the amount of soil needed is $90 in ^3$

Step-by-step explanation:

N/B: the width is 8 in

this problem bothers on the mensuration of solid shapes, rectangular prism

Given data

length l= 15 in

width w= 8 in

height h= 3/4 in

The amount of soil need in the flower box is equivalent to the volume of the box for the given data.

$volume- of- soil= L*W*H\\\volume- of- soil= 15*8*(3/4)\\\volume- of- soil= \frac{360}{4} \\\volume- of- soil=90 in^3$

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3 years ago

How do I do this?*Look at the directions in the photo*​

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