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2 years ago

I am trying to help my daughter out and I have no clue

Mathematics

1 answer:

kobusy [5.1K]2 years ago

5 0

M=(y2-y1)/(x2-x1)

M=(2-0)/(2-(-3)

M=(2)/(5)

Slope is 2/5
I have no idea if that’s right I look algebra 1 last semester

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he value of y varies directly with the value of x. When y = 32.5, x = 5. What is the value of y when x = 10? A) 1.5 B) 2.2 C) 37

xz_007 [3.2K]

Answer:

D) 65

Step-by-step explanation:

The formula for direct variation is

y = kx

where k is the constant of variation.

If we know y=32.5 and x=5, we can substitute in to calculate k

32.5 = k * 5

Divide each side by 5

32.5/5 = k

6.5 = k

So the equation is

y = 6.5k

Given x =10

y = 6.5*10

y = 65

7 0

3 years ago

Read 2 more answers

What IS 20 divided by 13 ?

goldenfox [79]

$Solution,\mathrm{Convert\:improper\:fractions\:to\:mixed\:numbers}:\quad \frac{20}{13}=1\frac{7}{13}$

$Steps:$

$\frac{20}{13}$

$\mathrm{Long\:Division}\:\frac{20}{13}:\quad 1\quad \mathrm{Remainder}\quad \:7$

$Convert\:to\:mixed\:number:\:Quotient\frac{Remainder}{Divisor},\\\frac{20}{13}=1\frac{7}{13}$

$Hope\: This\: Helps!!!$

4 0

3 years ago

Read 2 more answers

How do I do this?<br>*Look at the directions in the photo*

lora16 [44]

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:

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

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

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

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

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