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3 years ago

A tree is 6 feet 7 inches tall. How tall is it in inches?

Mathematics

1 answer:

matrenka [14]3 years ago

6 0

Answer:

79 inches

Step-by-step explanation:

1 foot equals 12 inches

6 feet = 6*12 = 72 inches

72+7=79

79 inches

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quester [9]

Answer:

Step-by-step explanation:

do it your self

6 0

3 years ago

A cable company charges $70 per month for cable service. Please help ASAP Check screenshot pleasee (20 points) Thank youu! :)) &

LiRa [457]

Answer:

for #1: 70x = y

for #2: 70x + 100 = y

Step-by-step explanation:

you pay $70 for x amount of time with will equal your total therefore it's 70x = y

if you have to pay a one-time fee of $100.00 then you have to add that to your equation giving you 70x + 100 = y

hope this is right

hope it helps too :)

8 0

3 years ago

Choose the simplified form of the fifth term of 6C4(2x)2(-y2)4.

Katarina [22]

You didn't give us the choices. It doesn't matter. I think you're trying to write

$\displaystyle {6 \choose 4} (2x)^2 (-y^2)^4$

$= \dfrac{6!}{4! 2!} ( 4x^2 y^8)$

$= \dfrac{6(5)}{2} ( 4x^2 y^8)$

$=60x^2 y^8$

Answer: 60 x² y⁸

4 0

4 years ago

Electric charge is distributed over the disk x2 + y2 ≤ 16 so that the charge density at (x, y) is rho(x, y) = 2x + 2y + 2x2 + 2y

professor190 [17]

Answer:

Required total charge is $256\pi$ coulombs per square meter.

Step-by-step explanation:

Given electric charge is dristributed over the disk,

$x^2=y^2\leq 16$ so that the charge density at (x,y) is,

$\rho (x,y)=2x+2y+2x^2+2y^2$

To find total charge on the disk let Q be the total charge and $x=r\cos\theta,y=r\sin\theta$ so that,

$Q={\int\int}_Q\rho(x,y) dA$ where A is the surface of disk.

$=\int_{0}^{2\pi}\int_{0}^{4}(2x+2y+2x^2+2y^2)dA$

$=\int_{0}^{2\pi}\int_{0}^{4}(2r\cos\theta+2r\sin\theta+2r^2 \cos^{2}\theta+2r^2\sin^2\theta)rdrd\theta$

$=2\int_{0}^{2\pi}\int_{0}^{4}r^2(\cos\theta+\sin\theta)drd\theta+2\int_{0}^{2\pi}\int_{0}^{4}r^3drd\theta$

$=\frac{2}{3}\int_{0}^{2\pi}(\sin\theta+\cos\theta)\Big[r^3\Big]_{0}^{4}d\theta+2\int_{0}^{2\pi}\Big[\frac{r^4}{4}\Big]d\theta$

$=\frac{128}{3}\int_{0}^{2\pi}(\sin\theta+\cos\theta)d\theta+128\int_{0}^{2\pi}d\theta$

$=\frac{128}{3}\Big[\sin\theta-\cos\theta\Big]_{0}^{2\pi}+128\times 2\pi$

$=\frac{128}{3}\Big[\sin 2\pi-\cos 2\pi-\sin 0+\cos 0\Big]+256\pi$

$=256\pi$

Hence total charge is $256\pi$ coulombs per square meter.

3 0

3 years ago

PLEASE HELP

GREYUIT [131]

Answer:

Step-by-step explanation:

Polygon A translated 5 blocks right.

4 0

3 years ago