3 years ago

In the equation y=x/3 , what is the unit rate?

Mathematics

1 answer:

Vlada [557]3 years ago

8 0

$\bf y=\cfrac{x}{3}\implies y=\stackrel{rate}{\cfrac{1}{3}}x$

You might be interested in

Help 13 points!!!!!!!!

DIA [1.3K]

Replace x with a^2:

7(a^2)^2 - 8

Multiply the exponents:

7a^4 - 8

3 0

3 years ago

the bases of a trapezoid are 11 meters and 14 meters .The heigth measures 8 meters. What is the area of the trapezoid. SHOW YOUR

Bogdan [553]

11x8=88
1.5x8=12
12/2=6
1.5x8=12
12/2=6

88+6+6=100

8 0

3 years ago

Which number is an integer?

Ne4ueva [31]

Answer:

Step-by-step explanation:

0 is an integer.

plz mark as brainliest

6 0

3 years ago

Assume that y varies directly as x.

Anon25 [30]

75/15=5
3*5=15
y would equal 15

Hope this helps :)

4 0

3 years ago

Read 2 more answers

Find the volume and surface area of the composite figure. Give your answer in terms of π.

irakobra [83]

Answer:

V = 240π cm^3 , S= 168π cm^2

Step-by-step explanation:

The given figure is a combination of hemi-sphere and a cone

<u>Volume:</u>

For volume

r = 6 cm

h = 8 cm

$Volume\ of\ cone = \frac{1}{3}\pi r^2h\\= \frac{1}{3}\pi (6)^2*8\\=\frac{1}{3}\pi *36*8\\=\frac{288}{3}\pi\\=96\pi cm^3 \\\\Volume\ of\ hemisphere = \frac{2}{3}\pi r^3\\=\frac{2}{3}*\pi * (6)^3\\=\frac{2}{3}*\pi *216\\=\frac{432}{3}\\=144\pi cm^3 \\\\Total\ Volume= Volume\ of\ cone + Volume\ of\ hemisphere\\= 96\pi +144\pi \\=240\pi cm^3$

<u>Surface Area:</u>

For this particular figure we have to consider the lateral area of the cone shape and surface area of the hemisphere

We have to find the lateral height

$l = \sqrt{r^2+h^2}\\ l = \sqrt{(6)^2+(8)^2} \\l= \sqrt{36+64}\\ l = \sqrt{100}\\l = 10cm\\\\Surface\ area\ of\ cone = \pi rl\\= \pi (6)(10)\\=\pi *60\\=60 \pi\ cm^2\\\\Surface\ area\ of\ hemisphere = 2\pi r^2\\= 2 \pi * (6)\\= 2 \pi *36\\= 72 \pi\ cm^2\\\\Total\ surface\ Area = Surface\ area\ of\ cone + Surface\ area\ of \ hemisphere\\= 60 \pi + 72 \pi\\=132 \pi\ cm^2$

Hence the first option is correct ..

3 0

3 years ago