Ask question

Ask question

All categories

olya-2409 [2.1K]

1 year ago

11

Mat played tennis 35 times in 7 weeks. sue played tennis 15 times in 21 days. are their rates equivalent?

1 answer:

morpeh [17]1 year ago

5 0

Yes they are because you convert 7 weeks to days. Then you divide 49 by 21 because you have to see if they go in with each other and do the same thing with 35 and 15.

You might be interested in

Solve using the standard algorithm.

Vinvika [58]

Answer:

A= 1.12

B= 1.11

Step-by-step explanation:

A. 0.82 B. 1.03

+ 0.3 + 0.08

______ _______

1.12 1.11

8 0

2 years ago

Movie tickets are $4.00 for children and $8.00 for adults. The total amount of ticket sales in one night was $1,176. There was 1

Liono4ka [1.6K]

Answer: Cant answer this question because the question doesnt say/contain how many children were there and how many parents were there. if you list this, i can help

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Help will give branliest !!!

Nataly [62]

Answer:c

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Answer this volume based Question. I will make uh brainliest + 50 points

Harlamova29_29 [7]

Answer:

$\huge{\purple {r= 2\times\sqrt[3]3}}$

$\huge 2\times \sqrt [3]3 = 2.88$

Step-by-step explanation:

For solid iron sphere:
radius (r) = 2 cm (Given)

Formula for $V_{sphere}$ is given as:

$V_{sphere} =\frac{4}{3}\pi r^3$

$\implies V_{sphere} =\frac{4}{3}\pi (2)^3$

$\implies V_{sphere} =\frac{32}{3}\pi \:cm^3$

For cone:
r : h = 3 : 4 (Given)
Let r = 3x & h = 4x

Formula for $V_{cone}$ is given as:

$V_{cone} =\frac{1}{3}\pi r^2h$

$\implies V_{cone} =\frac{1}{3}\pi (3x)^2(4x)$

$\implies V_{cone} =\frac{1}{3}\pi (36x^3)$

$\implies V_{cone} =12\pi x^3\: cm^3$

It is given that: iron sphere is melted and recasted in a solid right circular cone of same volume
$\implies V_{cone} = V_{sphere}$

$\implies 12\cancel{\pi} x^3= \frac{32}{3}\cancel{\pi}$

$\implies 12x^3= \frac{32}{3}$

$\implies x^3= \frac{32}{36}$

$\implies x^3= \frac{8}{9}$

$\implies x= \sqrt[3]{\frac{8}{3^2}}$

$\implies x={\frac{2}{ \sqrt[3]{3^2}}}$

$\because r = 3x$

$\implies r=3\times {\frac{2}{ \sqrt[3]{3^2}}}$

$\implies r=3\times 2(3)^{-\frac{2}{3}}$

$\implies r= 2\times (3)^{1-\frac{2}{3}}$

$\implies r= 2\times (3)^{\frac{1}{3}}$

$\implies \huge{\purple {r= 2\times\sqrt[3]3}}$
Assuming log on both sides, we find:

$log r = log (2\times \sqrt [3]3)$

$log r = log (2\times 3^{\frac{1}{3}})$

$log r = log 2+ log 3^{\frac{1}{3}}$

$log r = log 2+ \frac{1}{3}log 3$

$log r = 0.4600704139$

Taking antilog on both sides, we find:

$antilog(log r )= antilog(0.4600704139)$

$\implies r = 2.8844991406$

$\implies \huge \red{r = 2.88\: cm}$

$\implies 2\times \sqrt [3]3 = 2.88$

8 0

1 year ago

Please I need help!

Sidana [21]

With what
With what
With what
With what

6 0

2 years ago