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3 years ago

8

Please answer ASAP: Determine which of the following situations can be modeled by the ratio 7/2. Select all situations that appl

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For every twenty-eight cats, there are 4 dogs
Sara runs an average of fourteen miles every four days.
An online clothing store charges $7.50 to ship ten clothing items
A recipe calls for 3 cups of flour for every 1 cup of sugar.
Bottled water is on sale this week for a rate of $1.50 for two bottles

1 answer:

wlad13 [49]3 years ago

3 0

the only one that applies is:

Sara runs 14 miles every 4 days = 7 miles every 2 days = 7/2

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OverLord2011 [107]

20/40 = 1/2 or 0.5 as a decimal

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Estimate how many 12 -inch rulers will be about the same length as a sheet of paper

Zinaida [17]

1 would be the answer

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4 years ago

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Jamie has saved $42 of her allowance money to buy books. If she buys 5 books at d dollars per book, she will have 42-5d of her a

valentina_108 [34]

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4 years ago

PLS HELP ME!!! The figure is made up of a cone and a hemisphere. To the nearest whole number, what is the approximate volume of

Pachacha [2.7K]

Hello!

The figure is made up of a cone and a hemisphere. To the nearest whole number, what is the approximate volume of this figure? Use 3.14 to approximate π . Enter your answer in the box. cm³

Data: (Cone)

h (height) = 12 cm

r (radius) = 4 cm (The diameter is 8 being twice the radius)

Adopting: $\pi \approx 3.14$

V (volume) = ?

Solving: (Cone volume)

$V = \dfrac{ \pi *r^2*h}{3}$

$V = \dfrac{ 3.14 *4^2*\diagup\!\!\!\!\!12^4}{\diagup\!\!\!\!3}$

$V = 3.14*16*4$

$\boxed{V = 200.96\:cm^3}$

Note: Now, let's find the volume of a hemisphere.

Data: (hemisphere volume)

V (volume) = ?

r (radius) = 4 cm

Adopting: $\pi \approx 3.14$

If: We know that the volume of a sphere is $V = 4* \pi * \dfrac{r^3}{3}$ , but we have a hemisphere, so the formula will be half the volume of the hemisphere $V = \dfrac{1}{2}* 4* \pi * \dfrac{r^3}{3} \to \boxed{V = 2* \pi * \dfrac{r^3}{3}}$

Formula: (Volume of the hemisphere)

$V = 2* \pi * \dfrac{r^3}{3}$

Solving:

$V = 2* \pi * \dfrac{r^3}{3}$

$V = 2*3.14 * \dfrac{4^3}{3}$

$V = 2*3.14 * \dfrac{64}{3}$

$V = \dfrac{401.92}{3}$

$\boxed{ V_{hemisphere} \approx 133.97\:cm^3}$

What is the approximate volume of this figure?

Now, to find the total volume of the figure, add the values: (cone volume + hemisphere volume)

Volume of the figure = cone volume + hemisphere volume

Volume of the figure = 200.96 cm³ + 133.97 cm³

$\boxed{\boxed{\boxed{V = 334.93\:cm^3 \to Volume\:of\:the\:figure \approx 335\:cm^3 }}}\end{array}}\qquad\quad\checkmark$

Answer:

The volume of the figure is approximately 335 cm³

_______________________

I Hope this helps, greetings ... Dexteright02! =)

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3 years ago

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Find the indicated nth partial sum of the arithmetic sequence.<br> −7, −3, 1, 5, . . ., n = 30

TiliK225 [7]

Step-by-step explanation:

Given the Arithmetic sequence

$-7, -3, 1, 5, . . .$

An arithmetic sequence has a constant difference d and is defined by

$a_n=a_1+\left(n-1\right)d$

$\mathrm{Compute\:the\:differences\:of\:all\:the\:adjacent\:terms}:\quad \:d=a_{n+1}-a_n$

$-3-\left(-7\right)=4,\:\quad \:1-\left(-3\right)=4$

$\mathrm{The\:difference\:between\:all\:of\:the\:adjacent\:terms\:is\:the\:same\:and\:equal\:to}$

$d=4$

$\mathrm{The\:first\:element\:of\:the\:sequence\:is}$

$a_1=-7$

as

$a_n=a_1+\left(n-1\right)d$

$\mathrm{Therefore,\:the\:}n\mathrm{th\:term\:is\:computed\:by}\:$

$a_n=4\left(n-1\right)-7$ ∵ $d=4$

$\mathrm{Arithmetic\:sequence\:sum\:formula:}$

$n\left(a_1+\frac{d\left(n-1\right)}{2}\right)$

$\mathrm{Plug\:in\:the\:values:}$

$n=30,\:\space a_1=-7,\:\spaced=4$

$=30\left(-7+\frac{4\left(30-1\right)}{2}\right)$

$=30\left(58-7\right)$ ∵ $\frac{4\left(30-1\right)}{2}=58$

$=30\cdot \:51$

$=1530$ ∵ $\mathrm{Multiply\:the\:numbers:}\:30\cdot \:51=1530$

Therefore, the indicated nth partial sum of the arithmetic sequence is 1530.

ANOTHER METHOD

as

$a_n=4\left(n-1\right)-7$

n = 30

$\sum _{n=1}^{30}\:4\left(n-1\right)-7$

$=\sum _{n=1}^{30}4n-11$

$\mathrm{Apply\:the\:Sum\:Rule}:\quad \sum a_n+b_n=\sum a_n+\sum b_n$

$=\sum _{n=1}^{30}4n-\sum _{n=1}^{30}11$

as

$\sum _{n=1}^{30}4n=1860$

and

$\sum _{n=1}^{30}11=330$

so

$=1860-330$

$=1530$

7 0

4 years ago