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3 years ago

Is 12 posters for 36 students equivalent to 21 posters to 63 student

Mathematics

2 answers:

Zanzabum3 years ago

7 0

Answer:

12 posters for 36 students equivalent to 21 posters to 63 student

Step-by-step explanation:

we need to check whether 12 posters for 36 students equivalent to 21 posters to 63 student

12 posters for 36 students can be written as $\frac{12}{36} = \frac{1}{3}$

Now 21 posters to 63 student can be written as

$\frac{21}{63}$ divide top and bottom by 3

$\frac{21}{63} = \frac{7}{21}=\frac{1}{3}$

So, 12 posters for 36 students equivalent to 21 posters to 63 student

Dmitry [639]3 years ago

3 0

yes.

this is division. 36 divided by 12 is 3. 63 divided by 21 is 3.

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A geometric sequence can be used to describe the population of rabbits on a

vlabodo [156]

Answer:

step 1

Find the average growth per year of the populations of rabbits farm A

Over the first 2 years

for t=0

numbers of rabbits=5 (is not exact)

for t=2

numbers of rabbits=40

average=[40-5]/2---------> 17.5

step 2

Find the average growth per year of the populations of rabbits farm B

Over the first 2 years

for t=0

numbers of rabbits=5 (is not exact)

for t=2

numbers of rabbits=30

average=[30-5]/2---------> 12.5

step 3

Compare the average growth per year of the populations of rabbits on both farms

farm A=17.5

farm B=12.5

the average rate of population growth of rabbits in farm A is greater than average rate of population growth of rabbits in farm B by about 5 rabbits per year.

therefore

the answer is the option C)

the average rate of population growth of rabbits in farm A is greater than average rate of population growth of rabbits in farm B by about 6 rabbits per year.

Step-by-step explanation:

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2 years ago

Does the order in which you add two integers with the same sign affect the sum

kondor19780726 [428]

If you add positive 5 and positive 5, you'll get positive 10.
If you add positive 5 and negative 5, you'll get positive 0.
If you add negative 5 and negative 5, you'll get negative 10.

-5 + 5 = -5 + 5
5 + 5 = 5 + 5
- 5 + -5 = -5 + -5

Does that help your question?

6 0

3 years ago

Can anybody help me with this ?

frutty [35]

f(x) = (3/4)ˣ - 4

As x gets really negative (3/4)ˣ grows positively without bound (because it's (4/3)⁻ˣ and 4/3 > 1).

As x gets really positive (3/4)ˣ tends to zero but never quite gets there.

So (3/4)ˣ > 0 for all x

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3 years ago

Solve step by step solution then only i can do it plxx

Anuta_ua [19.1K]

Answer:

<h3><u>Let's</u><u> </u><u>understand the concept</u><u>:</u><u>-</u></h3>

Here angle B is 90°

So $\triangle ABC$ and $\triangle ABD$ Are right angled triangle

So we use Pythagoras thereon for solution

<h3><u>Required Answer</u><u>:</u><u>-</u></h3>

First in triangle ABC

perpendicular=p=8cm

Hypontenuse =h =10cm

We need to find base=b

According to Pythagoras thereon

${\boxed{\sf b^2=h^2-p^2}}$

Substitutethe values

$\longrightarrow$ $\sf b^2=10^2-p^2$

$\longrightarrow$ $\sf b={\sqrt {10^2-8^2}}$

$\longrightarrow$ $\sf b={\sqrt{100-64}}$

$\longrightarrow$ $\bf b={\sqrt {36}}$

$\longrightarrow$ $\sf b=6$

$\therefore$ $\overline{BC}=6cm$

BD=BC+CD

$\longrightarrow$ $BD=9+6$

$\longrightarrow$ $BD=15cm$

Now in $\triangle ABD$

Perpendicular=p=8cm

Base =b=15cm

We need to find Hypontenuse =AD(x)

According to Pythagoras thereon

${\boxed {\sf h^2=p^2+b^2}}$

Substitute the values

$\longrightarrow$ $\sf h^2=8^2+15^2$

$\longrightarrow$ $\sf h={\sqrt {8^2+15^2}}$

$\longrightarrow$ $\sf h={\sqrt {64+225}}$

$\longrightarrow$ $\sf h={\sqrt {289}}$

$\longrightarrow$ $\sf h=17cm$

$\therefore$ ${\underline{\boxed{\bf x=17cm}}}$

3 0

3 years ago

The distance between city E and F is 678.35 miles and the distance between city F and G is 156.8 miles. What is the distance bet

Free_Kalibri [48]

E to F = 678.35 miles
F to G = 156.8 miles
G to H = x
Total distance = 2,457 miles
E to G = 678.35 + 156.8 miles
E to G = 835.15
G to H = 2,457 - 835.15
G to H = 1621.85 miles

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3 years ago

Read 2 more answers