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2 years ago

9

Derrick plays football. During practice the ratio of quarterbacks to wide receivers is 5:12. If there were 36 wide receivers, ho

w many quarterbacks would there be?

1 answer:

kirill [66]2 years ago

5 0

Answer:

I believe there would be 15.

Step-by-step explanation:

well 12×3=36

5×3=15 so 15:36

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If two cylinder are similar and the ratio between the lengths of their edges is 4;3 what is the ratio of their volumes

Tems11 [23]

Answer:

64 : 27

Step-by-step explanation:

Given 2 similar figures with

ratio of lengths = a : b, then

ratio of volumes = a³ : b³

For 2 cylinders with ratio of lengths = 4 : 3, then

ratio of volumes = 4³ : 3³ = 64 : 27

3 0

2 years ago

Read 2 more answers

Scott baked a batch of rolls. He gave a bag of 10 rolls to each of 7 friends he kept 1 roll for himself. How many rolls did he b

Serjik [45]

7 times 10 is 70, so in total the friends got that amount, but we aren’t done, because we still need to add the roll that he kept for himself, so it’s 71. Notice that it didn’t say he left a whole bag for himself, just one roll.

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2 years ago

Read 2 more answers

The no. Of adel beads to that of claire's beads was the ratio 7 :4 .When adel gave 68 beads to claire the ratio become 25 :41 .

Novosadov [1.4K]

Answer:

37 beads

Step-by-step explanation:

Given that:

Adel : Claire = 7 : 4

Let Adel be = A; &

Let Claire be = C

A: C = 7 : 4

$\dfrac{A}{C} =\dfrac{7}{4}$

7C = 4A

$C = \dfrac{4A}{7}$ --- (1)

When Adel gave 68 beads to claire, we have:

C + 68 = 25:41

$\dfrac{A}{C+68}= \dfrac{25}{41}$

25(C + 68) = 41A

C + 68 = $\dfrac{41}{25}A$

$C = \dfrac{41}{25}A - 68$ --- (2)

Equating (1) and (2) together;

$\dfrac{4 * A}{7} = \dfrac{41 *A}{25} - 68$

$\dfrac{4 * A}{7}-\dfrac{41 *A}{25} = - 68$

$\dfrac{100A - 287 A}{175} =- 68$

$\dfrac{-187A}{175} = -68$

- 187 A = 175 × (-68)

-187 A = - 11900

A = - 11900/ -187

A ≅ 64

From; $C = \dfrac{4A}{7}$

$C = \dfrac{4*64}{7}$

C = 36.57

C ≅ 37 beads

4 0

2 years ago

WHOEVER IS RIGHT GETS BRAINLIST

pochemuha

The answer to the first question is b
I don’t know the answer to the second one sorry

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2 years ago

HELP HELP HELP<br> HELP HELP <br> HELP <br> (STEP BY STEP)

Marizza181 [45]

With some simple rearrangement, we can rewrite the numerator as

$2x^3 - 3x^2 - x + 4 = 2(x^3 - x) - 3x^2 + x + 4 \\\\ ~~~~~~~~ = 2x(x^2-1) - 3(x^2 - 1) + x + 1 \\\\ ~~~~~~~~ = (2x-3)(x^2-1) + x+1$

Then factorizing the difference of squares, $x^2-1=(x-1)(x+1)$ , we end up with

$\dfrac{2x^3 - 3x^2 - x + 4}{x^2 - 1} = \dfrac{(2x-3)(x-1)(x+1) + x+1}{(x-1)(x+1)} \\\\ ~~~~~~~~ = \boxed{2x-3 + \dfrac1{x-1}}$

3 0

1 year ago