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3 years ago

14

NGO has a collection of dimes and quarters with a total of $3.50. The number of dimes is 7 more than the number of quarters. How

many of each coin does he have?

1 answer:

Tpy6a [65]3 years ago

8 0

Answer: $24.50

Step-by-step explanation: $3.50 times 7 equals $24.50

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Prove that sin3a-cos3a/sina+cosa=2sin2a-1

Sloan [31]

Answer:

$\frac{sin(3a)-cos(3a)}{sin(a)+cos(a)} =2sin(2a)-1$

Step-by-step explanation:

we are given

$\frac{sin(3a)-cos(3a)}{sin(a)+cos(a)} =2sin(2a)-1$

we can simplify left side and make it equal to right side

we can use trig identity

$sin(3a)=3sin(a)-4sin^3(a)$

$cos(3a)=4cos^3(a)-3cos(a)$

now, we can plug values

$\frac{(3sin(a)-4sin^3(a))-(4cos^3(a)-3cos(a))}{sin(a)+cos(a)}$

now, we can simplify

$\frac{3sin(a)-4sin^3(a)-4cos^3(a)+3cos(a)}{sin(a)+cos(a)}$

$\frac{3sin(a)+3cos(a)-4sin^3(a)-4cos^3(a)}{sin(a)+cos(a)}$

$\frac{3(sin(a)+cos(a))-4(sin^3(a)+cos^3(a))}{sin(a)+cos(a)}$

now, we can factor it

$\frac{3(sin(a)+cos(a))-4(sin(a)+cos(a))(sin^2(a)+cos^2(a)-sin(a)cos(a)}{sin(a)+cos(a)}$

$\frac{(sin(a)+cos(a))[3-4(sin^2(a)+cos^2(a)-sin(a)cos(a)]}{sin(a)+cos(a)}$

we can use trig identity

$sin^2(a)+cos^2(a)=1$

$\frac{(sin(a)+cos(a))[3-4(1-sin(a)cos(a)]}{sin(a)+cos(a)}$

we can cancel terms

$=3-4(1-sin(a)cos(a))$

now, we can simplify it further

$=3-4+4sin(a)cos(a))$

$=-1+4sin(a)cos(a))$

$=4sin(a)cos(a)-1$

$=2\times 2sin(a)cos(a)-1$

now, we can use trig identity

$2sin(a)cos(a)=sin(2a)$

we can replace it

$=2sin(2a)-1$

so,

$\frac{sin(3a)-cos(3a)}{sin(a)+cos(a)} =2sin(2a)-1$

7 0

3 years ago

Read 2 more answers

If x plus one upon X equal to 11 find the value of x square + 1 upon x square

zubka84 [21]

119 <answer (check the solution in above pic)

hey friend i hope this answer is helpful

marl as brainliest if i deserve

6 0

3 years ago

Find the volume of a square pyramid with base edge = 9 cm

kondaur [170]

16 cm

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4 0

3 years ago

Jeff gave 1/4 of a sum of money to his wife. Then he divided the remainder equally among his 4 children.

Semenov [28]

Answer:

A) The fraction of sum of money did each child receive is $\frac{3 x}{16}$

B) The sum of money did Jeff have $ 3200

Step-by-step explanation:

Given as :

Let The sum of money did Jeff have = $ x

The fraction of money did Jeff's wife get = $\frac{1}{4}$ of $ x

The remaining money Jeff will have = $ x - $\frac{1}{4}$ of $ x

I.e The remaining money Jeff will have = $\frac{4 x - x}{4}$

= $\frac{3 x}{4}$

A ) The remaining amount of money is divided equally among 4 children

So, The fraction of sum of money did each child receive = $\frac{\frac{3 x}{4}}{4}$

I.e The fraction of sum of money did each child receive = $\frac{3 x}{16}$

B ) If each child will receive $ 600

∴, $\frac{3 x}{16}$ = $ 600

Or, 3 x = $ 600 × 16

Or, 3 x = $ 9600

∴ x = $\frac{9600}{3}$

I.e x = $ 3200

So, The sum of money did Jeff have $ 3200

Hence ,

A) The fraction of sum of money did each child receive is $\frac{3 x}{16}$

B) The sum of money did Jeff have $ 3200 Answer

8 0

3 years ago

Need help solving this please

Aloiza [94]

Answer:

A) x=39

Step-by-step explanation:

117 = 3x

117 divided by 3 = 39

3x = 117

3(39) = 117

117 = 117

A) x = 39

4 0

3 years ago