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3 years ago

Please help me with 25-29!!

Mathematics

1 answer:

vredina [299]3 years ago

4 0

25-29 ok so first do 9-5 equals 4

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000

grigory [225]

Answer:

Step-by-step explanation:

4 0

3 years ago

A cash box contains $74 made up of quarters, half-dollars, and one-dollar

Strike441 [17]

Answer:

25 one-dollar coins, 16 half-dollar coins, and 164 quarters

Step-by-step explanation:

First, set up equations based on the information given:

$0.25q+0.50h+1.00d=74$

$\displaystyle{h=\frac{3}{5}d+1}$

$q=4(d+h)$

Then, substitute q in the first equation with the expression from the third equation:

$0.25[4(d+h)]+0.50h+1.00d=74\\1d+1h+0.50h+1.00d=74\\2d+1.5h=74$

Next, substitute h in that equation with the expression from the second equation:

$\displaystyle{2d+1.5(\frac{3}{5}d+1)=74}$

$2d+0.9d+1.5=74\\2.9d+1.5=74$

Solve for d, the number of one-dollar coins:

$2.9d+1.5=74\\2.9d=72.5\\d=25$

Substitute 25 for d in the second equation to find h, the number of half-dollar coins:

$\displaystyle{h=\frac{3}{5}d+1}$

$\displaystyle{h=\frac{3}{5}(25)+1}$

$h=15+1\\h=16$

Substitute 25 for d and 16 for h in the third equation to find q, the number of quarters:

$q=4(d+h)\\q=4(25+16)\\q=4(41)\\q=164$

Then, verify that the coins total $74:

$0.25(164)+0.50(16)+1.00(25)=74\\41+8+25=74\\74=74\\\text{Check.}$

Next, verify that the number of half-dollar coins is one more than three-fifths of the number of one-dollar coins:

$\displaystyle{h=\frac{3}{5}d+1}$

$\displaystyle{16=\frac{3}{5}(25)+1}$

$16 = 15 + 1\\16 = 16\\\text{Check.}$

Finally, verify that the number of quarters is four times the number one-dollar and half-dollar coins together:

$q=4(d+h)\\164=4(25+16)\\164=4(41)\\164=164\\\text{Check.}$

6 0

3 years ago

Which of the following options have the same value as 75%, percent of 96?

dexar [7]

The choices should be B, C, and E. Hope this helps

6 0

3 years ago

Altitudes AA1 and BB1 are drawn in acute △ABC. Prove that A1C·BC=B1C·AC

Sophie [7]

Answer:

See the attached figure which represents the problem.

As shown, AA₁ and BB₁ are the altitudes in acute △ABC.

△AA₁C is a right triangle at A₁

So, Cos x = adjacent/hypotenuse = A₁C/AC ⇒(1)

△BB₁C is a right triangle at B₁

So, Cos x = adjacent/hypotenuse = B₁C/BC ⇒(2)

From (1) and (2)

∴ A₁C/AC = B₁C/BC

using scissors method

∴ A₁C · BC = B₁C · AC

7 0

3 years ago

Which of the following equations shows how substitution can be used to solve the following system of equations?

klasskru [66]

Answer:

i have no idea does anyone else know

Step-by-step explanation:

5 0

2 years ago