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2 years ago

How can you find 75% of any number?

Mathematics

2 answers:

Elena-2011 [213]2 years ago

8 0

Answer:

Multiply the number you want to find 75% of by 0.75

Step-by-step explanation:

Aleks04 [339]2 years ago

5 0

Answer:

Step-by-step explanation:

Multiply it by 0.75, or find 3/4 of that number. Whichever you prefer! The first one always works though the second one sometimes does but sometimes 3/4 of the number isn't that obvious.

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What is 25% of $8? $ Submit

vekshin1

Answer:

2 dollars?

Step-by-step explanation:

8 divided by 4

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2 years ago

I know the answers to this problem. I will give brainiest to the first one who ANSWER IS CORRECT.

const2013 [10]

Answer:

24240300

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2 years ago

What is the surface area of a rectangular prism that has these dimensions? 2 cm x 3 cm x 5 cm?

algol [13]

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3 years ago

An illusionist needs up to 10 volunteers for a show. She needs no fewer than 4 female volunteers. Let x represent the number of

Sliva [168]

The correct answers are:

B- x + y ≤ 10 ; and D- x ≥ 4

Explanation:

The illusionist needs at least 4 female volunteers. This means he could use 4 or more; this means x, the number of female volunteers, is greater than or equal to 4:

x ≥ 4

The total of number of volunteers needed is up to 10. This means the illusionist can use 10 or fewer people; this means the total is less than or equal to 10.

The number of female volunteers is x and the number of male volunteers is y; together they create the expression x+y. This gives us

x+y ≤ 10

7 0

3 years ago

Read 2 more answers

1+secA/sec A = sin^2 A / 1-cos A

Fofino [41]

Answer: see proof below

Step-by-step explanation:

$\dfrac{1+\sec A}{\sec A}=\dfrac{\sin^2 A}{1-\cos A}$

Use the following Identities:

sec Ф = 1/cos Ф

cos² Ф + sin² Ф = 1

Proof LHS → RHS

$\text{LHS:}\qquad \qquad \dfrac{1+\sec A}{\sec A}$

$\text{Identity:}\qquad \qquad \dfrac{1+\frac{1}{\cos A}}{\frac{1}{\cos A}}$

$\text{Simplify:}\qquad \qquad \dfrac{\frac{\cos A+1}{\cos A}}{\frac{1}{\cos A}}\\\\\\.\qquad \qquad \qquad =\dfrac{1+\cos A}{1}$

$\text{Multiply:}\qquad \qquad \dfrac{1+\cos A}{1}\cdot \bigg(\dfrac{1-\cos A}{1-\cos A}\bigg)\\\\\\.\qquad \qquad \qquad =\dfrac{1-\cos^2 A}{1-\cos A}$

$\text{Identity:}\qquad \qquad \dfrac{\sin^2 A}{1-\cos A}$

$\text{LHS = RHS:}\quad \dfrac{\sin^2 A}{1-\cos A}=\dfrac{\sin^2 A}{1-\cos A}\quad \checkmark$

3 0

3 years ago