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2 years ago

28 is what percent of 64?

Mathematics

2 answers:

bezimeni [28]2 years ago

8 0

Solution:

Note that:

64 × x/100 = 28

Simplifying the LHS:

64 × x/100 = 28
=> 64x/100 = 28

Using cross multiplication:

64x/100 = 28
=> 64x = 2800

Divide 64 both sides.

=> x = 2800/64 = 43.75%

28 is 43.75% of 64

andriy [413]2 years ago

7 0

28 is 43.75% of 64
you can calculate this by making a fraction of 28/64 and timing this by 100!

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Use Pythagorean theorem

kogti [31]

Answer:

Step-by-step explanation:

a^2+b^2=c^2

y^2+8^2=17^2

Step 1: Simplify both sides of the equation.

y^2+64=289

Step 2: Subtract 64 from both sides.

y^2+64−64=289−64

y^2=225

Step 3: Take square root.

y=15

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2 years ago

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Troyanec [42]

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2 years ago

Add. 5/12 + 7/8 Write your answer as a fraction in simplest form.

Anastasy [175]

5/12=10/24 and 7/8= 21/24, add those and you will get 31/24 subtract 31 from24 and get 7 your new fraction is 1 7/24 that is 5/12 + 7/8 in simplest form

6 0

2 years ago

Alonzo works as a tutor for $7 an hour and as a waiter for $8 an hour this month he worked a combined total dollar total dollar

docker41 [41]

Answer: -1t + 720

Step-by-step explanation:

He worked at both jobs for 90 hours for this month.

He worked t hours as a tutor which means that as a waiter he worked;

= 90 - t hours

The amount earned is therefore;

= 7t + (8 * (90 - t))

= 7t + 720 - 8t

= -1t + 720

6 0

2 years ago

Given:

diamong [38]

Answer:

10-5 $\sqrt{2}$

Step-by-step explanation:

As per the attached figure, right angled $\triangle MDL$ has an inscribed circle whose center is $I$ .

We have joined the incenter $I$ to the vertices of the $\triangle MDL$ .

Sides MD and DL are equal because we are given that $\angle M = \angle L = 45 ^\circ.$

Formula for area of a $\triangle = \dfrac{1}{2} \times base \times height$

As per the figure attached, we are given that side a = 10.

Using pythagoras theorem, we can easily calculate that side ML = 10 $\sqrt{2}$

Points P,Q and R are at $90 ^\circ$ on the sides ML, MD and DL respectively so IQ, IR and IP are heights of $\triangle$ MIL, $\triangle$ MID and $\triangle$ DIL.

Also,

$\text {Area of } \triangle MDL = \text {Area of } \triangle MIL +\text {Area of } \triangle MID+ \text {Area of } \triangle DIL$

$\dfrac{1}{2} \times 10 \times 10 = \dfrac{1}{2} \times r \times 10 + \dfrac{1}{2} \times r \times 10 + \dfrac{1}{2} \times r \times 10\sqrt2\\\Rightarrow r = \dfrac {10}{2+\sqrt2} \\\Rightarrow r = \dfrac{5\sqrt2}{\sqrt2+1}\\\text{Multiplying and divinding by }(\sqrt2 +1)\\\Rightarrow r = 10-5\sqrt2$

So, radius of circle = $10-5\sqrt2$

8 0

2 years ago