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3 years ago

How can you use proportions to solve percent problems

Mathematics

2 answers:

Xelga [282]3 years ago

6 0

Fractions can be easily converted to percents such as:
1/4 = 25%

mestny [16]3 years ago

5 0

Convert the fractions to percenta

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I need help :(((((((

Effectus [21]

Answer:Allen is correct

Step-by-step explanation: Allen is correct because angle 2 has more information than angle 1 and angle 1 is only 30 degrees.

3 0

2 years ago

Read 2 more answers

What is the missing constant term in the perfect square that starts with x2 – 16x?

andre [41]

Answer:

Step-by-step explanation:

the expression is

x² - 16x + _

the constant term is the square of the number residing with x divided by 2

the number number x = 16

therefore the constant term is (16/ 2)²

= 8²

= 64

the formula being used here is

(x + a)² = x² + 2ax + a²

4 0

3 years ago

What is the area of the figure shown delow?

Mariulka [41]

Answer:

160mm2 is the correct answer

Step-by-step explanation:

7 0

2 years ago

PLEASE HELP VERY CONFUESED

Eva8 [605]

V=(4/3)pir^3
r=3
V=(4/3)pi3^3
V=36pi
v=113.04 cubic feet

7 0

3 years ago

Read 2 more answers

Cos^2x+cos^2(120°+x)+cos^2(120°-x)<br>i need this asap. pls help me

o-na [289]

Answer:

$\frac{3}{2}$

Step-by-step explanation:

Using the addition formulae for cosine

cos(x ± y) = cosxcosy ∓ sinxsiny

---------------------------------------------------------------

cos(120 + x) = cos120cosx - sin120sinx

= - cos60cosx - sin60sinx

= - $\frac{1}{2}$ cosx - $\frac{\sqrt{3} }{2}$ sinx

squaring to obtain cos² (120 + x)

= $\frac{1}{4}$ cos²x + $\frac{\sqrt{3} }{2}$ sinxcosx + $\frac{3}{4}$ sin²x

--------------------------------------------------------------------

cos(120 - x) = cos120cosx + sin120sinx

= -cos60cosx + sin60sinx

= - $\frac{1}{2}$ cosx + $\frac{\sqrt{3} }{2}$ sinx

squaring to obtain cos²(120 - x)

= $\frac{1}{4}$ cos²x - $\frac{\sqrt{3} }{2}$ sinxcosx + $\frac{3}{4}$ sin²x

--------------------------------------------------------------------------

Putting it all together

cos²x + $\frac{1}{4}$ cos²x + $\frac{\sqrt{3} }{2}$ sinxcosx + $\frac{3}{4}$ sin²x + $\frac{1}{4}$ cos²x - $\frac{\sqrt{3} }{2}$ sinxcosx + $\frac{3}{4}$ sin²x

= cos²x + $\frac{1}{2}$ cos²x + $\frac{3}{2}$ sin²x

= $\frac{3}{2}$ cos²x + $\frac{3}{2}$ sin²x

= $\frac{3}{2}$ (cos²x + sin²x) = $\frac{3}{2}$

5 0

3 years ago