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4 years ago

What is 75% of 17000

1 answer:

6 0

$\displaystyle 75\% \ of \ 17000 \\ \\ \\ \frac{75}{1\not0\not0} \cdot 170\not0\not0= \\ \\ \\ 75 \cdot 170= \\ \\ \\ 12.750 \\ \\$

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Jim is painting the inside of a circular helicopter pad. How many cans of paint would he need if the pad is 50 feet across and e

Inessa [10]

Answer:

Number of can needed to paint inside of the circular helicopter pad = 1

Step-by-step explanation:

The circular pad is given to be 50 feet across.

⇒ Circumference of the circular pad = 50 feet

⇒ 2π·r = 50

⇒ π·r = 25

⇒ r = 7.96 feet

Now, Area of the circular pad = π·r²

⇒ Area = 3.14 × 7.96²

⇒ Area = 198.96 square feet

Now, one can of paint covers 200 feet² area of the pad

Hence, 198 feet² area can also be painted by using one can.

Therefore, number of can needed to paint inside of the circular helicopter pad = 1

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3 years ago

I WILL GIVE BRAINLIEST Solve the equation using inverse operations. Check your solutions. In your final answer, include all of y

oee [108]

Answer:

2.5 = X

Step-by-step explanation:

2.5^3 = 15.7 rounded up to 16

2 + 16 = 18

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3 years ago

What is the slope of a line on a standard xy-plane that passes through the point (1,3) and (4,-3)

nexus9112 [7]

Answer:

Slope(m) = -2

Step-by-step explanation:

The line passes through (1,3) and (4,-3)

Slope(m) = change in y ÷ change in x

m = $\frac{-3 - 3}{4 - 1}$ = $\frac{-6}{3}$ = -2

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4 years ago

Parker grew

vladimir1956 [14]

Answer:

Step-by-step explanation:

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3 years ago

Read 2 more answers

The expression (secx + tanx)2 is the same as _____.

trapecia [35]

<u>Answer:</u>

The expression $\bold{(\sec x+\tan x)^{2} \text { is same as } \frac{1+\sin x}{1-\sin x}}$

<u>Solution:</u>

From question, given that $\bold{(\sec x+\tan x)^{2}}$

By using the trigonometric identity $(a + b)^{2} = a^{2} + 2ab + b^{2}$ the above equation becomes,

$(\sec x+\tan x)^{2} = \sec ^{2} x+2 \sec x \tan x+\tan ^{2} x$

We know that $\sec x=\frac{1}{\cos x} ; \tan x=\frac{\sin x}{\cos x}$

$(\sec x+\tan x)^{2}=\frac{1}{\cos ^{2} x}+2 \frac{1}{\cos x} \frac{\sin x}{\cos x}+\frac{\sin ^{2} x}{\cos ^{2} x}$

$=\frac{1}{\cos ^{2} x}+\frac{2 \sin x}{\cos ^{2} x}+\frac{\sin ^{2} x}{\cos ^{2} x}$

On simplication we get

$=\frac{1+2 \sin x+\sin ^{2} x}{\cos ^{2} x}$

By using the trigonometric identity $\cos ^{2} x=1-\sin ^{2} x$ ,the above equation becomes

$=\frac{1+2 \sin x+\sin ^{2} x}{1-\sin ^{2} x}$

By using the trigonometric identity $(a+b)^{2}=a^{2}+2ab+b^{2}$

we get $1+2 \sin x+\sin ^{2} x=(1+\sin x)^{2}$

$=\frac{(1+\sin x)^{2}}{1-\sin ^{2} x}$

$=\frac{(1+\sin x)(1+\sin x)}{1-\sin ^{2} x}$

By using the trigonometric identity $a^{2}-b^{2}=(a+b)(a-b)$ we get $1-\sin ^{2} x=(1+\sin x)(1-\sin x)$

$=\frac{(1+\sin x)(1+\sin x)}{(1+\sin x)(1-\sin x)}$

$= \frac{1+\sin x}{1-\sin x}$

Hence the expression $\bold{(\sec x+\tan x)^{2} \text { is same as } \frac{1+\sin x}{1-\sin x}}$

8 0

3 years ago