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Karolina [17]
3 years ago
11

Pa help po pls thank you​

Mathematics
1 answer:
maxonik [38]3 years ago
4 0

Step-by-step explanation:

1.

Let's say she starts on a Monday(first day).

She walked 10min on Monday

She then starts increasing the number of minutes each day by "2".

So

Tuesday --- 12min

Wednesday--- 14min

Thursday --- 16min

Friday --- 18min

Saturday --- 20min

Sunday ---- 22min

Total = 10 + 12 + 14 + 16 + 18 + 20 + 22

= 112minutes

She walks 112minutes in one week.

2.

Since each 8 cans are consumed per day

...

200cans would be consumed in

200/8 = 25days.

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Find the $x$-intercept of the line $3x - 5y = 7.$
kotegsom [21]

The x-intercept of the line 3x - 5y = 7 is at point ( 7/3, 0 ).

<h3>What is the x-intercept of the line?</h3>

Given the function;

  • 3x - 5y = 7
  • x-intercept ?

X-intercept occurs on the graph when y=0. Hence, to determine x-intercept, we substitute in 0 for y and solve for x.

3x - 5y = 7

3x - 5(0) = 7

3x - 0 = 7

3x = 7

x = 7/3

Hence, we have a point ( 7/3, 0 ).

The x-intercept of the line 3x - 5y = 7 is at point ( 7/3, 0 ).

Learn more about x and y-intercepts here: brainly.com/question/12791065

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1 year ago
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cricket20 [7]
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3 0
3 years ago
Rewrite with only sin x and cos x.
Annette [7]

Option A

\cos 3 x=\cos x-4 \cos x \sin ^{2} x

<em><u>Solution:</u></em>

Given that we have to rewrite with only sin x and cos x

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cos 3x = cos(x + 2x)

We know that,

\cos (a+b)=\cos a \cos b-\sin a \sin b

Therefore,

\cos (x+2 x)=\cos x \cos 2 x-\sin x \sin 2 x  ---- eqn 1

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\sin 2 x=2 \sin x \cos x

\cos 2 x=\cos ^{2} x-\sin ^{2} x

Substituting these values in eqn 1

\cos (x+2 x)=\cos x\left(\cos ^{2} x-\sin ^{2} x\right)-\sin x(2 \sin x \cos x)  -------- eqn 2

We know that,

\cos ^{2} x-\sin ^{2} x=1-2 \sin ^{2} x

Applying this in above eqn 2, we get

\cos (x+2 x)=\cos x\left(1-2 \sin ^{2} x\right)-\sin x(2 \sin x \cos x)

\begin{aligned}&\cos (x+2 x)=\cos x-2 \sin ^{2} x \cos x-2 \sin ^{2} x \cos x\\\\&\cos (x+2 x)=\cos x-4 \sin ^{2} x \cos x\end{aligned}

\cos (x+2 x)=\cos x-4 \cos x \sin ^{2} x

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\cos 3 x=\cos x-4 \cos x \sin ^{2} x

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