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kipiarov [429]
3 years ago
14

12r r=1/6 Please show your work

Mathematics
1 answer:
Flauer [41]3 years ago
6 0

Answer:

Step-by-step explanation:

If m = 4, z = 9 and r = 1/6

1. 3 + m = 3 + 4 = 7

2. z - m = 9 - 4 = 5

3. 12r = 12 × 1/6 = 2

4. 60r - 4 = 60(1/6) - 4 = 10-4 = 6

5. 4m - 2 = 4(4) - 2 = 16-2 = 14

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A moving company charges $0.7 per pound for a move from New York to Florida.A family estimates that their belongings weigh about
DanielleElmas [232]

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Same

Step-by-step explanation:

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4 years ago
A recipe calls for 1/2 lb of flour for 1 batch. How many batches can be made with 1/4 lb?
zhenek [66]

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1/2 batch

Step-by-step explanation:

1/2=1

1/4=1/2

3 0
3 years ago
Read 2 more answers
Is it ever possible that cos(A−B)=cos ⁡A − cos ⁡B?
son4ous [18]

Answer:

yes

Step-by-step explanation:

I'm going to give you a slightly different answer, but it's going to make sense :-)

First, let's review what "sin" and "cos" really mean. They are functions that take as an input an angle, which we call theta. They output the base (cos) and height (sin) of a triangle which as a hypotenuse of length 1.    

Now, let's pick some examples. If we happen to set theta to 45 degress, you will get a triangle that looks like this:

In this case, both sin(theta) and cos(theta) are the same number, the square root of 1/2. So cos(theta) + cos (theta) is 2 times the square tool of 1/2.

Now imagine that we now want to find cos (theta + theta). Remember that theta was 45 degrees, so this will be cos (45 + 45), or cos (90).

But remember that cos is the base of a triangle where theta is the angle with the base. Well, that's not a triangle at all, is it? It's just a vertical line. In fact, cos(90) will be zero.

5 0
3 years ago
Which of the following is the graph of f(x) = cos(x+<br> +32
Anit [1.1K]

Answer:

B

Step-by-step explanation:

pi / 2 is equal to 180°/2 which is 90°.

Cos (90°) = 0

if cos90 = 0 then what kind of graph will we have at 90° to the x axis ?

a graph passing through the origin. (0,0)

7 0
3 years ago
The sum of 30 times (1/3)^(n-1) from 1 to infinity
sergeinik [125]

Let

S_n=\displaystyle1+\frac13+\frac1{3^2}+\cdots+\frac1{3^n}

Then

\dfrac13S_n=\displaystyle\frac13+\frac1{3^2}+\frac1{3^3}+\cdots+\frac1{3^{n+1}}

and

S_n-\dfrac13S_n=\dfrac23S_n=1-\dfrac1{3^{n+1}}\implies S_n=\dfrac32-\dfrac1{2\cdot3^n}

and as n\to\infty, we end up with

\displaystyle\lim_{n\to\infty}S_n=\lim_{n\to\infty}\sum_{i=1}^{n+1}\frac1{3^{i-1}}=\lim_{n\to\infty}\left(\frac32-\frac1{2\cdot3^n}\right)=\frac32

So we have

\displaystyle\sum_{n=1}^\infty30\left(\frac13\right)^{n-1}=30\cdot\frac32=45

8 0
3 years ago
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