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8_murik_8 [283]
4 years ago
7

Please save my life from fail grades

Mathematics
1 answer:
Harlamova29_29 [7]4 years ago
5 0

1/x is smallest when x is the largest it can be, so it is 1/3

1/y is smallest when y is the largest it can be, so it is 1/5

The average of 1/3 is 1/5 can be found by adding them together and dividing by 2:

1/3 + 1/5 = 5/15 + 3/15 = 8/15

Divide it by 2:

4/15

The answer is C.

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Answer:

sqrt(x-3)+3

Step-by-step explanation:

Consider the graph of a function f(x), if we want to move the graph right by a units we have f(x-a), to then move it up by b units we have f(x-a)+b. In this case you replace the x with x-a to get sqrt(x-a)+b

Notice how the graph goes through (3,3) and (4,4) like the original graph goes through (0,0) and (1,1). This means we want to translate it 3 up and 3 right giving us the answer.

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Answer. To verify that the decimals are lined up correctly, we must ensure that they are both lined up at the same point. The reason this is done is so that whole numbers will be subtracted from other whole numbers, not decimals.
Extra. Hope this helped, if it’s correct let me know so others can use it :)
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6 0
3 years ago
Read 2 more answers
270 divided by 17 need help
kifflom [539]
About 15.88 hope that answers
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3 years ago
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70. Machine Shop Calculations A steel plate has the
3241004551 [841]

The coordinates of a point is the location of the point in a plane.

<em>The coordinates of the centers of holes are: (48.5, 28) and (28, 48.5)</em>

Given

\theta_1 = \theta_2 = \theta_3 = 30^o

R = 60

r = 56

I've added an attachment as an illustration

<u />

<u>Considering </u>(x_1,y_1)<u />

To solve for x1, we make use of cosine ratio.

So, we have:

\cos(\theta_1) =\frac{x}{r}

Make x the subject

x_1 = r \times \cos(\theta_1)

x_1 = 56 \times \cos(30^o)

x_1 = 48.5

To solve for y1, we make use of sine ratio.

So, we have:

\sin(\theta_1) =\frac{y_1}{r}

Make y the subject

y_1 = r \times \sin(\theta_1)

y_1 = 56 \times \sin(30^o)

y_1 = 28

So, we have:

(x_1,y_1) = (48.5,28)

<u>Considering </u>(x_2,y_2)<u />

To solve for x2, we make use of cosine ratio.

So, we have:

\cos(\theta_1+\theta_2) =\frac{x_2}{r}

Make x the subject

x_2 = r \times \cos(\theta_1+\theta_2)

x_2 = 56 \times \cos(30+30)

x_2 = 56 \times \cos(60^o)

x_2 = 28

To solve for y1, we make use of sine ratio.

So, we have:

\sin(\theta_1+\theta_2) =\frac{y_2}{r}

Make y the subject

y_2 = r \times \sin(\theta_1+\theta_2)

y_2 = 56 \times \sin(30+30)

y_2 = 56 \times \sin(60)

y_2 = 48.5

So, we have:

(x_2,y_2) = (28,48.5)

<em>Hence, the coordinates of the centers of the holes are: (48.5, 28) and (28, 48.5)</em>

Read more about coordinate geometry at:

brainly.com/question/8121530

6 0
3 years ago
Find the numbers such that fifferent is greater than the half of thier qoutient.​
Grace [21]
1/15 is greatest the 3/5
5 0
4 years ago
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