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Airida [17]
3 years ago
6

16 2/3 divided 5 5/7

Mathematics
2 answers:
Stells [14]3 years ago
8 0

16\frac{2}{3}  : 5\frac{5}{7}  = \\ \\ = \frac{16 \times 3 + 2}{3} : \frac{5 \times 7 + 5}{7} = \\ \\ = \frac{50}{3} : \frac{40}{7} = \\ \\ = \frac{50}{3} \times \frac{7}{40} = \\ \\ = \frac{350}{120} = 2.91(6)

Marizza181 [45]3 years ago
4 0

Answer: 2 and 11/12

Step-by-step explanation: To divide mixed numbers, first write the mixed numbers as improper fractions.

We can write a mixed number as an improper fraction by multiplying the denominator by the whole number and then adding the numerator. Finally, we put our new number over original denominator.

So here, we can write 16 and 2/3 as the improper fraction 50/3 and we can write 5 and 5/7 as the improper fraction 40/7.

So here we have 50/3 ÷ 40/7.

Dividing by a fraction is the same as multiplying by the reciprocal of that fraction. So here, 50/3 ÷ 40/7 is the same as 50/3 × 7/40.

So now we are simply multiplying fractions.

To multiply fractions, we multiply across the numerators and multiply across the denominators.

\frac{50}{3} x \frac{7}{40} = \frac{350}{120}

Notice however that 350/120 is not in lowest terms so we divide the numerator and the denominator by 10 and we get the equivalent fraction 35/12. The fraction 35/12 can be converted into a mixed number by dividing the denominator into the numerator.

35 ÷ 12 = 2 with a remainder of 11 or 2 and 11/12.

Therefore, 16 and 2/3 ÷ 5 and 5/7 = 2 and 11/12.

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Which angle is supplementary to 65 degrees​
Alexeev081 [22]

Answer:

115°

Step-by-step explanation:

Supplementary angles equal 180.

180-65=115°

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3 years ago
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If the if a cylinder is 7 in tall and has a volume of 63 Pi inches what is the area of the cross section it goes right down the
vitfil [10]

Answer: 9\pi \ in.^2

Step-by-step explanation:

Given

The height of the cylinder is h=7\ in.

The volume of the cylinder is V=63\pi \ in.^3

The volume of the cylinder is the product of area and height

\Rightarrow V=\pi r^2h\\

Insert the values

\Rightarrow 63\pi =A\times 7\\\\\Rightarrow A=\dfrac{63\pi}{7}\\\\\Rightarrow A=9\pi\ in.^2

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7 0
2 years ago
The answer is 288, but i don’t know how to get the answer. please give me an explanation.
andriy [413]
Sorry I don’t know the answer and also sorry about that other comment they are very rude I hope you find someone who actually knows the answer!! ❤️
3 0
2 years ago
Can you please explain this?​
Leni [432]

Answer:

The question is giving you pairs of points in space which can be used to define lines. It is then asking you to determine if the lines defined by those points are parallel, perpendicular, or neither.

Step-by-step explanation:

Two key things you need to know to solve this is that the lines will be parallel if their slopes are the same, and perpendicular if one slope is the negative reciprocal of the other (i.e. s_{1} = -s_{2}^{-1})

Let's start with question 11. You are given two pairs of points, each of which describes a distinct line:

(3,5)-(1,1) and (0, 2)-(5, 12)

To find the slope of each pair, take the vertex with the lesser x co-ordinate, and subtract it from the vertex with the greater x co-ordinate.  That will give us a valid Δx and Δy to get the slope.

In the first pair, 3 > 1, so we'll subtract the second point from the first:

s = \Delta y / \Delta x\\s = \frac{5 - 1}{3 - 1}\\s = 4/2\\s = 2

So the first pair of vectors describe a line with a slope of 2.  Let's look at the other pair:

s = \Delta y / \Delta x\\s = (12 - 2) / (5 - 0)\\s = 10 / 5\\s = 2

That also gives us a slope of 2, meaning that the two lines are parallel.

This same process will need to be done for the other three questions.  We can't answer questions 12 or 14 here, as the last point is cut off on the edge of the image.  For question 3 though, one line has a slope of 7/3, and the other 3/7. That puts them in the "neither" category, as one is not the negative reciprocal of the other, but instead the positive reciprocal.

6 0
3 years ago
Y+3=7(x-2) in standard form
uysha [10]
7x- y = 17 is the standard form
4 0
3 years ago
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