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Maurinko [17]
3 years ago
9

Choose the fraction that goes in the blank 7/8 > - > 3/5

Mathematics
1 answer:
Mama L [17]3 years ago
3 0

Answer:

The fraction is \frac{3}{4}

Step-by-step explanation:

* Lets explain how to solve the problem

- The fraction m/n has m as a numerator and n as a denominator

- To compare between fractions we must to put all of them with

  same denominator

- We can make that by find the lowest common multiple (L.C.M) of

 the all denominators

- Ex: The L.C.M of the denominators of these fraction

 \frac{1}{2},\frac{3}{4},\frac{5}{6} is 12 because 12 is the first

 common multiple of 2 , 4 , 6

- The fractions will be change:

# \frac{1}{2}=\frac{6}{12} ⇒ we multiplied up and down by 6

  to make the denominator = 12

# \frac{3}{4}=\frac{9}{12} ⇒ we multiplied up and down by 3

  to make the denominator = 12

# \frac{5}{6}=\frac{10}{12} ⇒ we multiplied up and down by 2

  to make the denominator = 12

- The fractions will be \frac{6}{12},\frac{9}{12},\frac{10}{12}

* Lets solve the problem

- To chose a fraction between \frac{7}{8},\frac{3}{5}, we must

 find The L.C.M for the denominators 8 and 5

∵ The least common multiple of 8 and 5 is 40

∴ The fraction \frac{7}{8}=\frac{35}{40} ⇒ we multiplied up

  and down by 5 to make the denominator = 40

∴ The fraction \frac{3}{5}=\frac{24}{40} ⇒ we multiplied up

  and down by 8 to make the denominator = 40

- Lets find any fraction between \frac{35}{40} and \frac{24}{40}

∵ The fraction \frac{30}{40} is between the fractions

    \frac{35}{40} and  \frac{24}{40}

∴ \frac{35}{40}>\frac{30}{40}>\frac{24}{40}

- Lets reduce the fraction to its simplest form

∵ The simplest form of \frac{30}{40} is \frac{3}{4}

   by dividing up and down by 10

∴ \frac{7}{8}>\frac{3}{4}>\frac{3}{5}

* The fraction is \frac{3}{4}

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Which values of PP and QQ result in an equation with no solutions? -60x+32=Qx+P−60x+32=Qx+P Choose all answers that apply: Choos
Inga [223]
<h3>Answer:</h3>

  A and C

<h3>Step-by-step explanation:</h3>

Given:

  -60x+32 = Qx+P

Find:

Which values of P and Q result in an equation with no solutions? Choose all answers that apply:

(Choice A) A Q=-60 P=60

(Choice B) B Q=32 P=60

(Choice C) C Q=-60 P=−32

(Choice D) D Q=32 P=−60

Solution:

The equation will have no solution if it reduces to ...

  0 = (non-zero constant)

If we add 60x-32 to both sides, we get

  0 = 60x +Qx + P-32

  0 = (Q+60)x +(P-32)

The x-term must be zero, so Q+60 = 0, or Q = -60.

The constant term must be non-zero, so P-32 ≠0, or P ≠ 32.

The appropriate answer choices are those with Q=-60 and P≠32, A and C.

8 0
3 years ago
A cylinder has diameter of 4ft and a height of 9ft. Explain whether halving the diameter has the same effect on the surface area
elena55 [62]

Answer:

See Below

Step-by-step explanation:

The surface area of cylinder is given by the formula:

SA=2\pi r^2 + 2\pi r h

Where

r is radius ( diameter is 4, so radius is 4/2 = 2)

h is height ( h = 9)

Lets find original surface are:

SA=2\pi r^2 + 2\pi r h\\SA=2\pi (2)^2 + 2\pi (2) (9)\\SA=8\pi +36\pi\\SA=44\pi

<u>Halving diameter:</u>

diameter would be 4/2 = 2, so radius would be 2/2 = 1

So, SA would be:

SA=2\pi r^2 + 2\pi r h\\SA=2\pi (1)^2 + 2\pi (1) (9)\\SA=2\pi +18\pi\\SA=20\pi

<u>Halving height:</u>

Height is 9, halving would make it 9/2 = 4.5

Now, calculating new SA:

SA=2\pi r^2 + 2\pi r h\\SA=2\pi (2)^2 + 2\pi (2) (4.5)\\SA=8\pi + 18\pi\\SA= 26\pi

Original SA is 44\pi,

Halving diameter makes it 20\pi

Halving height makes it 26\pi

So, halving diameter does not have same effect as halving height.

5 0
3 years ago
Please help asap plz
butalik [34]

Answer:

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Step-by-step explanation:

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On a coordinate plane, solid circles appear at the following points: (negative 3, 2), (negative 3, 3), (1, 4), (1, negative 4),
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Answer:

3 points

Step-by-step explanation:

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So the relation can be represented with points (x, y), which means that point x is being mapped into point y.

Such that the needed condition is that each element of the domain is mapped into only one element in the domain.

So for example, if for a relationship we have the points (a, b) and (a, c)

The point a is being mapped into two different outputs.

Now let's go to our problem, we have the points:

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If we have repeated inputs, we need to remove points until we have one of each.

We can see that the inputs -3, 1, and 2 are repeated one time.

Then we need to remove one of each, for example if we remove the second, fourth and sixth points, the set becomes:

(-3, 2), (1, 4), and (2, -4)

This is a function.

Then we need to remove 3 points.

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