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3 years ago

Can you help me find the factors?

Mathematics

1 answer:

Scilla [17]3 years ago

6 0

Answer:

1) 1,2,3, 6

2) 1,7

3) 1, 3, 5, 9, 15, 45

4) 1, 2, 4, 5, 10, 20

Step-by-step explanation:

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What is 3 / 2/3 * 12

Mekhanik [1.2K]

3 × 2/3x12 will = 6/9 12

7 0

3 years ago

there are 32 students in 4th grade class.Each table in the classroom seats 6 students.how many tables will be needed

vagabundo [1.1K]

This question wants you to divide 32 tables by 6 students which is 5.3333. Since 5 tables is not enough, 6 tables are needed.

3 0

3 years ago

What are the solutions to the following system?

bija089 [108]

The solutions are $(\sqrt{2},-1) \text { and }(-\sqrt{2},-1)$ .

Solution:

Given equation:

$-2 x^{2}+y=-5$

Add $2 x^{2}$ on both sides.

$-2 x^{2}+y+2 x^{2}=-5+2 x^{2}$

$y=-5+2 x^{2}$ -------- (1)

$y=-3 x^{2}+5$ (given) -------- (2)

Equate (1) and (2).

$-5+2 x^{2}=-3 x^{2}+5$

Add $3 x^{2}$ on both sides.

$-5+2 x^{2}+3 x^{2}=-3 x^{2}+5 +3 x^{2}$

$-5+5x^{2}=5$

Add 5 on both sides.

$-5+5x^{2}+5=5+5$

$5x^{2}=10$

Divide by 5 on both sides, we get

$x^{2}=2$

Taking square root on both sides, we get

$x=\sqrt{2}, x=-\sqrt{2}$

Substitute $x=\sqrt{2}$ in (1).

$y=-5+2(\sqrt{2})^2$

$y=-5+4$

$y=-1$

Substitute $x=-\sqrt{2}$ in (1).

$y=-5+2(-\sqrt{2})^2$

$y=-5+4$

$y=-1$

Therefore the solutions are $(\sqrt{2},-1) \text { and }(-\sqrt{2},-1)$ .

Option C is the correct answer.

6 0

3 years ago

There are 28 nonfiction books and 35 fiction books in the teacher's collection. Which of the following expresses the ratio of no

Lostsunrise [7]

The answer is B) 4:5

First write the ratio out. To find the simplified expression, divide by the greatest common factor (GCF). In this case, the GCF of 28 and 35 is 7.

28:35
--- ----
7 7
=4:5

4:5 is the simplest form because you cannot simplify it further.

5 0

3 years ago

Read 2 more answers

Use implicit differentiation to find the points where the parabola defined by x2−2xy+y2+4x−8y+20=0 has horizontal and vertical t

Komok [63]

Answer:

The parabola has a horizontal tangent line at the point (2,4)

The parabola has a vertical tangent line at the point (1,5)

Step-by-step explanation:

Ir order to perform the implicit differentiation, you have to differentiate with respect to x. Then, you have to use the conditions for horizontal and vertical tangent lines.

-To obtain horizontal tangent lines, the condition is:

$\frac{dy}{dx}=0$ (The slope is zero)

--To obtain vertical tangent lines, the condition is:

$\frac{dy}{dx}=\frac{1}{0}$ (The slope is undefined, therefore the denominator is set to zero)

Derivating respect to x:

$\frac{d(x^{2}-2xy+y^{2}+4x-8y+20)}{dx} = \frac{d(x^{2})}{dx}-2\frac{d(xy)}{dx}+\frac{d(y^{2})}{dx}+4\frac{dx}{dx}-8\frac{dy}{dx}+\frac{d(20)}{dx}$ = $2x -2(y+x\frac{dy}{dx})+2y\frac{dy}{dx}+4-8\frac{dy}{dx}= 0$