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Vladimir79 [104]
3 years ago
8

Help me please,Thanks

Mathematics
1 answer:
Oksana_A [137]3 years ago
7 0
3 cubes is most likely the answer.
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48 é dividido por 4?​
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The product of two rational numbers is (-28/81). If one of them is (-2/3), the find the other.
KatRina [158]

Answer:

The other rational number is \dfrac{14}{27}.

Step-by-step explanation:

First rational number is \dfrac{-2}{3}

Let other rational number is \dfrac{x}{y}

The product of two rational numbers is \dfrac{-28}{81}

According to question,

\dfrac{x}{y}\times \dfrac{-2}{3}=\dfrac{-28}{81}

Multiplying both sides by (-3/2) such that,

\dfrac{x}{y}\times \dfrac{-2}{3}\times \dfrac{-3}{2}=\dfrac{-28}{81}\times \dfrac{-3}{2}\\\\\dfrac{x}{y}=\dfrac{14}{27}

So, the other rational number is \dfrac{14}{27}.

6 0
3 years ago
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NO LINKS!!
dexar [7]

Answer: Anything between 0 and 10, excluding both endpoints.

In terms of symbols we can say 0 < w < 10 where w is the width.

===================================================

Explanation:

You could do this with two variables, but I think it's easier to instead use one variable only. This is because the length is dependent on what you pick for the width.

w = width

2w = twice the width

2w-5 = five less than twice the width = length

So,

  • width = w
  • length = 2w-5

which lead to

area = length*width

area = (2w-5)*w

area = 2w^2-5w

area < 150

2w^2 - 5w < 150

2w^2 - 5w - 150 < 0

To solve this inequality, we will solve the equation 2w^2-5w-150 = 0

Use the quadratic formula. Plug in a = 2, b = -5, c = -150

w = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\w = \frac{-(-5)\pm\sqrt{(-5)^2-4(2)(-150)}}{2(2)}\\\\w = \frac{5\pm\sqrt{1225}}{4}\\\\w = \frac{5\pm35}{4}\\\\w = \frac{5+35}{4} \ \text{ or } \ w = \frac{5-35}{4}\\\\w = \frac{40}{4} \ \text{ or } \ w = \frac{-30}{4}\\\\w = 10 \ \text{ or } \ w = -7.5\\\\

Ignore the negative solution as it makes no sense to have a negative width.

The only practical root is w = 10.

If w = 10 feet, then the area = 2w^2-5w results in 150 square feet.

----------------------

Based on that root, we need to try a sample value that is to the left of it.

Let's say we try w = 5.

2w^2 - 5w < 150

2*5^2 - 5*5 < 150

25 < 150 ... which is true

This shows that if 0 < w < 10, then 2w^2-5w < 150 is true.

Now try something to the right of 10. I'll pick w = 15

2w^2 - 5w < 150

2*15^2 - 5*15 < 150

375 < 150 ... which is false

It means w > 10 leads to 2w^2-5w < 150  being false.

Therefore w > 10 isn't allowed if we want 2w^2-5w < 150 to be true.

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3 years ago
Which is the simplified form of the expression
ki77a [65]

Are you gonna put the expression or what?

8 0
4 years ago
In the equation below, identify the constant ; 4(x+1) -7x = -11
castortr0y [4]

Answer:

It will be A because of if you round it it would be -7

3 0
3 years ago
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