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Volgvan
3 years ago
11

I will pick brainliest first come, first serve! I need help! Give me an answer that makes sense!

Mathematics
1 answer:
nordsb [41]3 years ago
3 0

Step-by-step explanation:

Total amount of apples in one bag:

x + y + z

Hence, total amount of apples in two bags:

2(x + y + z)

or

2x + 2y + 2z

or

x + x + y + y + z +  z

Hence, B, D and E are correct.

A is incorrect because you are adding 2 more apples on top of the total amount of apples in one bag, and you cannot assume that there are 2 apples in each bag.

C is incorrect because you are multiplying the apples together, but you are supposed to add them up to find the total.

F is incorrect because you are finding 1/2 a bag.

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weeeeeb [17]
Itd Take Her 105 Minutes Start By Dividing 30 By 18 Then Multiplying That by 63
3 0
3 years ago
Read 2 more answers
Find the output, y, when the input x, is 30.<br> y = 14 – 0.5x<br> y =
Mumz [18]

Answer:

y=-1

Step-by-step explanation:

14-(.5*30) same as dividing 30 by 2

14-15=-1   combine

6 0
3 years ago
How do I solve this question ??????
Vlada [557]
Since you have 4 people total splitting the money, you can find the total amount made in the two months by: 184(4)=736
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3 0
3 years ago
Please write the answer​
Travka [436]

\sf \dfrac{3^x - 5 \times 3^{(x-2)}}{3^{(x-3)}} \\ \\ \longrightarrow \sf \dfrac{ {3}^{x} }{ {3}^{(x - 3)}} - \frac{5}{ {3}^{(x - 3)} } \times {3}^{(x - 2)}  \\ \\ \longrightarrow \sf {3}^{[x - (x - 3)]} - \dfrac{5}{ {3}^{(x - 3)} } \times {3}^{(x - 2)} \\ \\ \longrightarrow \sf {3}^{3} - \dfrac{5}{ {3}^{(x - 3)} } \times \dfrac{ {3}^{x}}{9} \\ \\ \longrightarrow \sf {3}^{3} - \dfrac{5}{ \bigg(\dfrac{ {3}^{x} }{ {3}^{3} }\bigg) } \times \dfrac{ {3}^{x}}{9}\\ \\ \longrightarrow \sf {3}^{3} - \dfrac{ ({3}^{3})( 5)}{ {3}^{x} } \times \dfrac{ {3}^{x}}{9}\\ \\ \sf \longrightarrow {3}^{3} - (5 \times 3) \\  \\ \longrightarrow \sf \: 27 - 15 \\  \\ \longrightarrow \leadsto{\underline{\boxed{\sf{ \pink{ 12}}}}}

6 0
2 years ago
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.

4 0
2 years ago
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