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snow_tiger [21]
3 years ago
6

What is y + 1/5 = 3x in standard form?

Mathematics
1 answer:
vazorg [7]3 years ago
6 0
5y + 1 = 15x
- 15x + 5y = - 1
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Red and grey bricks were used to build a decorative wall. The number of red bricks and grey bricks was 5/2. There were 140 brick
ozzi

Answer:

100 red bricks

Step-by-step explanation:

Let's say red bricks is equal to 5x

Gray bricks is equal to 2x

We have an equation

2x + 5x = 140

= 7x = 140

Divide through by 7 to get the value of x

X = 140/7

X = 20

Red bricks = 5(x)

= 5(20)

= 100

Gray bricks = 2(x)

= 2(20)

= 40

Therefore in conclusion the number of red bricks is 100.

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2 years ago
You and your friends are going to a local concert. Each ticket costs $14. The total amount of ticket sales for the concert was $
serious [3.7K]

Answer: 628

Step-by-step explanation:

8792/14 = 628

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3 years ago
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Mamont248 [21]

Answer:

Your answer will be 69

Step-by-step explanation:

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2 years ago
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Answer:

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Step-by-step explanation:

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3 years ago
Read 2 more answers
You throw a ball up and its height h can be tracked using the equation h=2x^2-12x+20.
postnew [5]

<em><u>This problem seems to be wrong because no minimum point was found and no point of landing exists</u></em>

Answer:

1) There is no maximum height

2) The ball will never land

Step-by-step explanation:

<u>Derivatives</u>

Sometimes we need to find the maximum or minimum value of a function in a given interval. The derivative is a very handy tool for this task. We only have to compute the first derivative f' and have it equal to 0. That will give us the critical points.

Then, compute the second derivative f'' and evaluate the critical points in there. The criteria establish that

If f''(a) is positive, then x=a is a minimum

If f''(a) is negative, then x=a is a maximum

1)

The function provided in the question is

h(x)=2x^2-12x+20

Let's find the first derivative

h'(x)=4x-12

solving h'=0:

4x-12=0

x=3

Computing h''

h''(x)=4

It means that no matter the value of x, the second derivative is always positive, so x=3 is a minimum. The function doesn't have a local maximum or the ball will never reach a maximum height

2)

To find when will the ball land, we set h=0

2x^2-12x+20=0

Simplifying by 2

x^2-6x+10=0

Completing squares

x^2-6x+9+10-9=0

Factoring and rearranging

(x-3)^2=-1

There is no real value of x to solve the above equation, so the ball will never land.

This problem seems to be wrong because no minimum point was found and no point of landing exists

3 0
3 years ago
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