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saveliy_v [14]
3 years ago
13

What is the result of subtracting the second equation from the first? -2x+y=0 7x+3y=2

Mathematics
1 answer:
ikadub [295]3 years ago
4 0

Answer:

5x -2y = -2

Step-by-step explanation:

-2x+y=0

-7x+3y=2

Subtract the second equation from the first

-2x+y=0

7x-3y=-2

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

5x -2y = -2

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A function is given: f (x) = 2x + 3<br> What is the value of f(-2)?
Nadusha1986 [10]

Answer:

f(-2) = -1

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3 years ago
Four times the sum of a number and 15 is at least 120. Find all possible solutions for x
zloy xaker [14]
First translate the English phrase "Four times the sum of a number and 15 is at least 120" into a mathematical inequality.

"Four times..." means we're multiplying something by 4.

"... the sum of a number and 15..." means we're adding an unknown and 15 and then multiplying the result by 4.

"... is at least 120" means when we substitute the unknown for a value, in order for that value to be in the solution set, it can only be less than or equal to 120.

So, the resulting inequality is 4(x + 15) ≤ 120.

Simplify the inequality.

4(x + 15) ≤ 120
4x + 60 ≤ 120 <-- Using the distributive property
4x ≤ 60 <-- Subtract both sides by 60
x ≤ 15 <-- Divide both sides by 4

Now that we have the inequality in a simplified form, we can easily see that in order to be in the solution set, the variable x can be no bigger than 15.

In interval notation it would look something like this:

[15, ∞)

In set builder notation it would look something like this:

{x | x ∈ R, x ≤ 15}

It is read as "the set of all x, such that x is a member of the real numbers and x is less than or equal to 15".
7 0
3 years ago
4) A window is in the form of a rectangle surmounted by a semicircle. The rectangle is of clear glass, whereas the semicircle is
lisov135 [29]

Answer:

l=0.1401P\\

w =0.2801P

where P = perimeter

Step-by-step explanation:

Given that a window is in the form of a rectangle surmounted by a semicircle.

Perimeter of window =2l+\pid/2+w

P= 2l+3.14 (w/2)+w

Or P = 2l+2.57w\\l = \frac{P-2.57w}{2}

To allow maximum light we must have maximum area

Area = area of rectangle + area of semi circle where rectangle width = diameter of semi circle

A=lw +\pi \frac{w^2}{8}

A=lw +\pi \frac{w^2}{8}\\A=w*\frac{P-2.57w}{2}+0.3925w^2\\2A= Pw-2.57w^2+(0.785w^2)\\2A' = P-5.14w+1.57 w\\2A" =-5.14+1.57

Hence we get maximum area when i derivative is 0

i.e. P-5.14w+1.57 w=0\\3.57w =P\\w = \frac{P}{3.57} =0.2801P

l = \frac{P-2.57w}{2}\\l = 0.1401P

Dimensions can be

l=0.1401P\\w =0.2801P

5 0
2 years ago
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