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Aliun [14]
2 years ago
12

Which of the following values for x and y make the equation 4x+2y+3=19 true?

Mathematics
1 answer:
Aleksandr-060686 [28]2 years ago
7 0

Answer:

x = 3, y = 2

Step-by-step explanation:

Substitute the values of x from each alternative in the previous <em>question</em>, then solve for y. This is depicted below.

  • <em>A. y = -7, x = 5, 4(5) + 2y + 3 = 19</em>
  • <em>B. y = 0 and x = 4, 4(4) + 2y + 3 = 19</em>
  • <em>C. y = 2 and x = 3, 4(3) + 2y + 3 = 1</em>
  • <em>D. y = 4, x = 2, 4(2) + 2y + 3 = 19</em>

Only the third option has a y value that is <em>comparable</em> to the value in the computations, according to the calculations.

As a result, the letter C is the correct response.

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A page should have perimeter of 42 inches. The printing area within the page
Tamiku [17]

Overall  dimensions of the page in order to maximize the printing area is  page should be 11 inches wide and 10 inches long .

<u>Step-by-step explanation:</u>

We have , A page should have perimeter of 42 inches. The printing area within the page  would be determined by top and bottom margins of 1 inch from each side, and the  left and right margins of 1.5 inches from each side. let's assume  width of the page be x inches  and its length be y inches So,

Perimeter = 42 inches

⇒ 2(x+y) = 42\\x+y = 21\\y = 21-x

width of printed area = x-3  & length of printed area = y-2:

area = length(width)

area = (x-3)(y-2)\\area = (x-3)(21-x-2)\\area = (x-3)(19-x)\\area = -x^{2} + 22x -57

Let's find \frac{d(area)}{dx}:

\frac{d(area)}{dx} = \frac{d(-x^{2}+22x-57)}{dx} = -2x +22 , for area to be maximum \frac{d(area)}{dx}= 0

⇒ -2x+22 = 0\\2x =22\\x=11 inches

And ,

y = 21-x\\y = 21-11\\y = 10 inches

∴ Overall  dimensions of the page in order to maximize the printing area is  page should be 11 inches wide and 10 inches long .

7 0
3 years ago
Solve for x:<br> -3(x-2)^2+17=0 <br> round your answer to the nearest hundredth
kicyunya [14]

Answer:

-3(x-2)×2+17=0

  1. -3x+6×2+17=0
  2. -3x+12+17=0
  3. -3x+29=0
  4. -3x=0-29=-29
  5. -3x=-29
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8 0
2 years ago
Help with 21 would be much appreciated.
umka2103 [35]

Answer:

\large\boxed{x=2\sqrt{21}\ and\ y=4\sqrt3}

Step-by-step explanation:

Look at the picture.

ΔADC and ΔCDB are similar. Therefore the sides are in proportion:

\dfrac{AD}{CD}=\dfrac{CD}{DB}

We have

AD=14-8=6\\CD=y\\DB=8

Substitute:

\dfrac{6}{y}=\dfrac{y}{8}         <em>cross multiply</em>

y^2=(6)(8)

y^2=48\to y=\sqrt{48}\\\\y=\sqrt{16\cdot3}\\\\y=\sqrt{16}\ cdot\sqrt3\\\\\boxed{y=4\sqrt3}

For x use the Pythagorean theorem:

x^2=6^2+(4\sqrt3)^2\\\\x^2=36+48\\\\x^2=84\to x=\sqrt{84}\\\\x=\sqrt{4\cdot21}\\\\x=\sqrt4\cdot\sqrt{21}\\\\\boxed{x=2\sqrt{21}}

3 0
3 years ago
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