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Alexxx [7]
3 years ago
6

A rectangle has a length this is three times its width. If the area of the rectangle is 27 square feet, what are the dimensions

of the rectangle
Mathematics
1 answer:
yuradex [85]3 years ago
6 0

Answer:

The width of the rectangle is 3 and the width is 9. Those are reasonable measurements.

Step-by-step explanation:

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Which ordered pair identifies a point in Quadrant IV? A) (2, 5) B) (0, 1) C) (-1, -1) D) (12, -5)
MariettaO [177]
Correct Answer:
Option D (12, -5)

In Quadrant IV, the x-component of the point is positive and the y-component is negative. From the given points, option D contains the point with a positive x-component i.e.12 and negative y-component i.e -5. Therefore, this point lies in Quadrant IV.
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Once I lower the points because I'm going broke and won't be able to ask any questions , NOBODY ANSWERS.
Georgia [21]

Answer:

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Step-by-step explanation:

8 0
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3
Vaselesa [24]

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6 0
2 years ago
Multiply. −2(−5)(−3)
Dominik [7]

The multiplication of −2(−5)(−3) is -30

The multiplication of numbers in brackets with negative signs takes a method whereby the signs are taken into account when performing the arithmetic operation.

From the given question, we are to multiply:

= - 2 × (-5) × (-3)

= +10 × (-3)

= -30

Learn more about the multiplication of numbers here:

brainly.com/question/1755985

4 0
3 years ago
Read 2 more answers
What is the result of substituting for y in the bottom equation<br> y=x+3<br> y=x^2+2x-4
8_murik_8 [283]

The solutions are (2.1925, 5.1925) and (-3.1925, -0.1925)

<em><u>Solution:</u></em>

Given that,

y = x + 3 ------- eqn 1\\\\y = x^2 + 2x - 4  ----- eqn 2

<em><u>We have to substitute eqn 1 in eqn 2</u></em>

x + 3 = x^2 + 2x - 4

\mathrm{Switch\:sides}\\\\x^2+2x-4=x+3\\\\\mathrm{Subtract\:}3\mathrm{\:from\:both\:sides}\\\\x^2+2x-4-3=x+3-3\\\\\mathrm{Simplify}\\\\x^2+2x-7=x\\\\\mathrm{Subtract\:}x\mathrm{\:from\:both\:sides}\\\\x^2+2x-7-x=x-x\\\\\mathrm{Simplify}\\\\x^2+x-7=0

\mathrm{Solve\:with\:the\:quadratic\:formula}\\\\\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\mathrm{For\:}\quad a=1,\:b=1,\:c=-7\\\\x =\frac{-1\pm \sqrt{1^2-4\cdot \:1\left(-7\right)}}{2\cdot \:1}

x = \frac{-1 \pm \sqrt{ 1 + 28}}{2}\\\\x = \frac{ -1 \pm \sqrt{29}}{2}

x = \frac{ -1 \pm 5.385 }{2}\\\\We\ have\ two\ solutions\\\\x = \frac{ -1 + 5.385 }{2}\\\\x = 2.1925

Also\\\\x = \frac{ -1 - 5.385 }{2}\\\\x = -3.1925

Substitute x = 2.1925 in eqn 1

y = 2.1925 + 3

y = 5.1925

Substitute x = -3.1925 in eqn 1

y = -3.1925 + 3

y = -0.1925

Thus the solutions are (2.1925, 5.1925) and (-3.1925, -0.1925)

6 0
3 years ago
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