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yarga [219]
3 years ago
15

Solve for the missing x coordinate above

Mathematics
1 answer:
Cloud [144]3 years ago
7 0
The unknown vertex is on the perpendicular bisector of the opposite side.  So the x coordinate is the same as that of the midpoint, x=(18+52)/2=35.


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ziro4ka [17]

Answer: one

Step-by-step explanation:

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Will mark brainlest! <br>formula of (x+1)cube <br>​
ANTONII [103]

Answer:

(x+1)^3=x^3+3x^2+3x+1

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3 years ago
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Which of the following terms can be combined in this expression? -13 - x/5 + 6.2y + 20.5x - 3/8y
dsp73
C. 6.2g can be combined with -3/8y. This is because they both end in ‘y’.
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3 years ago
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Find the volume of a right circular cone that has a height of 3.9 ft and a base with a circumference of 16.3 ft. Round your answ
MissTica

Answer:

The answer to your question is Volume = 27.5 ft³

Step-by-step explanation:

Data

Volume = ?

height = 3.9 ft

Circumference = 16.3 ft

Process

1.- Calculate the radius of the Cone

   Circumference = 2πr = 16.3 ft

-Solve for r

                                    r = 16.3/2π

                                    r = 16.3/6.28

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2.- Calculate the volume

Volume = 1/3πr²h

-Substitution

Volume = 1/3(3.14)(2.5955)²(3.9)

-Simplification

Volume = 1/3(82.4966)

-Result

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8 0
3 years ago
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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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