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timama [110]
3 years ago
15

Which of the following is a solution to the equation x2 + 25x = 0

Mathematics
1 answer:
GalinKa [24]3 years ago
6 0

Answer:

x = 0

Or

x = -25

Step-by-step explanation:

x^2 + 25x = 0

Factorise

x( x + 25)=0

x = 0

Or

x + 25 = 0

x = -25

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Find the tangent of angle Θ in the triangle below.
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Step-by-step explanation:

By the Pythagorean theorem, the unknown side has length

\sqrt{9^{2}-4^{2}}=\sqrt{65}

Therefore,

\tan \theta=\boxed{\frac{\sqrt{65}}{4}}

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An expression that is equivalent to (-8a-1)-(-7a+5)
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Step-by-step explanation:

(-8a - 1)- (-7a + 5) = -8a - 1 - 7a*(-1) + 5*(-1) { (-1) is disturbed to all the terms in (-7a +5)}

                         = -8a - 1 + 7a - 5

                         = -8a + 7a -1 - 5 {Combine like terms}

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https://www.emathhelp.net/calculators/algebra-1/parallel-and-perpendicular-line-calculator/

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Read 2 more answers
A solid is formed by adjoining two hemispheres to the ends of a right circular cylinder. An industrial tank of this shape must h
mestny [16]

Answer:

Radius =6.518 feet

Height = 26.074 feet

Step-by-step explanation:

The Volume of the Solid formed  = Volume of the two Hemisphere + Volume of the Cylinder

Volume of a Hemisphere  =\frac{2}{3}\pi r^3

Volume of a Cylinder =\pi r^2 h

Therefore:

The Volume of the Solid formed

=2(\frac{2}{3}\pi r^3)+\pi r^2 h\\\frac{4}{3}\pi r^3+\pi r^2 h=4640\\\pi r^2(\frac{4r}{3}+ h)=4640\\\frac{4r}{3}+ h =\frac{4640}{\pi r^2} \\h=\frac{4640}{\pi r^2}-\frac{4r}{3}

Area of the Hemisphere =2\pi r^2

Curved Surface Area of the Cylinder =2\pi rh

Total Surface Area=

2\pi r^2+2\pi r^2+2\pi rh\\=4\pi r^2+2\pi rh

Cost of the Hemispherical Ends  = 2 X  Cost of the surface area of the sides.

Therefore total Cost, C

=2(4\pi r^2)+2\pi rh\\C=8\pi r^2+2\pi rh

Recall: h=\frac{4640}{\pi r^2}-\frac{4r}{3}

Therefore:

C=8\pi r^2+2\pi r(\frac{4640}{\pi r^2}-\frac{4r}{3})\\C=8\pi r^2+\frac{9280}{r}-\frac{8\pi r^2}{3}\\C=\frac{9280}{r}+\frac{24\pi r^2-8\pi r^2}{3}\\C=\frac{9280}{r}+\frac{16\pi r^2}{3}\\C=\frac{27840+16\pi r^3}{3r}

The minimum cost occurs at the point where the derivative equals zero.

C^{'}=\frac{-27840+32\pi r^3}{3r^2}

When \:C^{'}=0

-27840+32\pi r^3=0\\27840=32\pi r^3\\r^3=27840 \div 32\pi=276.9296\\r=\sqrt[3]{276.9296} =6.518

Recall:

h=\frac{4640}{\pi r^2}-\frac{4r}{3}\\h=\frac{4640}{\pi*6.518^2}-\frac{4*6.518}{3}\\h=26.074 feet

Therefore, the dimensions that will minimize the cost are:

Radius =6.518 feet

Height = 26.074 feet

5 0
3 years ago
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