All categories

3 years ago

Enter the value for x that makes the equation 3x – 5 = 7x - 21 true.

Mathematics

2 answers:

Nitella [24]3 years ago

8 0

Answer:

Step-by-step explanation:

3x-5=7x-21

3x-7x=5-21

-4x=-16 ( divide by -4 both sides)

x= 4

AlexFokin [52]3 years ago

8 0

Answer:4

Step by step solution:

3x-5=7x-21
-5=4x-21
16=4x
x=4

You might be interested in

PLEASE HELP 23 point

Lubov Fominskaja [6]

Answer:

Step-by-step explanation:

7 0

3 years ago

Find the volume of a cylinder with a diameter of 8 inches and a height that is three times the radius round to the nearest hundr

patriot [66]

volume for cylinder = pi x r^2 x h

radius = half the diameter = 8/2 =4

height = 3 times the radius = 3*4 =12

using 3.14 for pi

volume = 3.14 x 4^2 x 12 = 602.88 cubic inches

5 0

3 years ago

Use the mid-point rule with n = 4 to approximate the area of the region bounded by y = x3 and y = x. (10 points)

USPshnik [31]

See the graph attached.

The midpoint rule states that you can calculate the area under a curve by using the formula:
$M_{n} = \frac{b - a}{2} [ f(\frac{x_{0} + x_{1} }{2}) + f(\frac{x_{1} + x_{2} }{2}) + ... + f(\frac{x_{n-1} + x_{n} }{2})]$

In your case:
a = 0
b = 1
n = 4
x₀ = 0
x₁ = 1/4
x₂ = 1/2
x₃ = 3/4
x₄ = 1

Therefore, you'll have:
$M_{4} = \frac{1 - 0}{4} [ f(\frac{0 + \frac{1}{4} }{2}) + f(\frac{ \frac{1}{4} + \frac{1}{2} }{2}) + f(\frac{\frac{1}{2} + \frac{3}{4} }{2}) + f(\frac{\frac{3}{4} + 1} {2})]$
$M_{4} = \frac{1}{4} [ f(\frac{1}{8}) + f(\frac{3}{8}) + f(\frac{5}{8}) + f(\frac{7}{8})]$

Now, to evaluate your f(x), you need to look at the graph and notice that:
f(x) = x - x³

Therefore:
$M_{4} = \frac{1}{4} [(\frac{1}{8} - (\frac{1}{8})^{3}) + (\frac{3}{8} - (\frac{3}{8})^{3}) + (\frac{5}{8} - (\frac{5}{8})^{3}) + (\frac{7}{8} - (\frac{7}{8})^{3})]$

$M_{4} = \frac{1}{4} [(\frac{1}{8} - \frac{1}{512}) + (\frac{3}{8} - \frac{27}{512}) + (\frac{5}{8} - \frac{125}{512}) + (\frac{7}{8} - \frac{343}{512})]$

M₄ = 1/4 · (2 - 478/512)
= 0.2666

Hence, the <span>area of the region bounded by y = x³ and y = x</span> is approximately 0.267 square units.

6 0

3 years ago

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

WILL MARK BRAINLIEST ASAP!!

7nadin3 [17]

$\dfrac{x^2+ax-4}{x+2}+\dfrac{x+b}{x+2}=\dfrac{x^2+ax-4+x+b}{x+2}=\dfrac{x^2+(a+1)x+(b-4)}{x+2}\\\\\dfrac{x^2+(a+1)x+(b-4)}{x+2}=x+1\\\\\dfrac{x^2+(a+1)x+(b-4)}{x+2}=\dfrac{(x+1)(x+2)}{x+2}\\\\\dfrac{x^2+(a+1)x+(b-4)}{x+2}=\dfrac{(x)(x)+(x)(2)+(1)(x)+(1)(2)}{x+2}\\\\\dfrac{x^2+(a+1)x+(b-4)}{x+2}=\dfrac{x^2+2x+x+2}{x+2}\\\\\dfrac{x^2+(a+1)x+(b-4)}{x+2}=\dfrac{x^2+3x+2}{x+2}\iff a+1=3\ \wedge\ b-4=2\\\\\boxed{a=2\ \wedge\ b=6}$

3 0

3 years ago

Read 2 more answers