All categories

3 years ago

PLS HELP DUE SOON.....

Mathematics

2 answers:

melisa1 [442]3 years ago

5 0

Answer: For a translation, you need a distance and a direction.

For a rotation, you need a point that is the center of the rotation and the degree (and sometimes direction) of rotation.

For reflection you need the axis or a defined line that the triangle is reflected across.

Step-by-step explanation:

mars1129 [50]3 years ago

3 0

What he said lol sidedjdjejejjek

You might be interested in

I need to know how to solve this

dedylja [7]

Answer:

Step-by-step explanation:

I think you have to call a = 14/32 and b = 7/4

3a = 3 * 14/32 = 42/32

b = 1 3/4 = (4 + 3)/4 = 7/4

42/32 // 7/4 Invert and multiply

42/32 * 4/7 Cancel 4 into 32 and 7 into 42

6/8 = 3/4

I make the answer C

3 0

3 years ago

Read 2 more answers

ILL GIVE BRAINLIEST PLS HELP ME

luda_lava [24]

One solution

The two (or more) equations would only intersect once

No solution

Lines never intersect

The lines are parallel

Infinite solutions

The lines are the exact same

That’s all the characteristics I know

3 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:

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