1.6x 10^5 in standard notation.

Mathematics

1 answer:

Gekata [30.6K]3 years ago

3 0

The answer is 160,000. Since it's positive,move the decimal point 5 times to the right.

ANSWER: 160,000

You might be interested in

Use substitution to determine if the solution is correct (PINK HIGHLIGHTED ONE!!!)

mart [117]

Answer:

It is not correct.

Step-by-step explanation:

If you plug 4 in for x, you get -2*4+5=13. This simplifies to: -8+5=13. This is an untrue statement since -8+5 is actually -3, not 13.

Hope this is helpful! :)

4 0

3 years ago

Read 2 more answers

Pls answer show work, At a store, notebooks are on sale for 78% off the regular price. The cost of a notebook is represented by

Assoli18 [71]

Answer:

Step-by-step explanation:

<h3> dont subtract the cost of the notebook by 0.78 while multiping it with the cost.</h3>

5 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:

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