Ask question

Ask question

All categories

3 years ago

15

What is 8.038×10 to the eighth power in standard form

1 answer:

Vlad1618 [11]3 years ago

4 0

The answer is 803,800,000

You might be interested in

a telephone pole casts a 10 food shadow. A horse that stands 5.5 feet tall casts a 2.5 foot shadow. How tall is the pole?

Fittoniya [83]

Answer:

72

i think i remember doing this

8 0

3 years ago

Is 7/10 more than 1/2

Bingel [31]

7/10 is bigger because it is more than half

5 0

3 years ago

A rectangular box with a square base is to be constructed from material that costs $9 per ft2 for the bottom, $5 per ft2 for the

kramer

Answer:

$18.73ft^3$

Step-by-step explanation:

Let

Side of square base=x

Height of rectangular box=y

Area of square base=Area of top= $(side)^2=x^2$

Area of one side face= $l\times b=xy$

Cost of bottom=$9 per square ft

Cost of top=$5 square ft

Cost of sides=$4 per square ft

Total cost=$204

Volume of rectangular box= $V=lbh=x^2y$

Total cost= $9(x^2)+5x^2+4(4xy)=14x^2+16xy$

$204=14x^2+16xy$

$204-14x^2=16xy$

$y=\frac{204-14x^2}{16x}=\frac{102-7x^2}{8x}$

Substitute the values of y

$V(x)=x^2\times \frac{102-7x^2}{8x}=\frac{1}{8}(102x-7x^3)$

Differentiate w.r.t x

$V'(x)=\frac{1}{8}(102-21x^2)=0$

$V'(x)=0$

$\frac{1}{8}(102-21x^2)=0$

$102-21x^2=0$

$102=21x^2$

$x^2=\frac{102}{21}=4.85$

$x=\sqrt{4.85}=2.2$

It takes positive because side length cannot be negative.

Again differentiate w.r. t x

$V''(x)=\frac{1}{8}(-42x)$

Substitute the value

$V''(2.2)=-\frac{42}{8}(2.2)=-11.55$

Hence, the volume of box is maximum at x=2.2 ft

Substitute the value of x

$y=\frac{102-7(2.2)^2}{8(2.2)}=3.87 ft$

Greatest volume of box= $x^2y=(2.2)^2\times 3.87=18.73 ft^3$

5 0

3 years ago

Help plsss A sign in a bakery gives these options: 12 cupcakes for $29, 24 cupcakes for $56, and 50 cupcakes for $129 a) Find ea

xxTIMURxx [149]

Answer:

A) $2.42, $2.33, $2.58

B) The best deal is 24 cupcakes for $56

Step-by-step explanation:

5 0

3 years ago

Leviafan [203]

$\dfrac{\mathrm dy}{\mathrm dt}=ky\iff\dfrac{\mathrm dy}y=k\,\mathrm dt$

Integrating both sides, we get

$\ln|y|=kt+C$
$\implies y=e^{kt+C}=e^{kt}e^C=Ce^{kt}$

When $t=0$ , we have $y=10000$ , so that

$10000=Ce^{0k}\implies C=10000$

When $t=4$ , we have $y=8000$ , which means

$8000=10000e^{4k}\implies k=\dfrac14\ln\dfrac{8000}{10000}\approx-0.0558$

5 0

3 years ago