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bogdanovich [222]

3 years ago

15

There where x cookies to the beginning of party.By the end of party 16 of then had been eaten.Using x write an expression for th

e number of cookies that were left

1 answer:

guajiro [1.7K]3 years ago

8 0

Answer:

x-16= # of cookies left.

Step-by-step explanation:

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4 less than 6x, cubed

inn [45]

<h3>✽ - - - - - - - - - - - - - - - ~<u>Hello There</u>!~ - - - - - - - - - - - - - - - ✽</h3>

➷ You would write this as:

6x^3 - 4

<h3><u>✽</u></h3>

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

3 0

2 years ago

Read 2 more answers

How do you solve for the segment area in a circle?

Aleks04 [339]

The area of the entire sector of DEF = 60 / 360 * PI * radius^2
sector area = 1 / 6 * 3.14159265... * 20^2
sector area = <span> <span> <span> 209.4395102393 </span> </span> </span>

segment area = sector area - triangle DEF Area
triangle DEF Area = (1/2) * 20 * sine 60 * 20
triangle DEF Area = (1/2) * 0.86603 * 400
triangle DEF Area = <span><span><span>(1/2) * 346.412 </span> </span> </span>
triangle DEF Area = <span> <span> <span> 173.206 </span> </span> </span>

segment area = <span> <span> 209.4395102393 </span> -173.206
</span>
segment area = <span> <span> <span> 36.2335102393 </span> </span> </span>
segment area = 36.23 m

Source:
http://www.1728.org/circsect.htm

6 0

3 years ago

RATHER THAN USE DECIMALS-convert it to cents..

stepan [7]

Answer:

<h2>there are 22 dimes</h2><h2>there are 44 nickles</h2><h2 />

Step-by-step explanation:

D be dimes and N be nickles

2 nickles as dime

nickle = 5 cents and dime =10 cents

N=2D

0.10 D + 0.05 (N)=4.4 substitute N=2D

0.10 D +0.05(2D)=4.4

0.10D+0.10D=4.4

0.20 D=4.4

D=4.4/0.2 = 22

<h2>there are 22 dimes</h2>

N=2D

N=2(22)

N=44

<h2>there are 44 nickles</h2>

check : 0.10(22)+0.05(44)= 4.4 (correct)

5 0

3 years ago

Let S be the solid beneath z = 12xy^2 and above z = 0, over the rectangle [0, 1] × [0, 1]. Find the value of m > 1 so that th

jonny [76]

Answer:

The answer is $\sqrt{\frac{6}{5}}$

Step-by-step explanation:

To calculate the volumen of the solid we solve the next double integral:

$\int\limits^1_0\int\limits^1_0 {12xy^{2} } \, dxdy$

Solving:

$\int\limits^1_0 {12x} \, dx \int\limits^1_0 {y^{2} } \, dy$

$[6x^{2} ]{{1} \atop {0}} \right. * [\frac{y^{3}}{3}]{{1} \atop {0}} \right.$

Replacing the limits:

$6*\frac{1}{3} =2$

The plane y=mx divides this volume in two equal parts. So volume of one part is 1.

Since m > 1, hence mx ≤ y ≤ 1, 0 ≤ x ≤ $\frac{1}{m}$

Solving the double integral with these new limits we have:

$\int\limits^\frac{1}{m} _0\int\limits^{1}_{mx} {12xy^{2} } \, dxdy$

This part is a little bit tricky so let's solve the integral first for dy:

$\int\limits^\frac{1}{m}_0 [{12x \frac{y^{3}}{3}}]{{1} \atop {mx}} \right.\, dx =\int\limits^\frac{1}{m}_0 [{4x y^{3 }]{{1} \atop {mx}} \right.\, dx$

Replacing the limits:

$\int\limits^\frac{1}{m}_0 {4x(1-(mx)^{3} )\, dx =\int\limits^\frac{1}{m}_0 {4x-4x(m^{3} x^{3} )\, dx =\int\limits^\frac{1}{m}_0 ({4x-4m^{3} x^{4}) \, dx$

Solving now for dx:

$[{\frac{4x^{2}}{2} -\frac{4m^{3} x^{5}}{5} ]{{\frac{1}{m} } \atop {0}} \right. = [{2x^{2} -\frac{4m^{3} x^{5}}{5} ]{{\frac{1}{m} } \atop {0}} \right.$

Replacing the limits:

$\frac{2}{m^{2} }-\frac{4m^{3}\frac{1}{m^{5}}}{5} =\frac{2}{m^{2} }-\frac{4\frac{1}{m^{2}}}{5} \\ \frac{2}{m^{2} }-\frac{4}{5m^{2} }=\frac{10m^{2}-4m^{2} }{5m^{4}} \\ \frac{6m^{2} }{5m^{4}} =\frac{6}{5m^{2}}$

As I mentioned before, this volume is equal to 1, hence:

$\frac{6}{5m^{2}}=1\\m^{2} =\frac{6}{5} \\m=\sqrt{\frac{6}{5} }$

3 0

3 years ago

Hate math to be honest

KatRina [158]

Could you take a little bit clear picture? :)

3 0

3 years ago

Read 2 more answers