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2 years ago

Give me an example of how you would solve 1+1 in the best way possible

Mathematics

2 answers:

sergey [27]2 years ago

7 0

You have one dog, you get another dog, you now have two dogs

TEA [102]2 years ago

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Hold two fingers up and put them together

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TIMED PLEASE HURRY HELP WITH 2 EASY QUESTIONS

Alex_Xolod [135]

<h2> Question # 1</h2><h2>Which statements are true?</h2><h2 /><h3><u>Analyzing and solving the first statement:</u></h3>

$4g^2-g=g^2\left(4-g\right)$

Solving the expression

$4g^2-g$

$\mathrm{Apply\:exponent\:rule}:\quad \:a^{b+c}=a^ba^c$

$g^2=gg$

So,

$4gg-g$

$\mathrm{Factor\:out\:common\:term\:}g$

$g\left(4g-1\right)$

So,

$4g^2-g:\quad g\left(4g-1\right)$

Therefore, the statement $4g^2-g=g^2\left(4-g\right)$ is NOT CORRECT.

<h3><u>Analyzing and solving the second statement:</u></h3>

$35g^5-25g^2=\:5g^2\left(7g^3-5\right)$

Solving the expression

$35g^5-25g^2$

$\mathrm{Apply\:exponent\:rule}:\quad \:a^{b+c}=a^ba^c$

$g^5=g^3g^2$

So,

$35g^3g^2-25g^2$

$\mathrm{Rewrite\:}25\mathrm{\:as\:}5\cdot \:5$

$\mathrm{Rewrite\:}35\mathrm{\:as\:}5\cdot \:7$

$5\cdot \:7g^3g^2-5\cdot \:5g^2$

$\mathrm{Factor\:out\:common\:term\:}5g^2$

$5g^2\left(7g^3-5\right)$

So,

$35g^5-25g^2=\:5g^2\left(7g^3-5\right)$

Therefore, the statement $35g^5-25g^2=\:5g^2\left(7g^3-5\right)$ is CORRECT.

<h3><u>Analyzing and solving the third statement:</u></h3>

$24g^4+18g^2=\:6g^2\left(4g^2+3g\right)$

<h3 />

Solving the expression

<h3> $24g^4+18g^2$ </h3><h3> $24g^2g^2+18g^2$ </h3><h3> $\mathrm{Rewrite\:}18\mathrm{\:as\:}6\cdot \:3$ </h3><h3> $\mathrm{Rewrite\:}24\mathrm{\:as\:}6\cdot \:4$ </h3><h3> $6\cdot \:4g^2g^2+6\cdot \:3g^2$ </h3><h3> $\mathrm{Factor\:out\:common\:term\:}6g^2$ </h3><h3> $6g^2\left(4g^2+3\right)$ </h3>

So,

<h3> $24g^4+18g^2=6g^2\left(4g^2+3\right)$ </h3>

Therefore, the statement $24g^4+18g^2=\:6g^2\left(4g^2+3g\right)$ is CORRECT.

<h3><u>Analyzing and solving the fourth statement:</u></h3>

$9g^3+12=\:3\left(3g^3+4\right)$

Solving the expression

$9g^3+12$

$\mathrm{Rewrite\:}12\mathrm{\:as\:}3\cdot \:4$

$\mathrm{Rewrite\:}9\mathrm{\:as\:}3\cdot \:3$

$3\cdot \:3g^3+3\cdot \:4$

$\mathrm{Factor\:out\:common\:term\:}3$

$3\left(3g^3+4\right)$

So,

$9g^3+12=\:3\left(3g^3+4\right)$

Therefore, the statement $9g^3+12=\:3\left(3g^3+4\right)$ is CORRECT.

<h2> Question # 2</h2><h2>Which expressions are completely factored?</h2>

<u>Solving first expression</u>

Considering the expression

$30a^6-24a^2$

$30a^6-24a^2$

$30a^4a^2-24a^2$

$\mathrm{Rewrite\:}24\mathrm{\:as\:}6\cdot \:4$

$\mathrm{Rewrite\:}30\mathrm{\:as\:}6\cdot \:5$

$6\cdot \:5a^4a^2-6\cdot \:4a^2$

$\mathrm{Factor\:out\:common\:term\:}3a^2$

$3a^2\left(10a^4-8\right)$

Thus, the expression $30a^6-24a^2=3a^2\left(10a^4-8\right)\:$ is completely factored.

<u>Solving second expression</u>

Considering the expression

$12a^3-8a$

$12a^3-8a$

$\mathrm{Apply\:exponent\:rule}:\quad \:a^{b+c}=a^ba^c$

$a^3=a^2a$

So,

$12a^2a-8a$

$\mathrm{Rewrite\:}8\mathrm{\:as\:}4\cdot \:2$

$\mathrm{Rewrite\:}12\mathrm{\:as\:}4\cdot \:3$

$4\cdot \:3a^2a-4\cdot \:2a$

$\mathrm{Factor\:out\:common\:term\:}4$

$4\left(3a^3-2a\right)$

Thus, the expression $12a^3-8a=\:4\left(3a^3-2a\right)$ is completely factored.

<u>Solving third expression</u>

$16a^5-20a^3\:\:\:\:\:\:\:\:\:\:\:\:$

$16a^5-20a^3$

$\mathrm{Apply\:exponent\:rule}:\quad \:a^{b+c}=a^ba^c$

$a^5=a^2a^3$

So,

$16a^2a^3-20a^3$

$\mathrm{Rewrite\:}20\mathrm{\:as\:}4\cdot \:5$

$\mathrm{Rewrite\:}16\mathrm{\:as\:}4\cdot \:4$

$4\cdot \:4a^2a^3-4\cdot \:5a^3$

$\mathrm{Factor\:out\:common\:term\:}4a^3$

$4a^3\left(4a^2-5\right)$

Thus, the expression $16a^5-20a^3\:=4a^3\left(4a^2-5\right)$ is completely factored.

<u>Solving fourth expression</u>

$24a^4+18$

$24a^4+18$

$\mathrm{Rewrite\:}18\mathrm{\:as\:}6\cdot \:3$

$\mathrm{Rewrite\:}24\mathrm{\:as\:}6\cdot \:4$

$6\cdot \:4a^4+6\cdot \:3$

$\mathrm{Factor\:out\:common\:term\:}6$

$6\left(4a^4+3\right)$

Thus, the expression $24a^4+18=6\left(4a^4+3\right)$ is completely factored.

Keywords: expression, factoring

Learn more about expression factoring from brainly.com/question/14051207

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8 0

3 years ago

Let event A= the place is a city let event B= the place in north America which outcomes are in A or B?

fredd [130]

Answer:

Step-by-step explanation:

Given two events which are

A = the event that the place is a city

and

B = the event that the place is in north America.

We have to find outcomes which are either A or B

This can be denoted as event AUB which implies A or B or both

The event A or B is defined by the place is either a city or it is in north America or a place in north America.

Hence AUB = the place is a city or place is in North America

6 0

3 years ago

Read 2 more answers

How many degrees are in an angle that turns through 3/4 of a circle?

Zepler [3.9K]

The answer is 270 degrees.

Since a circle has a full degree measure of 360 degrees, we can multiply 360 by 3/4 to result in 270 degrees.

Hope this helps.

8 0

3 years ago

Age at BBM 2019 Question Group 'A Brief answer questions; 1. Integrate the following: S(ax? + 2bx + c)3/2(2xa + 2b) dx 2.

worty [1.4K]

If the integral is

$\displaystyle \int (ax^2+2bx+c)^{3/2}(2ax+2b)\,\mathrm dx$