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2 years ago

HELP ME, seriously no one has helped me yet

Mathematics

2 answers:

SOVA2 [1]2 years ago

6 0

Answer:

1. Lunch ($16)

2. Coffee ($4)

3. how much money did she have left?

she had ($12) left

Step-by-step explanation:

1. To get the amount of money she used on lunch you have to look at the word problem it says "she spent 1/2 of her money on lunch" her total is $32 dollars and 1/2 is .5 as a decimal and 50% as a percentage so 50% of 32 is (16) so she spent $16 dollars on lunch alone.

2. To find how much money she used on Coffee you need to read the context of the problem its asking for "how much did she use of the money that is left from her lunch" so she has $16 dollars she used from lunch so now it wants you to find 1/4 of $16 dollars or 25% of 16 and 25% of 16 is (4)

so she spent 4 dollars of what was left so the answer is 4.

3. To find how much is left add up 16 and 4 and you get 20 then subtract it by the original number which is 32 so it should look like this 32 - 20 = 12

so the money she has left is $12 dollars.

larisa [96]2 years ago

3 0

Answer: Lunch:$16 Coffee: $4 Julie had $12 left.

Step-by-step explanation:

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Find the diameter of the object

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Answer:

Step-by-step explanation:

The diameter of a circle is 2 times the radius. (the diameter is the length from an outer point in the circle, to one directly across)

In this case, the radius is 2.

So, 2 x 2 = 4.

The diameter is 4.

7 0

2 years ago

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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