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3 years ago

50 POINTS ANSWER FAST PLEASE

Mathematics

2 answers:

Sholpan [36]3 years ago

6 0

Answer:

Question 2.) 16

144÷9=16

Question 3.) 16

16÷12=1.33... and all the others equal 1.33... as well

Question 4.) 20

20÷16=1.25 and all tho others equal 1.25 as well

Question 5.)12

8÷12=0.66... and all the others equal 0.66.... as well

I hope this is good enough for you:

Lelu [443]3 years ago

4 0

Answer:

The first one is 16

The second one is 14

*not sure about the second one*

Step-by-step explanation:

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Suppose you take out a loan for $9,000, at 12% ordinary interest. If the amount of interest is $762.00, what is the time period?

DaniilM [7]

For the answer to the question above,
Let's figure out the formula first
FV = PV(1+i/n)^(nt)
9762= 9000(1+0.12/12)^(12x)
9762= 9000(1.01)^12x
9762/9000=1.01^12x
1.08467=1.01^12x
ln1.08467/12 x ln1.01=0.681 years (8.17 months) same as my calculator
So,
0.681x360=
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3 years ago

best answer will be awarded the brainliest and 5 stars please evaluate 6/2 and remember to show your work

Svet_ta [14]

6/2 is the number 6 being divided in half by 2; to figure this out just count your 2 table factors! Shown work:

2 x 2 = 4
2 x 3 = 6!

So 6 divided by 2 (6/2) is just the oppoposite of 2 x 3 = 6

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3 years ago

Please help 10 point question!

Vsevolod [243]

Answer:

not sure about 1

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3 years ago

To simplify the following expression, which operation would you complete second?

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3 years ago

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Fittoniya [83]

<h3>Answer:</h3>

$\boxed{\pink{\sf \leadsto Yes \ there \ is \ a \ solution \ of \ the \ given \ inequality .}}$

<h3>Step-by-step explanation:</h3>

A inequality is given to us and we need to convert it into standard form and see whether if it has a solution . So let's solve the inequality.

The inequality given to us is :-

$\bf\implies |2y + 3 | - 1 \leq 0 \\\\\bf\implies |2y+3|\leq 1 \\\\\bf\implies (|2y+3|)^2 \leq 1^2 \\\\\bf\implies (2y+3)^2 \leq 1 \\\\\bf\implies (2y)^2+3^2+2(2y)(3) \leq 1 \\\\\bf\implies 4y^2+9+12y - 1 \leq 0 \\\\\bf\implies 4y^2+12y+8 \leq 0 \\\\\bf\implies 4( y^2 + 3y + 2 ) \leq 0 \\\\\bf\implies y^2+3y +2 \leq 0 \:\:\bigg\lgroup \purple{\bf Standard \ form \ of \ inequality }\bigg\rgroup \\\\\bf\implies y^2y+2y+y+2 \leq 0 \\\\\bf\implies y(y+2)+1(y+2)\leq 0 \\\\\bf\implies ( y+2)(y+1)\leq 0 \\\\\bf\implies \boxed{\red{\bf y \leq (-2) , (-1) }}$

Let's plot a graph to see its interval . Graph attached in attachment .

Now we can see that the Interval notation of would be ,

$\boxed{\boxed{\orange \tt \purple{\leadsto}y \in [-2,-1] }}$

<h3>Hence the standard form of inequality is y²+3y +2 ≤ 0 and the Solution set of the inequality is [ -2 , -1 ] .</h3>

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3 years ago