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antiseptic1488 [7]
4 years ago
5

Name a point that is 2 away from (-1,5).

Mathematics
1 answer:
pashok25 [27]4 years ago
3 0

Answer:[7,2]

Step-by-step explanation:

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A triangle has side lengths of 3 meters, 3 meters, and 3 meters. What kind of triangle is it?
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Answer:

I am pretty sure its an equilateral triangle

Step-by-step explanation:

im so sorry if its wrong .

7 0
3 years ago
If a the base on a logarithum is blank is it 1,0,or 10
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0 is the answer of to question

4 0
3 years ago
students collected 600 cans for the canned food drive.That was 80% of their goal.How many more cans do they need to reach their
algol13

Answer:

C = 75.

Step-by-step explanation:

Set up our eq:

600 x

___ = ___

80 100

Cross multiply:

6 x 10 = 60 (add two zero's) = 6000

now, 80 * x. 80x

Now we divide:

6000

_____

80

C = 75

6 0
3 years ago
Oliver interviewed 30% of the 9th grade class and 70% of the 10th grade class at his school. Jenny interviewed 75% of the 9th gr
mixer [17]

Answer:

A . 36

Step-by-step explanation:

We are given a total of 176 interviewed by Oliver and a total of 140 interviewed by Jenny. To find how many more 10th graders than 9th graders were interviewed, subtract the totals given

176 - 140 = 36

This is how we came to the answer:

We are given 70% of the 10th-grade and 30% of the 9th-grade with a total of 176 for Oliver.

While we're given 75% of the 9th-grade class and 25% of the 10th-grade with a total of 140 interviewed by Jenny

<u>Oliver's Interviewees</u>

  • 10-graders

Firstly, let's find what the number of 9th-graders was interviewed by Oliver; find the percentage of the 9th-graders by the total;

70% of 176 =

\frac{70}{100} * \frac{176}{1}

Cross multiply

123.2 were 10-graders interviewed by Oliver

  • 9th-graders

Now, to find the number of 9th-graders was interviewed by Oliver; find the percentage of the 9th-graders by the total;

30% of 176 =

\frac{30}{100} * \frac{176}{1}

Cross multiply

52.8 were 9th-graders interviewed by Oliver

<u>Jenny's Interviewees</u>

  • 9th-graders

Firstly, let's find what the number of 9th-graders was interviewed by Jenney; find the percentage of the 9th-graders by the total;

75% of 140 =

\frac{75}{100} * \frac{140}{1}

Cross multiply

105 students were 9th-graders interviewed by Jenney.

  • 10th-graders

Now, to find the number of 10th-graders was interviewed by Jenney; find the percentage of the 10th-graders by the total;

25% of 140 =

\frac{25}{100} * \frac{140}{1}

Cross multiply

35 students were 10th-graders interviewed by Jenney.

<u />

<u>Total calculation</u>

Use the results and sum them up by 9th-grade plus 9th-grade and 10th-grade plus 10-grade. Then subtract the amount gotten from 9th-grade away from the amount gotten from 10th-grade;

Oliver's 9th-grade = 52.8

Jenny's 9th-grade = 105

105 + 52.8 = 157.8

Oliver's 10th-grade = 123.2

Jenny's 10th-grade = 35

123.2 + 35 = 158.2

Total calculation: 158. 2 - 157.8 = 0.4

<h2>Therefore, there are 36 more 10th than 9th.</h2>

<u />

<h3><u>Extra Info:</u></h3>

<u>Oliver's Interviewees Percentage</u>

Since we are given 30% of the 9th-grade class and 70% of the 10th-grade class, first, let's add the percentages. To do so, set it up as a fraction;

30% = \frac{30}{100} while, 70% = \frac{70}{100}

Now solve it;

\frac{30}{100} + \frac{70}{100}

Simplify; cancel bottom zero's;

\frac{30}{1} + \frac{70}{1}

Add the remaining numerators;

30 + 70 = 100

Which is 100%

<u>Jenny's Interviewees Percentage</u>

Since we're given 75% of the 9th-grade class and 25% of the 10th-grade, it will end up the same answer. I'll show you how; first, let's add the percentages. To do so, set it up as a fraction;

25% = \frac{25}{100} and, 75% = \frac{75}{100}

Now solve it;

\frac{25}{100} + \frac{75}{100}

Simplify; cancel bottom zero's

\frac{25}{1} + \frac{75}{1}

Add the remaining numerators;

25 + 75 = 100

Meaning 100%

5 0
3 years ago
Abc is a triangle, ac=19cm bc=17cm and angle bac=60 degrees. calculate the size of angle abc? answer
Sphinxa [80]
\frac{|AC|}{sin\angle ABC}=\frac{|BC|}{sin\angle BAC}\\\\|AC|=19cm;\ |BC|=17cm;\ |\angle BAC|=60^o;\ |\angle ABC|=x\\\\\frac{19}{sinx}=\frac{17}{sin60^o}\\\\cross\ multiply\\\\17sinx=19sin60^O\\\\17sinx=19\cdot\frac{\sqrt3}{3}\ \ \ \ /:17\\\\sinx=\frac{19\sqrt3}{51}\\\\sinx\approx0.6453\Rightarrow x\approx40^o11'\\\\Answer:|\angle ABC|\approx40^o11'
3 0
3 years ago
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