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3 years ago

I need too know what 2838x13 is right now or else ill flunk my test

Mathematics

2 answers:

alukav5142 [94]3 years ago

4 0

Answer:

36894

Step-by-step explanation:

hope it helps

sattari [20]3 years ago

3 0

Answer:

36,894

Step-by-step explanation:

Multiply by 13

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What percentage is equal to 2/5?

nikdorinn [45]

40%

Because, 2/5 equals 4/10. 10x10=100, so 4x10=40, so 40%.

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2 years ago

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The school sold 200 tickets to play. Student tickets are $4 and adult tickets are $6. If they made a total $930 how many of each

r-ruslan [8.4K]

$200 + \div 4 =$
$50$
$200 \div 6 =$
$33 \infty$
There was 50 student tickets sold for the play.
There was 33 infinity adult tickets sold for the play

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3 years ago

On a map, the distance between two historical markers is 4.5 inches, and 2.5 inches represents 10 miles.

Morgarella [4.7K]

Answer:

The actual distance between the historical markers is 18 miles.

Step-by-step explanation:

Given : On a map, the distance between two historical markers is 4.5 inches, and 2.5 inches represents 10 miles.

To find : What is the actual distance between the historical markers?

Solution :

We have given that,

2.5 inches represents 10 miles.

i.e, 1 inch represent $\frac{10}{2.5}$ miles.

1 inch = 4 miles.

Now, The distance between two historical markers is 4.5 inches.

The actual distance between the historical markers in 4.5 inch is

$4.5 \text{ inch}=4.5\times 4\text{ miles}$

$4.5 \text{ inch}=18\text{ miles}$

Therefore, The actual distance between the historical markers is 18 miles.

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3 years ago

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Determine the zeroes of the polynomial

Anna71 [15]

Answer:

3,7/6

Step-by-step explanation:

$(\sqrt{x^2-4x+3} )+(\sqrt{x^{2} -9} )-(\sqrt{4x^2-14x+6} )\\=(\sqrt{x^2-x-3x+3} )+(\sqrt{(x^2-3^2})-(\sqrt{4x^2-2x-12x+6})\\ =(\sqrt{x(x-1)-3(x-1)} )+\sqrt{(x+3)(x-3)}-\sqrt{2x(2x-1)-6(2x-1)} \\=\sqrt{(x-1)(x-3)}+\sqrt{(x+3)(x-3)} -\sqrt{2(2x-1)(x-3)} \\=\sqrt{x-3} (\sqrt{x-1} +\sqrt{x+3} -\sqrt{2(2x-1)} )\\$

$\sqrt{x-3} =0~gives~x=3\\or~\sqrt{x-1} +\sqrt{x+3} -\sqrt{2(2x-1)} =0\\or~ \sqrt{x-1} +\sqrt{x+3} =\sqrt{2(2x-1)} \\squaring\\x-1+x+3+2\sqrt{x-1} \sqrt{x+3} =2(2x-1)\\2x+2+2\sqrt{(x-1)(x+3)} =4x-2\\2\sqrt{x^2-x+3x-3} =2x-4\\\sqrt{x^2+2x-3} =x-2\\again ~squaring\\x^2+2x-3=x^2-4x+4\\\\2x+4x=4+3\\6x=7\\x=\frac{7}{6}$

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2 years ago

Please help with both questions AND SHOW ALL WORK please thank you. :)

Svet_ta [14]

Question 1

angle 1=angle 3
angle 3=37
therefore angle 1=37

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3 years ago

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