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Ahat [919]
3 years ago
6

HELP ASAP Select the real-world problem that could be solved using a proportion.

Mathematics
1 answer:
sergeinik [125]3 years ago
5 0

Answer: D

Step-by-step explanation:

D can be solved using 300/6 = x/10 making it a proportion

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Simplify expression X^2+x-2/x^3-x^2+2x-2
professor190 [17]

Answer:


=\frac {x+2}{x^2+2} is the simplest form of given expression.


Step-by-step explanation:

The given question is \frac{x^2+x-2}{x^3-x^2+2x-2}


To solve the problem we have to group or split middle term and then factorise


\frac{x^2+(2x-x)-2}{(x^3-x^2)+(2x-2)}


Taking x^2 common from first two terms of denominator and 2 from next two terms


= \frac{x^2+2x-x-2}{x^2(x-1)+2(x-1)}


Now,taking x common from first two terms of numerator and -1 from next two terms and in denominator taking(x-1) common from both terms


= \frac{ x(x+2)-1(x+2)}{(x-1)(x^2+2)}



=\frac{(x+2)(x-1)}{(x-1)(x^2+2)}


Now cancel out x-1 from both numerator and denominator we get


=\frac {x+2}{x^2+2} is the required simplest form.



7 0
3 years ago
Helen's house is located on a rectangular lot that is1 1-8 miles by 9/10 mile.Estimate the distance around the lot
klemol [59]

Answer: The distance around the lot is given by

4\frac{1}{20}\ miles

Step-by-step explanation:

Since we have given that

Length of rectangular lot is given by

1\frac{1}{8}\\\\=\frac{9}{8}\ miles

Width of rectangular lot is given by

\frac{9}{10}\ mile

We need to find the distance around the lot.

As we know the formula for "Perimeter of rectangle":

Perimeter=2(Length+Width)\\\\Perimeter=2(\frac{9}{8}+\frac{9}{10})\\\\Perimeter=2(\frac{45+36}{40})\\\\Perimeter=2\times \frac{81}{40}\\\\Perimeter=\frac{81}{20}\\\\Perimeter=4\frac{1}{20}\ miles

Hence, the distance around the lot is given by

4\frac{1}{20}\ miles


3 0
3 years ago
Read 2 more answers
The 12th term in a sequence with a common difference of-9 is -106. which of the following formulas can be used to represent this
Effectus [21]

Answer:

T12=a-99=106

Step-by-step explanation:

that's the answer

6 0
2 years ago
A group of students wanted to investigate the claim that the average number of text messages sent yesterday by
Likurg_2 [28]
First we need to write the null and alternate hypothesis for this case.

Let x be the average number of text message sent. Then

Null hypothesis: x = 100
Alternate hypothesis: x > 100

The p value is 0.0853

If p value > significance level, then the null hypothesis is not rejected. If p value < significance level, then the null hypothesis is rejected.

If significance level is 10%(0.10), the p value will be less than 0.10 and we reject the null hypothesis and CAN conclude that:
The mean number of text messages sent yesterday was greater than 100.

If significance level is 5%(0.05), the p value will be greater than 0.05 and we cannot reject the null hypothesis and CANNOT conclude that:
The mean number of text messages sent yesterday was greater than 100.
7 0
3 years ago
Which are acute, obtuse, or right?
qaws [65]
5,7 and 8 are right angles, and 6 is an acute angle this is because of u look at 9 and 10 you can see a little box looking thing under the h and if u do that to all of them 5, 7 and 8 can fit the little box but 6 you can’t because it is an acute angle. Hope this helped!
8 0
3 years ago
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