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3 years ago

13

What is the GCF of

dle" class="latex-formula">, - $10x^{7}$ , $2x^{8}$ ?? please help

1 answer:

jolli1 [7]3 years ago

5 0

Answer:

$2x^6$

Step-by-step explanation:

First, we find the greatest common factor 50, 10 and 2.

It is 2.

Then, we find the greatest common factor of $x^{6},x^{7}$ and $x^{8}$ which is $x^{6}$ .

Thus, the greatest common factor is $2x^6$

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5. Lauren is saving for gymnastics camp, Camp costs $225 to attend. She has 40 percent of the money saved. How much money has sh

barxatty [35]

To solve this problem you have to convert the percentage to a decimal which would
be .40.

40% =40/100=.40

After doing so all you need to do is divide the amount needed by the percentage which gives you the amount she has saved.

225/.40= $90

Lauren has $90 saved

3 0

3 years ago

−30=5(x+1)minus, 30, equals, 5, left parenthesis, x, plus, 1, right parenthesis<br> x =x=x, equals

ivann1987 [24]

Answer:

x = -7

Step-by-step explanation:

−30=5(x+1)

Divide each side by 5

-30/5=5/5(x+1)

-6 =x+1

Subtract 1 from each side

-6-1 = x+1-1

-7 =x

8 0

3 years ago

Help please? Questions 1-15 i really need it♡

Novosadov [1.4K]

Answer:

here:

1. -9

2. -28

3. 54

4. -100

5. -60

6. -0

7. 49

8. -135

9. -96

10. 300

11. -153

12. -192

THIS TOOK LONG BECAUSE I DID A LONG ONE BUT THEN I MADE IT SHORTER. :D

7 0

2 years ago

A tablet is on sale for 30% off. the regular price is $185. how much money will you save?

Burka [1]

Answer:

185×30÷100=55.5

HERES YOUR answer

8 0

2 years ago

A study found that a 95% confidence interval for the mean μ of a particular population was computed from a random sample of 1200

kotykmax [81]

Answer:

95% confidence interval for the mean μ is (6,14)

The Population mean μ lies between ( 6, 14 )

Step-by-step explanation:

<u><em>Explanation</em></u>:-

Given random sample 'n' = 1200

95% confidence interval for the mean μ is determined by

$(x^{-} -Z_{\alpha } \frac{S.D}{\sqrt{n} } , x^{-} +Z_{\alpha } \frac{S.D}{\sqrt{n} } )$

Level of significance = 95% 0r 0.05

Z₀.₀₅ = 1.96

$(x^{-} -Z_{\alpha } \frac{S.D}{\sqrt{n} } , x^{-} +Z_{\alpha } \frac{S.D}{\sqrt{n} } )$ = 10 ± 4

Mean of the small sample = 10

95% of confidence intervals are

( 10 ±4 )

( 10 -4 , 10+4)

( 6 , 14 )

95% confidence interval for the mean μ lies between ( 6, 14 )

8 0

2 years ago