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2 years ago

What are the factors of -30 and the sum of 4?

2 answers:

7 0

Answer: 30 = 1 x 30, 2 x 15, 3 x 10, or 5 x 6. Factors of 30: 1, 2, 3, 5, 6, 10, 15, 30. Prime facto...... 30 = 2 x 3 x 5.

sorry if im wrong and have a wonderful day!!!!!

Alex787 [66]2 years ago

4 0

Answer:

Answer: 30 = 1 x 30, 2 x 15, 3 x 10, or 5 x 6. Factors of 30: 1, 2, 3, 5, 6, 10, 15, 30. Prime factorization: 30 = 2 x 3 x 5.

Step-by-step explanation:

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What is root3(root3 - 1)

Alenkasestr [34]

Answer:

3 - root3 or 3 - √3

Step-by-step explanation:

root3(root3 - 1) =
√3(√3 - 1) =
√3(√3) - √3(1) =
√3² - √3 =
3 - √3
3 - root3

4 0

3 years ago

Read 2 more answers

Solve. x² – 8x - 12 = 36

professor190 [17]

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3 years ago

Find the value of each expression.

Damm [24]

I'm not sure

Step-by-step explanation:

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2 years ago

At the start of the month, the value of an investment was $73.42. By the end of the month, the value of the investment changed b

Aleonysh [2.5K]

Answer:

0.8%.

Step-by-step explanation:

Given:

At the start of the month, the value of an investment was $73.42.

By the end of the month, the value of the investment changed by a loss of $13.53.

Question asked:

What was the value, in dollars, of the investment at the end of the month?

What was the percent loss?

Solution:

At the start of the month, the value of an investment = $73.42.

Loss = $13.53.

The value, of the investment at the end of the month = The value of an investment at the start of the month - loss

The value, of the investment at the end of the month = $73.42 - $13.53

= $59.89

Thus, the value, of the investment at the end of the month is $59.89.

Now, we will find percent loss:-

$Percent\ loss = \frac{Loss}{value\ of \ investment\ at \ start}$

$=\frac{59.89}{73.42} \\=0.81\%$

Thus, percent loss on investment by the end of the month is 0.8%.

7 0

3 years ago

What is the distance between M(9, −5) and N(−11, 10)?

lesya692 [45]

Answer:

$\boxed{ \bold{ \huge{ \boxed{ \sf{25 \: units}}}}}$

Step-by-step explanation:

Let M ( 9 , -5 ) be ( x₁ , y₁ ) and N ( - 11 , 10 ) be ( x₂ , y₂ )

Finding the distance between these points

$\boxed{ \sf{distance = \sqrt{ {(x2 - x1)}^{2} + {(y2 - y1)}^{2} } }}$

$\longrightarrow{ \sf{ \sqrt{ {( - 11 - 9)}^{2} + {(10 - ( - 5))}^{2} } }}$

$\longrightarrow{ \sf{ \sqrt{ {( - 20)}^{2} + {(10 + 5)}^{2} } }}$

$\longrightarrow{ \sf{ \sqrt{ {( - 20)}^{2} + {(15)}^{2} } }}$

$\longrightarrow{ \sf{ \sqrt{400 + 225}}}$

$\longrightarrow{ \sf{ \sqrt{625}}}$

$\longrightarrow{ \sf{ \sqrt{ {(25)}^{2} } }}$

$\longrightarrow{ \sf{25 \: units}}$

Hope I helped!

Best regards! :D

7 0

3 years ago

What are the factors of -30 and the sum of 4?​

What are the factors of -30 and the sum of 4?