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Xelga [282]
3 years ago
15

When a product is raised to a power, it is equal to the product of each factor raised to that power. Show that this rule does no

t apply for a sum or difference to a power using the problem (5 – 3)2. You may indicate an exponent in your answer with ^. For example, 3x2 as 3x^2.
Mathematics
1 answer:
Airida [17]3 years ago
7 0

Answer:

In this instance you would do what is inside the parentheses first and then use the exponent. Therefore it would be 3^2 and your answer is 9.

Step-by-step explanation:

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Area of rectangle 12ft 6ft
Brilliant_brown [7]
12ft×6ft= area of 72ft
8 0
3 years ago
Help me with this tooo! It is the question that is not filled in
Kazeer [188]

you will have 8 tables so

72 chair / 8 tables = 9 chair per table

3 0
3 years ago
What is the value of y
kari74 [83]

Answer:

y= -6

Step-by-step explanation:

-2y - 5y= -7y

-7y=42

42/-7=

y= -6

5 0
2 years ago
Mason walked on Monday, Wensday, and Friday. Use these clues to find how far he walked each day. These distances were six-eights
krok68 [10]

Answer:

Monday he walked \frac{1}{4}miles

Wednesday he walked \frac{6}{8}miles

Friday he walked \frac{1}{6}miles

Step-by-step explanation:

Given Mason walked on Monday, Wednesday and Friday. These distances were six-eights mile, one-fourth mile, and one-sixth mile. He did not walk the farthest on Monday. He walked less on Friday than Monday. we have to find how far he walked each day.

distances are \frac{6}{8}, \frac{1}{4}, \frac{1}{6} that are 0.75, 0.25 and 0.17 respectively.

Now, he didn't walk farthest on Monday and also walked less on Friday than Monday.

∴ Less distance travelled is \frac{1}{6} which is on friday and then \frac{1}{4} on monday.

Rest distance which is \frac{6}{8} on wednesday.

Hence, Monday he walked \frac{1}{4}miles

Wednesday he walked \frac{6}{8}miles

Friday he walked \frac{1}{6}miles

4 0
3 years ago
Read 2 more answers
A college student is taking two courses. The probability she passes the first course is 0.67. The probability she passes the sec
andrezito [222]

Answer:

0.58 = 58% probability she passes both courses

Step-by-step explanation:

We can solve this question treating the probabilities as a Venn set.

I am going to say that:

Event A: She passes the first course.

Event B: She passes the second course.

The probability she passes the first course is 0.67.

This means that P(A) = 0.67

The probability she passes the second course is 0.7.

This means that P(B) = 0.7

The probability she passes at least one of the courses is 0.79.

This means that P(A \cup B) = 0.79

a. What is the probability she passes both courses

This is P(A \cap B).

We use the following relation:

P(A \cup B) = P(A) + P(B) - P(A \cap B)

So

P(A \cap B) = P(A) + P(B) - P(A \cup B) = 0.67 + 0.7 - 0.79 = 0.58

0.58 = 58% probability she passes both courses

5 0
3 years ago
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