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GREYUIT [131]
2 years ago
8

What is the solution to the equation x + 7 = 63? 9 16 56 70

Mathematics
1 answer:
mars1129 [50]2 years ago
4 0
The answer to your problem is 56.
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Multiply the binomials: (2c - d) (3c + d)
slega [8]

Answer:

6c^2-cd-d^2

Step-by-step explanation:

How you get the answer is you use the process called FOIL. The F is front, so you multiply the 2c and the 3c to get 6c^2. The O is the outer, so you multiply the 2c and the d(because they are on the outer sides of the parenthesis) to get 2cd. The I is the inner, so you multiply the -d and the 3c(because they are on the inner sides of the parenthesis) to get -3cd. The L is the last, so you multiply d and -d to get -d^2. The 2cd and the -3cd are the same so you have to minus them to get -cd. So the answer is:

6c^2-cd-d^2.

3 0
3 years ago
1/3(x+4)=20 need help
sukhopar [10]


1/3(x+4)=20

1/3x+4/3=20

1/3x=56/3

x= (56/3) / (1/3)

x= 56

5 0
3 years ago
Read 2 more answers
Which statement is true?
marshall27 [118]

Answer:

B, the second answer choice

Step-by-step explanation:

Hope this helped! Please mark me as brainliest if you can!

4 0
3 years ago
The library has at least 5,000 books.Which inequality represents the number of book,b,at the library
Scilla [17]
I need answer choices

3 0
3 years ago
For the following paired data, (1) Calculate the correlation coefficient, r, and (2) at the 0.05 significance level, determine w
expeople1 [14]

Answer:

Linear correlation exists

Step-by-step explanation:

Given the data :

X : | 2 4 5 6

Y : | 6 9 8 10

Using technology to fit the data and obtain the correlation Coefficient of the regression model,

The Correlation Coefficient, r is 0.886

To test if there exists a linear correlation :

Test statistic :

T = r / √(1 - r²) / (n - 2)

n = number of observations

T = 0.886 / √(1 - 0.886²) / (4 - 2)

T = 0.866 / 0.3535845

T = 2.449

Comparing Pvalue with α

If Pvalue < α ; Reject H0

Pvalue = 0.1143

α = 0.05

Pvalue > α ; We reject the null and conclude that linear correlation exists

7 0
3 years ago
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