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Ber [7]
3 years ago
5

Perform the indicated operation (x-2)(x-11))​

Mathematics
2 answers:
lawyer [7]3 years ago
6 0

Answer:

x^2 - 13x + 22

Step-by-step explanation:

( x - 2 ) ( x - 11)

Start off with the first variable which is the “x”

Then multiply it by the variables in the other binomial

x * x = x^2

x * - 11 = -11x

Then take the -2 and do the same thing

-2 * x = -2x

-2 * -11 = 22

Now the the formula that you would have is: x^2 - 11x - 2x + 22

Simplify and your answer is:

x^2 - 13x + 22

Juli2301 [7.4K]3 years ago
3 0

Answer:

x^2 - 22

Step-by-step explanation:

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Paisley is going to invest $28,000 and leave it in an account for 14 years. Assuming the interest is compounded monthly, what in
abruzzese [7]

Answer:

r=7

Step-by-step explanation:

74000=28000(1+r/12)^12(14)

divide 74000 by 28000, and cancel out 28000.

1.00580=1+r/12

(100) .06962=r

r = 6.96198

the 6 after the 9 rounds 9 to 0, which rounds the 6 to 7

r=7

5 0
3 years ago
Order the following values from least to greatest.
maks197457 [2]

Answer:

-6 1/2

-1

0

9

these are in order from least to greatest

3 0
3 years ago
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Write a rule for the function​
cestrela7 [59]

Answer:-18

Step-by-step explanation: i know evertyhing

3 0
3 years ago
What is the best approximation of the projection of (5,-1) onto (2,6)?
Hatshy [7]

Answer:

Hence, the scalar projection of \vec a onto \vec b= \frac{\sqrt{10} }{5}, and  the vector projection of \vec a onto \vec b = \frac{1}{5} \hat i+\frac{3}{5} \hat j.

Step-by-step explanation:

We have given two points  (5, -1) and (2, 6).

Let,     \vec a=5\hat {i}-\hat {j}  and  \vec b= 2\hat {i}+6\hat{j} .

and we have calculate the projection of \vec a onto \vec b.

Now,

For the calculation of projection, first we need to calculate the dot product of  \vec a  and \vec b.

\vec a.\vec b=(5\hat {i}-\hat{j}).(2\hat{i}+6\hat{j})

     =10-6

     =4

then, we have to calculate the magnitude of \vec b.

   \mid {\vec {b}}\mid = \sqrt{2^{2}+6^{2}  } = \sqrt{40} = 2\sqrt{10}.

Now, the scalar projection of \vec a onto \vec b = \frac{\vec a.\vec b}{\mid b\mid}

                                                                 = \frac{4}{2\sqrt{10} }\frac{2}{\sqrt{10} } \times\frac{\sqrt{10} }{\sqrt{10} } =\frac{2\sqrt{10} }{10} = \frac{\sqrt{10} }{5}

and the vector projection of \vec a onto \vec b = \frac{\vec a. \vec b}{\mid\vec b \mid^{2} } . \vec b

                                                               = \frac{4}{40} . (2\hat i+ 6\hat j)

                                                                = \frac{1}{5} \hat i+\frac{3}{5} \hat j

Hence, the scalar projection of \vec a onto \vec b= \frac{\sqrt{10} }{5}, and  the vector projection of \vec a onto \vec b = \frac{1}{5} \hat i+\frac{3}{5} \hat j.

                                                               

6 0
3 years ago
R to the 2nd power -6r-8=0
Natasha2012 [34]
R² - 6r - 8 = 0
r² = 7.12
r = -1.12
3 0
3 years ago
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