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2 years ago

What is the product of (x − 3) and (x + 7)?

Mathematics

2 answers:

tatuchka [14]2 years ago

6 0

Answer: x - 21

Step-by-step explanation: If you put a one in front of each part of the equation, you would get 1(x - 3) * 1(x + 7). From there you would get 1x - 3 * 1x + 7. Then you multiply -3 * 7, which would equal -21. Therefore, the answer is "x minus 21".

If this answer is correct, put this answer as the "Brainliest" so other people who refer to this question see this answer!

NeX [460]2 years ago

3 0

Answer:

The product of (x-3) and (x+7) is x² + 4x - 21

Step-by-step explanation:

The product of (x-3) and (x+7) is x² + 4x - 21

The product of (x - 3) and (x + 7) can be found as follows:

(x - 3)(x + 7)

open the bracket by multiplying

Therefore,

x(x) +x(7) - 3(x) - 3(7)

x² + 7x - 3x - 21

combine like terms

x² + 4x - 21

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Least to greatest<br> .<br> .<br> .<br> Please and thank you <3

gizmo_the_mogwai [7]

2.2, 5/2, square root of 10, 7/2, -4.8

5 0

2 years ago

Help please due tonight <br><br>50 points + brainliest

Leviafan [203]

Answer:

7. -37/36 or -1 1/36

8. -3/10

9. 1/24

10. -118/21 or -5 13/21

11. 97/12 or 8 1/12

12. 49/8 or 6 1/8

Step-by-step explanation:

To subtract fractions, they must have the same denominator. The common denominator will be the LCM (lowest common multiple) of the two denominators. Then, combine the fractions and simplify.

7. $-\frac{1}{4}-\frac{7}{9}$ Find a common denominator

$=-\frac{1*9}{4*9}-\frac{7*4}{9*4}$

$=-\frac{9}{36}-\frac{28}{36}$

$=\frac{-9-28}{36}$ Combine the fractions

$=\frac{-37}{36}$ Answer. Already in lowest terms.

$=-1\frac{1}{36}$ Keep a negative number. 37 goes into 36 one time, leaving 1 remainder (becomes numerator)

8. $\frac{3}{5}-\frac{9}{10}$ Find a common denominator

$=\frac{3*2}{5*2}-\frac{9}{10}$

$=\frac{6}{10}-\frac{9}{10}$

$=\frac{6-9}{10}$ Combine the fractions

$=\frac{-3}{10}$ Answer in simplest form.

9. $\frac{5}{8}-\frac{7}{12}$ Find common denominator

$=\frac{5*3}{8*3}-\frac{7*2}{12*2}$

$=\frac{15}{24}-\frac{14}{24}$

$=\frac{15-14}{24}$ Combine the fractions

$=\frac{1}{24}$ Answer in simplest form

10. $-1\frac{1}{3}-4\frac{2}{7}$ Translate to improper fraction

$=-\frac{4}{3}-\frac{30}{7}$ Multiplied each whole number with base, added to numerator

$=-\frac{28}{21}-\frac{90}{21}$ Found common denominator

$=\frac{-28-90}{21}$ Combined fractions

$=\frac{-118}{21}$ Answer in improper form

$=-5\frac{13}{21}$ Answer as a mixed number

11. $5\frac{5}{6}-(-2\frac{1}{4})$ Convert to improper fraction

$\frac{35}{6}-(-\frac{9}{4})$ Two minuses make a plus

$=\frac{35}{6}+\frac{9}{4}$ Find common denominator

$=\frac{35*2}{6*2}+\frac{9*3}{4*3}$

$=\frac{70}{12}+\frac{27}{12}$

$=\frac{70+27}{12}$ Combined fractions

$=\frac{97}{12}$ Answer as improper fraction

$=8\frac{1}{12}$ Answer as mixed fraction

12. $12\frac{1}{2}-6\frac{3}{8}$ Convert to improper fraction

$\frac{25}{2}-\frac{51}{8}$

$\frac{25*4}{2*4}-\frac{51}{8}$ Find common denominator

$\frac{100}{8}-\frac{51}{8}$

$\frac{100-51}{8}$ Combined fractions

$\frac{49}{8}$ Answer as improper fraction

$8\frac{1}{8}$ Answer as mixed number

5 0

3 years ago

( 0 0

ch4aika [34]

<h2>D. If subjects knew they were receiving an active treatment, researchers would not have known if any improvement was due to the new medication or to the expectation of <u>feeling </u>better. If the researchers knew which subjects received which treatments, they might have treated one group of subjects differently from the other group.</h2><h2 />

For your question... the answer would be D

7 0

3 years ago

Property for 1• (a + 3) = a +3

mina [271]

Answer: Infinitely Many Solutions

Step-by-step explanation is below.

5 0

3 years ago

Determine as a linear relation in x, y, z the plane given by the vector function F(u, v) = a + u b + v c when a = 2 i − 2 j + k,

Ostrovityanka [42]

Answer:

2x - y - 3z = 0

Step-by-step explanation:

Since the set

{i, j} = {(1,0), (0,1)}

is a base in $\mathbb{R}^2$

and F is linear, then

would be a base of the plane generated by F.

F(1,0) = a+b = (2i-2j+k)+(i+2j+k) = 3i+2k

F(0,1) = a+c = (2i-2j+k)+(2i+j+2k) = 4i-j+3k

Now, we just have to find the equation of the plane that contains the vectors 3i+2k and 4i-j+3k

We need a normal vector which is the cross product of 3i+2k and 4i-j+3k

(3i+2k)X(4i-j+3k) = 2i-j-3k

The equation of the plane whose normal vector is 2i-j-3k and contains the point (3,0,2) (the end of the vector F(1,0)) is given by

2(x-3) -1(y-0) -3(z-2) = 0

or what is the same

2x - y - 3z = 0

3 0

4 years ago