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Veseljchak [2.6K]
3 years ago
7

I need help please. and explanation will help for tomorrow in a test:)

Mathematics
1 answer:
Elenna [48]3 years ago
3 0
Answer : x^7

When you’re multiplying variables like the one in the problem above, instead of actually multiplying, you add. You only multiply in cases where there are coefficients.

Example : z^4 * z^2

Instead of multiplying, you would add the exponents together like so, z^6.
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Which expression gives the distance between the points
Alla [95]

Answer:

C

Step-by-step explanation:

The formula for finding the distance between 2 points (x1, y1) and (x2, y2) is

d = √[(x2 - x1)² + (y2 - y1)²]

Here (x2, y2) = (2, 3)  and (x1, y1) = (4, -3)

Plugging them gives us

d = √[(2 - 4)² + (3 - (-3))²]

  d = √[(2 - 4)² + (3 + 3)²]

8 0
3 years ago
How do you multiply fractions and express product in simplestform.
vagabundo [1.1K]
You just multiply top times top and bottom times bottom and then find a number you can divide into the number and you keep doing that until you can't simplify any more.
3 0
3 years ago
WORTH BRAINLIST!!!<br> What is the value of the expression 2 + 3^2 ⋅ (3 − 1)? 20 22 23 32
Wittaler [7]

Answer:

You have to use PEMDAS (Parentheses, Exponents, Multiplication, Division, Addition, Subtraction), and solve in that order. Your question is:

2 + 32 ⋅ (3 − 1)

First, solve the Parentheses part of the question. 3-1 is 2. Your question is now:

2+32*2

Now, solve the multiplication part of the question. 32*2 is 64. Your question is now:

2+64

This is now an easy problem to solve.

The answer is 66.

Hope this helps! :)

3 0
3 years ago
(5x-7)+(6+33)<br> solve the equation
Vinil7 [7]

Answer: 5x + 32

Step-by-step explanation:

4 0
3 years ago
7 to the negative 3rd porwer times 7 to the negative 2nd porwer times 2 the 0 power divided by 7 to negative 1st power times 7 t
Jobisdone [24]

oof

remember some rules

(a/b)(c/d)=(ac)/(bd)

\frac{x^a}{x^b}=x^{a-b}

x^0=1

(x^a)(x^b)=x^{a+b}

x^{-a}=\frac{1}{x^a}


\frac{(7^{-3})(7^{-2})(2^0)}{(7^{-1})(7^{-4})(2^3)}=

\frac{(7^{-3-2})(2^0)}{(7^{-1-4})(2^3)}=

\frac{(7^{-5})(2^0)}{(7^{-5})(2^3)}=

(\frac{7^{-5}}{7^{-5}})(\frac{2^0}{2^3})=

(7^{-5-(-5)})(2^{0-3})=

(7^0)(2^{-3})=

\frac{1}{2^3}=

\frac{1}{8}

7 0
3 years ago
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