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polet [3.4K]
2 years ago
13

Which do you think is quadrilaterals

Mathematics
2 answers:
deff fn [24]2 years ago
8 0

Answer:

2nd and last

Step-by-step explanation:

the ones with 4 sides are quadrilaterals

8090 [49]2 years ago
7 0

Answer: Only the second image

Step-by-step explanation:

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How did you compute sums of dollar amounts that were not whole numbers? For example, how did you compute the sum of$5.89 and$1.4
nignag [31]

Answer:

$7.34

Step-by-step explanation:

To compute sum of dollars that are not whole numbers. Using the sum of$5.89 and$1.45 as an illustration :

$5.89 + $1.45

Taking the whole numbers first:

$5 + $1 = $6

Take the sum of the decimals :

$0.89 + $0.45 = $1.34

Sum initial whole + whole of sum of decimal

$6 + $1 = $7

Remaining decimal : $1.34 - $1 = $0.34

$7 + $0.34 = $7.34

4 0
3 years ago
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Calculate the simple interest on each amount and the total amount at the end.
Butoxors [25]
Attached is a picture of my answers.

5 0
2 years ago
5 ( 3 squared x) + 9 ( 3 squared x)
yulyashka [42]

Answer:

42x^2

Step-by-step explanation:

5(3x^{2} )+9(3x^2)\\\\=(5+9)(3x^2)\\\\=14(3x^2)\\\\=42x^2

5 0
3 years ago
The side length of a cube is represented by s as shown in the equation below.
Gekata [30.6K]

Answer:

7 and -7

Step-by-step explanation:

4 0
2 years ago
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Help please!!! I dont understand these questions<br><br><br>currently attaching photos dont delete
Katyanochek1 [597]

Answer:

  1. b/a
  2. 16a²b²
  3. n¹⁰/(16m⁶)
  4. y⁸/x¹⁰
  5. m⁷n³n/m

Step-by-step explanation:

These problems make use of three rules of exponents:

a^ba^c=a^{b+c}\\\\(a^b)^c=a^{bc}\\\\a^{-b}=\dfrac{1}{a^b} \quad\text{or} \quad a^b=\dfrac{1}{a^{-b}}

In general, you can work the problem by using these rules to compute the exponents of each of the variables (or constants), then arrange the expression so all exponents are positive. (The last problem is slightly different.)

__

1. There are no "a" variables in the numerator, and the denominator "a" has a positive exponent (1), so we can leave it alone. The exponent of "b" is the difference of numerator and denominator exponents, according to the above rules.

\dfrac{b^{-2}}{ab^{-3}}=\dfrac{b^{-2-(-3)}}{a}=\dfrac{b}{a}

__

2. 1 to any power is still 1. The outer exponent can be "distributed" to each of the terms inside parentheses, then exponents can be made positive by shifting from denominator to numerator.

\left(\dfrac{1}{4ab}\right)^{-2}=\dfrac{1}{4^{-2}a^{-2}b^{-2}}=16a^2b^2

__

3. One way to work this one is to simplify the inside of the parentheses before applying the outside exponent.

\left(\dfrac{4mn}{m^{-2}n^6}\right)^{-2}=\left(4m^{1-(-2)}n^{1-6}}\right)^{-2}=\left(4m^3n^{-5}}\right)^{-2}\\\\=4^{-2}m^{-6}n^{10}=\dfrac{n^{10}}{16m^6}

__

4. This works the same way the previous problem does.

\left(\dfrac{x^{-4}y}{x^{-9}y^5}\right)^{-2}=\left(x^{-4-(-9)}y^{1-5}\right)^{-2}=\left(x^{5}y^{-4}\right)^{-2}\\\\=x^{-10}y^{8}=\dfrac{y^8}{x^{10}}

__

5. In this problem, you're only asked to eliminate the one negative exponent. That is done by moving the factor to the numerator, changing the sign of the exponent.

\dfrac{m^7n^3}{mn^{-1}}=\dfrac{m^7n^3n}{m}

3 0
3 years ago
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