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vlada-n [284]
3 years ago
10

Solve 9/10=z/8 . Round to the nearest tenth.

Mathematics
1 answer:
kompoz [17]3 years ago
4 0
\boxed {Question}
\frac{9}{10} =  \frac{z}{8}

\boxed {Cross \  multiply}
10z = 8 \times 9

\boxed {Evaluate}
10z = 72
z = 72 \div 10
z = 7.2

\Longrightarrow \bf \ Answer \ : \ z = 7.2

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Use the numbered angles in the diagram to answer the question.
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2 years ago
Which expression is equivalent to square root 55x^7y6/11x^11y^8
Novay_Z [31]

Answer:

\large\boxed{\sqrt{\dfrac{55x^7y^6}{11x^{11}y^8}}=\dfrac{\sqrt5}{x^2y}}

Step-by-step explanation:

\sqrt{\dfrac{55x^7y^6}{11x^{11}y^8}}=\sqrt{\dfrac{55}{11}\cdot\dfrac{x^7}{x^{11}}\cdot\dfrac{y^6}{y^8}}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\=\sqrt{5x^{7-11}y^{6-8}}=\sqrt{5x^{-4}y^{-2}}\qquad\text{use}\ \sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\\\\=\sqrt5\cdot\sqrt{x^{-4}}\cdot\sqrt{y^{-2}}=\sqrt5\cdot\sqrt{x^{(-2)(2)}}\cdot\sqrt{y^{(-1)(2)}}\qquad\text{use}\ (a^n)^m=a^{nm}\\\\=\sqrt5\cdot\sqrt{(x^{-2})^2}\cdot\sqrt{(y^{-1})^2}\qquad\text{use}\ \sqrt{a^2}=a

=\sqrt5\cdot x^{-2}\cdot y^{-1}\qquad\text{use}\ a^{-n}=\dfrac{1}{a^n}\to a^{-1}=\dfrac{1}{a}\\\\=\sqrt5\cdot\dfrac{1}{x^2}\cdot\dfrac{1}{y}=\dfrac{\sqrt5}{x^2y}

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3 years ago
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8 0
3 years ago
PLEASE HELP
Vladimir [108]

Answer:

5.5 inches

Step-by-step explanation:

1) the formula to find the area of a square is l^2= area and l=length

in this case we already know the area but not the length

so we have to find the square root of 30 as the opposite operation needs to be used to find the missing number

so

√30= 5.47

5.47 rounded= 5.5

the length of one side is 5.5 in

5 0
3 years ago
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