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3 years ago

Are these fractions equivalent to

Mathematics

2 answers:

Rina8888 [55]3 years ago

5 0

1: 9...
2: 36 yes
3: 18
4:24
5: 27

SIZIF [17.4K]3 years ago

5 0

Answer:

4/9 is equivalent to 8/18 and 12/27

Step-by-step explanation:

multiply 4 and 9 by 2 and you will get 8/18, therefore they are equivalent

multiply 4 and 9 by 3 and you will get 12/27, therefore they are equivalent

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Which ordered pair is a solution of the equation?

liq [111]

Answer:

y=39

Step-by-step explanation:

73-2y=-5

2y=73-(-5)

2y=73+5

2y=78

y=78/2

y=39

6 0

3 years ago

Find the area. HELP PLEASE!!!!!!

kaheart [24]

Answer:

616

Step-by-step explanation:

The equation to find the area of a trapezoid is: A = ½ (b $^1$ +b²) h.

b1=43

b2=45

h=14

Plug the variables in and solve.

$A=\frac{1}{2} (43+45)14$

$A=\frac{1}{2} (88)14$

$A=\frac{1}{2} (1232)$

$A=616$

3 0

2 years ago

Read 2 more answers

What is the sum of 1/9, 2/3, and 5/18?<br> A. 19/18<br> B.4/15<br> C. 12/9<br> D. 8/20

kari74 [83]

Answer:

19/18

Step-by-step explanation:

The GCF of 9, 3, and 18 is 18.

Each denominator must be multiplied to 18, and as a result what you multiply in the denominator MUST be multiplied in the numerator.

9 * 2 = 18

1 * 2 = 2

2/18

3 * 6 = 18

6 * 2 = 12

12/18

5/18

Add all three fractions together:

12/18 + 5/18 + 2/18 = 19/18

8 0

3 years ago

3x - y = 7

Nastasia [14]

Answer:

C) -3x=-3

Step-by-step explanation:

When you subtract the 3x-y=7 from 6x-y=10, you will receive the following equation:

3x-6x-y-(-y)=7-10

-3x-y+y=-3

-3x=-3

As we can see, this equation matches the one in option C, meaning that it's our answer!

6 0

3 years ago

Help me? idk the answer :P

Alexus [3.1K]

$\bf \left.\qquad \qquad \right.\textit{negative exponents}\\\\
a^{-{ n}} \implies \cfrac{1}{a^{ n}}
\qquad \qquad
\cfrac{1}{a^{ n}}\implies a^{-{ n}}
\qquad \qquad 
a^{{{ n}}}\implies \cfrac{1}{a^{-{{ n}}}}\\\\
-------------------------------\\\\
\left( 2^8\cdot 3^{-5}\cdot 6^0 \right)^{-2}\left( \cfrac{3^{-2}}{2^3} \right)^4\cdot 2^{28}\impliedby \textit{let's do the first group}
\\\\
-------------------------------\\\\
$

$\bf \left( 2^8\cdot \cfrac{1}{3^5}\cdot 1 \right)^{-2}\implies \left( \cfrac{2^8}{3^5} \right)^{-2}\implies \left( \cfrac{3^5}{2^8} \right)^{2}\implies \cfrac{3^{2\cdot 5}}{2^{2\cdot 8}}\implies \boxed{\cfrac{3^{10}}{2^{16}}}\\\\
-------------------------------\\\\
\textit{now the second group}\qquad \left( \cfrac{3^{-2}}{2^3} \right)^4\implies \left( \cfrac{\frac{1}{3^2}}{2^3} \right)^4\implies \left( \cfrac{1}{2^3\cdot 3^2} \right)^4$

$\bf \cfrac{1^4}{2^{4\cdot 3}\cdot 3^{4\cdot 2}}\implies \boxed{\cfrac{1}{2^{12}\cdot 3^8}}\\\\
-------------------------------\\\\
\textit{so we end up with\qquad }\cfrac{3^{10}}{2^{16}}\cdot \cfrac{1}{2^{12}\cdot 3^8}\cdot 2^{28}\implies \cfrac{3^{10}\cdot 2^{28}}{2^{16}\cdot 2^{12}\cdot 3^8}
\\\\\\
\cfrac{3^{10}\cdot 2^{28}}{2^{16+12}\cdot 3^8}\implies \cfrac{3^{10}\cdot 2^{28}}{2^{28}\cdot 3^8}\implies 3^{10}\cdot 2^{28}\cdot 2^{-28}\cdot 3^{-8}$

$\bf 3^{10-8}\cdot 2^{28-28}\implies 3^2\cdot 1\implies 3^2\implies 9$

7 0

3 years ago