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nlexa [21]
3 years ago
14

Multiplying fraction what is 5/8x6/15

Mathematics
2 answers:
tia_tia [17]3 years ago
4 0

\huge \sf \underline{Solution:-}

\frac{5}{8}  \times  \frac{6}{15}

=  \frac{30}{120}

=  \frac{1}{4}

Here, we just have to multiply the top numbers and then multiply the bottom numbers. Then we can simplify the fraction if needed.

So, correct answer is ¼.

Hope it helps you :)

BigorU [14]3 years ago
3 0

\text {Hello! Let's Solve this Problem!}

\text {When multiplying fractions you multiply numerator by numerator and } \text {demoninator by denominator. Simplifying is required when the answer comes} \text {out as fraction that can be simplified. Simplify until it's reduced to the lowest} \text {fraction it ca be.}

\text {\underline{Multiply}}

\text {5*6=30}

\text {8*15=120}

\text {Your Answer Would Be:}

\Huge\boxed {30/120}

\text {Since this is a fraction we can simplify we simplify it by 30}

\text {Q: What does simplify mean?}

\text {A: It means to divide. So if someone says simplify 2/4 that means divide.}

\text {\underline {Divide}}

\text {30/120 divided by 30 =}

\text {Your Simplified Fraction Answer Would Be:}

\Huge\boxed {1/4}

\text {Best of Luck!}

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Evgesh-ka [11]

Answer:

your TRASH NOOB LOSER

Step-by-step explanation:

so todya eat hesde

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7 0
3 years ago
Find the missing lengths of the sides.
Sati [7]

Answer:

sin 30°=8/c

1/2=8/c

c=16.

tan 60°=b/8

√3=b/8

b=8√3.

4 0
2 years ago
This question refers to unions and intersections of relations. Since relations are subsets of Cartesian products, their unions a
Mice21 [21]

Answer:

AXB= = {(x, y) ∈ A ✕ B| x ∈ A , y ∈  B}

R= {(x, y) ∈ A ✕ B| x R y ⇔ |x| = |y|}

S={(x, y) ∈ x A ✕ B | S y ⇔ x − y is even}

R ∪ S= {(x, y) ∈ A ✕ B | (x, y) ∈ R or (x, y) ∈ S}

R ∩ S = {(x, y) ∈ A ✕ B | (x, y) ∈ R and (x, y) ∈ S}

Step-by-step explanation:

Let A = {−4, 4, 7, 9} and B = {4, 7},

Then A X B= { (-4,4),(-4,7),(4,4),(4,7),(7,4),(7,7),(9,4),(9,7)}

AXB contains all elements of A and B such that x from A and y is from B.

AXB= = {(x, y) ∈ A ✕ B| x ∈ A , y ∈  B}

R= {(-4,4),(4,4),(7,7)}

R consists all ordered pairs where  |x| = |y|

R= {(x, y) ∈ A ✕ B| x R y ⇔ |x| = |y|}

S= { (-4,4),(4,4),(7,7)}

S={(x, y) ∈ x A ✕ B | S y ⇔ x − y is even}

S consists all ordered pairs where x-y is even.

R ∪ S, = { (-4,4),(4,4),(7,7)}

R US is a set containing subsets of both sets R and S

R ∪ S= {(x, y) ∈ A ✕ B | (x, y) ∈ R or (x, y) ∈ S}

R ∩ S=  {(-4,4),(4,4),(7,7)}

R ∩ Sis a set containing subsets only which are common between sets R and S

R ∩ S = {(x, y) ∈ A ✕ B | (x, y) ∈ R and (x, y) ∈ S}

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