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Ipatiy [6.2K]
3 years ago
13

How do you solve 4a/b divided by 2ac/b

Mathematics
2 answers:
joja [24]3 years ago
8 0
\frac{\frac{4a}{b}}{\frac{2ac}{b}}=?\\\\
To\ divide\ fraction\ multiply\ first\ fraction\ by\ inverse\ of\ the\ second:\\\\
\frac{\frac{4a}{b}}{\frac{2ac}{b}}=\frac{4a}{b}*\frac{b}{2ac}=?\\\\
Shortening\ fractions:\\\\
\frac{\frac{4a}{b}}{\frac{2ac}{b}}=\frac{4a}{b}*\frac{b}{2ac}=\frac{2}{c}\\\\Solution\ is\ \frac{2}{c}.
goblinko [34]3 years ago
4 0
\frac{4a}{b} :\frac{ 2ac}{b}= \frac{4a}{b} \cdot \frac{b}{ 2ac}= \frac{2}{c} \\ \\a \ and \ c \neq 0


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. Mr. View is 8 years older than Ms. Sun. In 7 years, the sum of their ages will be 110. How old are they now?​
Fofino [41]
Mr. View is now 63, while Ms. Sun is now 47. You’d have to divide 110 in half, this would give you 55. Then subtract 8 from 55, to give you Ms. Suns’ age. To find out Mr. Views age, you then subtract 47 from 110, and you’d then get 63.
5 0
3 years ago
Elena bought a bag of 75 glass beads. She used 7 beads to make a bracelet. She wants to make necklaces from the rest. Each neckl
Novosadov [1.4K]

Answer:

she can make 5 necklaces

Step-by-step explanation:

75-7=68

68/13=5.2

3 0
4 years ago
How many more pumpkins have a mass between 9 and 12 kg than between 0 and 3 kg?
KengaRu [80]

Answer:

According to the graph, there are 6 pumpkins with the mass of 9kg to 12kg and 4 pumpkins with the mass of 0kg to 3kg.

In order to find how many pumpkins more, you have to subtract it so it will be 6 pumpkins - 4 pumpkins = 2 pumpkins.

Therefore, there are 2 more pumpkins with the mass of 9 to 12kg compared to the pumpkins with the mass of 0 to 3kg.

5 0
4 years ago
PLZ HELP!!! I Will give brainliest. What is the value of x in sin(3x)=cos(6x) if x is in the interval of 0≤x≤π/2
sertanlavr [38]

Answer:

sin(2x)=cos(π2−2x)

So:

cos(π2−2x)=cos(3x)

Now we know that cos(x)=cos(±x) because cosine is an even function. So we see that

(π2−2x)=±3x

i)

π2=5x

x=π10

ii)

π2=−x

x=−π2

Similarly, sin(2x)=sin(2x−2π)=cos(π2−2x−2π)

So we see that

(π2−2x−2π)=±3x

iii)

π2−2π=5x

x=−310π

iv)

π2−2π=−x

x=2π−π2=32π

Finally, we note that the solutions must repeat every 2π because the original functions each repeat every 2π. (The sine function has period π so it has completed exactly two periods over an interval of length 2π. The cosine has period 23π so it has completed exactly three periods over an interval of length 2π. Hence, both functions repeat every 2π2π2π so every solution will repeat every 2π.)

So we get ∀n∈N

i) x=π10+2πn

ii) x=−π2+2πn

iii) x=−310π+2πn

(Note that solution (iv) is redundant since 32π+2πn=−π2+2π(n+1).)

So we conclude that there are really three solutions and then the periodic extensions of those three solutions.

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