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Alex787 [66]
3 years ago
14

Y is directly proportional to x when y=30, x=6

Mathematics
1 answer:
rosijanka [135]3 years ago
4 0

Answer:

a) y = 5x

b) y = 60

Step-by-step explanation:

Since, y is directly proportional to x when y=30, x=6

\therefore \:  y \:  \alpha \:  \: x \\ \therefore \:  y =  kx ...(1) \\ plug \: y = 30 \: and \: x = 6  \: in \: eq \: (1)\\ 30 = k \times 6 \\ k =  \frac{30}{6}  \\ k = 5 \\ plug \: k = 5 \: in \: (1) \\ y = 5x....(2) \: (required \: equation) \\ plug \: x = 12 \: in \: eq \: (2) \\ y = 5 \times 12 \\  \huge \red{ \boxed{y = 60}}

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Solve linear equation 1/4 x + 18 = X
Anton [14]

Answer:

24

Step-by-step explanation:

1 / 4 x + 18 = x

x / 4 + 18 = x

x = x / 4 + 18

x - x / 4 = 18

( 4x / 4 ) - ( x/4 ) = 18

( 4x - x ) / 4 = 18

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3 0
2 years ago
Could someone help me, please?
k0ka [10]

We can rewrite the expression under the radical as

\dfrac{81}{16}a^8b^{12}c^{16}=\left(\dfrac32a^2b^3c^4\right)^4

then taking the fourth root, we get

\sqrt[4]{\left(\dfrac32a^2b^3c^4\right)^4}=\left|\dfrac32a^2b^3c^4\right|

Why the absolute value? It's for the same reason that

\sqrt{x^2}=|x|

since both (-x)^2 and x^2 return the same number x^2, and |x| captures both possibilities. From here, we have

\left|\dfrac32a^2b^3c^4\right|=\left|\dfrac32\right|\left|a^2\right|\left|b^3\right|\left|c^4\right|

The absolute values disappear on all but the b term because all of \dfrac32, a^2 and c^4 are positive, while b^3 could potentially be negative. So we end up with

\dfrac32a^2\left|b^3\right|c^4=\dfrac32a^2|b|^3c^4

3 0
3 years ago
Can someone please help me?
Citrus2011 [14]

Answer:

A. 14

Step-by-step explanation:

Okay, so the square is 25 and the circle is 39.

For example, if you take out the square from the circle, that leaves the shaded areas.

39-25=14

14 is your answer.

Cheers!

4 0
3 years ago
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