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maks197457 [2]
2 years ago
14

Help needed!!! Serious answers only

Mathematics
1 answer:
timurjin [86]2 years ago
7 0

B is the answer I think but so t take my word for it

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Construct the graph of the equation y=kx if it is known that the point B belongs to the graph. The graph of which of these two e
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Answer:

The graph with B(2,-3) i.e) y=\frac{-3}{2} x goes through the point M(-10,15).

Step-by-step explanation:

Consider M(-10,15) and given that equation is y = kx.

Now, substitute M(-10,15) in the equation

⇒ 15 = k × -10

⇒ k = \frac{15}{-10} = \frac{-3}{2}

⇒ y = \frac{-3}{2} x

Now, check with the given points B(2,-3) and B(3\frac{1}{3} , -2)

1) B(2,-3)

y = \frac{-3}{2} x

⇒(-3) = \frac{-3}{2} × 2

⇒ -3 = -3 ⇒ LHS = RHS

⇒ B(2,-3) is the required point.

2) for b(3\frac{1}{3} , -2)

LHS ≠ RHS.

So,The graph with B(2,-3) i.e) y=\frac{-3}{2} x goes through the point M(-10,15).

8 0
2 years ago
If you invested $250 at 16% interest, how much will you have after 18 years?
Whitepunk [10]
After 1st year: 250$:100%=x$:116%, 250$*116%=x$*100%, x=(250*116)/100=290$. After 1st year I will have 290$
After 2nd year: 290$:100%=x$:116%, x=(290*116)/100=336.4$. After 2nd year I will have 336.4$
After 3rd year I will have (336.4*116)/100=390.224$
After 4th yr: (390.224*116)/100=452.65984$
After 5th yr: (452.65984*116)/100=525.085$
After- 6th yr: 609.1$, 7th yr: 706.556$, 8th yr: 819.605$, 9th yr: 950.742$
10th yr: 1102.86$, 11th yr: 1279.32$, 12th yr: 1484.01$, 13th yr: 1721.45$,
14th yr: 1996.88$, 15th: 2316.38$, 16th yr: 2687$, 17th yr: 3116.92$
After 18 years I will have 3615.63$.
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3 years ago
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Aleks04 [339]

74 / 100 hope this helps

5 0
2 years ago
Help me please! I really need help
Brums [2.3K]
If you looked it up online you could find it. look on safari
8 0
2 years ago
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If f(x)=-3+7 and g(x)=-7+3, what is the value of f(-3) - g(3)
Andreyy89
F(x) = -3 + 7 = 4
g(x) = -7 + 3 = -4

f(-3) - g(3) = 4 - (-4) = 4 + 4 = 8
6 0
3 years ago
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