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Len [333]
3 years ago
13

I need help please, I attached a photo.Graph y>1-3x

Mathematics
1 answer:
natali 33 [55]3 years ago
6 0
Your answer is the one on the bottom left
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I will give brainliest to the person who answers all questions
stellarik [79]

Answer: Hope this helps!

Left to right - Q1 - Q5

1,500, 1,250, 0.25, 105, 25%

Step-by-step explanation:

Q1 - 15,000 * 0.02 * 5 = 1,500

Q2 - 25,000 * 0.01 * 5 = 1,250

Q3 - 20 - ((6 * 1) + (3 * 0.75) + (10 * 0.35) + (4 * 2)) = 0.25, 25 cents

Q4 - 30x + 1,000 = 4,200 (X = 107) Approximately 105 jobs

Q5 - 1,584 / (32 * 66) = 0.75 (75%) 100% - 75% = 25%

4 0
2 years ago
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What numbers multiply to -30 and add to -3
krek1111 [17]
Let the numbers be x and y.  Then xy = -30 and  x+y = -3.

Solve xy = -30 for y:    y = -30/x

subst. -30/x for y in   x+y= -3:     x  -  30/x = -3

Multiply all 3 terms by x:  x^2 - 30 = -3x, so   x^2 + 3x - 10 = 0

Solve this quadratic equation for x.     x: {-5, 2}

If x = -5, then   x+y = -3 becomes -5 + y = -3, and y = 2.

You should check to determine whether x=2 is also correct.  If it is, what is the corresponding y value?
3 0
3 years ago
The owner of a bike shop sells unicycles and bicycles and keeps inventory by counting seats and wheels. One day, she counts 15 s
poizon [28]
Im going to guess its either b or c.
7 0
3 years ago
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One of the main contaminants of a nuclear accident, such as that at Chernobyl, is strontium-90, which decays exponentially at a
igor_vitrenko [27]

Answer: P=0.975^t


Step-by-step explanation:

We know that the general form of the exponential decay formula is


y=A(1-r)^t, where y is final amount remaining after t time, A is the original amount and r is the rate of decay


Now, the ratio of strontium-90 remaining, p , as a function of years, t , since the nuclear accident. P=\frac{A(1-0.02)^t}{A}=\frac{(0.975)^t}{1}

Hence, the ratio of remaining since the nuclear accident is P=0.975^t

3 0
3 years ago
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What is the sign of a.
posledela

As a = 0 , there is no sign . Its only 0. Thus the answer is (c)

6 0
3 years ago
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