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Alex73 [517]
3 years ago
8

Is this a relation a function

Mathematics
1 answer:
Archy [21]3 years ago
4 0
No, it is not since there is no constant change the points are all over the boards
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Write the standard form of the line that passes through the given points. Include your work in your final answer. Type your answ
Oksanka [162]
Standard form is ax+by=c

first we use
y-y1=m(x-x1)
a point is (x1,y1)
the slope is m


the slope between the points (x1,y1) and (x2,y2) is
(y2-y1)/(x2-x1)
given
(1,1) and (3,4)
slope=(4-1)/(3-1)=3/2

a oint is (1,1) and slope is 3/2
y-1=3/2(x-1)
y-1=3/2x-3/2
solve for ax+by=c form, a is normally positive and a and b and c are whole numbers
times 2 both sides
2y-2=3x-3
minus 2y
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add 3
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5 0
3 years ago
if one Circle has a diameter of 10 cm and a second Circle has a diameter of 20 cm what is the ratio of the area of the larger ci
Oxana [17]
Area of a circle is πr^2.

The radius of the smaller circle with diameter 10 cm is 5 cm.

The radius of the larger circle with diameter 20 cm.is 10 cm.

Area of larger circle : Area of smaller circle

Area of larger circle = 100π.

Area of smaller circle = 25π.

Ratio = 100π : 25π or 100π/25π.

7 0
3 years ago
Maria measured a centipede that was 1 1/12 inches long. Jerome measured a centipede that was 11/12 of an inch long. How much lon
4vir4ik [10]

Answer:

1/6

Step-by-step explanation:

The centipede that Maria measured was 1 1/12 inches long. If you convert this to an improper fraction it is equal to 13/12 inches. Now to find how much longer Maria's centipede was than Jeromes, you just take 11/12 and minus it from 13/12 which is equal to 2/12 of an inch, which then simplifies to 1/6 of an inch.

3 0
3 years ago
You have a bag of 14
AlekseyPX

Answer:

30% or 30/100

Step-by-step explanation:

14+15+9+12= 50 Jelly beans total

yllow- 15

15/50=30/100= 30%

7 0
3 years ago
HELP i’m having trouble with my homework assignments
Aleksandr-060686 [28]

Answer:

Collin: about $401 thousand

Cameron: about $689 thousand

Step-by-step explanation:

A situation in which doubling time is constant is a situation that can be modeled by an exponential function. Here, you're given an exponential function, though you're not told what the variables mean. That function is ...

P(t)=P_0(2^{t/d})

In this context, P0 is the initial salary, t is years, and d is the doubling time in years. The function gives P(t), the salary after t years. In this problem, the value of t we're concerned with is the difference between age 22 and age 65, that is, 43 years.

In Collin's case, we have ...

P0 = 55,000, t = 43, d = 15

so his salary at retirement is ...

P(43) = $55,000(2^(43/15)) ≈ $401,157.89

In Cameron's case, we have ...

P0 = 35,000, t = 43, d = 10

so his salary at retirement is ...

P(43) = $35,000(2^(43/10)) ≈ $689,440.87

___

Sometimes we like to see these equations in a form with "e" as the base of the exponential. That form is ...

P(t)=P_{0}e^{kt}

If we compare this equation to the one above, we find the growth factors to be ...

2^(t/d) = e^(kt)

Factoring out the exponent of t, we find ...

(2^(1/d))^t = (e^k)^t

That is, ...

2^(1/d) = e^k . . . . . match the bases of the exponential terms

(1/d)ln(2) = k . . . . . take the natural log of both sides

So, in Collin's case, the equation for his salary growth is

k = ln(2)/15 ≈ 0.046210

P(t) = 55,000e^(0.046210t)

and in Cameron's case, ...

k = ln(2)/10 ≈ 0.069315

P(t) = 35,000e^(0.069315t)

5 0
3 years ago
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