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dezoksy [38]
3 years ago
14

HELP ME WITH THIS PLEASE! What is the equation? OR GROWTH RATE

Mathematics
2 answers:
balu736 [363]3 years ago
8 0
To start, the first thing you need to do is to find the slope of your x and y values, or in other words, your rate of change. To do this, use the slope intercept formula or (y2-y1)/(x2-x1) and you can use any two pairs of x and y values.

Using the x and y values of (2, 96) and (3, 76.80), plug the values into the slope intercept formula. This would look like this:
(76.80-96)/(3-2)

When you simplify, you should get -19.2/1 or simply -19.2 as your slope. 

Then, using the linear equation y=mx+b, plug in a coordinate point into the x and y variables and plug your slope into the m value in order to solve for the value of your y-intercept, or b.
This would look like this:
96=-19.2(2)+b

When you simplify, you will get 96=-38.4+b and when you solve for b, you get that b=134.4

Now that you have your slope and your y-intercept, you can conclude that the equation to this input/output chart would be 
y=-19.2x+134.4





Gwar [14]3 years ago
7 0
It would help if you would post the complete question not just a few points in a table. Without all the information provided in the question ( initial value??); I can not be certain of the solution. Looking at the table, you can tell the rate is not constant, it changes between points. The data best fits an exponential decay graph. When plotted in excel it gives the equation:
y = 150 e^{-0.223x}

with a perfect R² = 1
This equation shows that whatever this data represents started at 150 and exponentially decayed/decreased at a rate of -0.223
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The volume of a right circular cone with radius r and height h is V = pir^2h/3.
Scorpion4ik [409]

The question is incomplete. The complete question is :

The volume of a right circular cone with radius r and height h is V = pir^2h/3. a. Approximate the change in the volume of the cone when the radius changes from r = 5.9 to r = 6.8 and the height changes from h = 4.00 to h = 3.96.

b. Approximate the change in the volume of the cone when the radius changes from r = 6.47 to r = 6.45 and the height changes from h = 10.0 to h = 9.92.

a. The approximate change in volume is dV = _______. (Type an integer or decimal rounded to two decimal places as needed.)

b. The approximate change in volume is dV = ___________ (Type an integer or decimal rounded to two decimal places as needed.)

Solution :

Given :

The volume of the right circular cone with a radius r and height h is

$V=\frac{1}{3} \pi r^2 h$

$dV = d\left(\frac{1}{3} \pi r^2 h\right)$

$dV = \frac{1}{3} \pi h \times d(r^2)+\frac{1}{3} \pi r^2 dh$

$dV = \frac{2}{3} \pi r h (dr)+\frac{1}{3} \pi r^2 dh$

a). The radius is changed from r = 5.9 to r = 6.8 and the height is changed from h = 4 to h = 3.96

So, r = 5.9  and dr = 6.8 - 5.9 = 0.9

     h = 4  and dh = 3.96 - 4 = -0.04

Now, $dV = \frac{2}{3} \pi r h (dr)+\frac{1}{3} \pi r^2 dh$

$dV = \frac{2}{3} \pi (5.9)(4)(0.9)+\frac{1}{3} \pi (5.9)^2 (-0.04)$

$dV=44.484951 - 1.458117$

$dV=43.03$

Therefore, the approximate change in volume is dV = 43.03 cubic units.

b).  The radius is changed from r = 6.47 to r = 6.45 and the height is changed from h = 10 to h = 9.92

So, r = 6.47  and dr = 6.45 - 6.47 = -0.02

     h = 10  and dh = 9.92 - 10 = -0.08

Now, $dV = \frac{2}{3} \pi r h (dr)+\frac{1}{3} \pi r^2 dh$

$dV = \frac{2}{3} \pi (6.47)(10)(-0.02)+\frac{1}{3} \pi (6.47)^2 (-0.08)$

$dV=-2.710147-3.506930$

$dV= -6.22$

Hence, the approximate change in volume is dV = -6.22 cubic units

8 0
2 years ago
Which fraction is equivalent to -7/8? A) 8 /- 7 b)7 /-8 c)-7/-8 d)7/ 8
lozanna [386]

Answer:

im pretty sure b

Step-by-step explanation:

because i took a test

6 0
3 years ago
Why is volume not a dose?
Ket [755]

Answer:

concentration

Step-by-step explanation:

Because concentration is a factor to how much of a volume to give

4 0
2 years ago
1-cot^2a+cot^4a=sin^2a(1+cot^6a) prove it.​
aliina [53]

Step-by-step explanation:

We have

1-cot²a + cot⁴a = sin²a(1+cot⁶a)

First, we can take a look at the right side. It expands to sin²a + cot⁶(a)sin²(a) = sin²a + cos⁶a/sin⁴a (this is the expanded right side) as cot(a) = cos(a)/sin(a), so cos⁶a = cos⁶a/sin⁶a. Therefore, it might be helpful to put everything in terms of sine and cosine to solve this.

We know 1 = sin²a+cos²a and cot(a) = cos(a)/sin(a), so we have

1-cot²a + cot⁴a = sin²a+cos²a-cos²a/sin²a + cos⁴a/sin⁴a

Next, we know that in the expanded right side, we have sin²a + something. We can use that to isolate the sin²a. The rest of the expanded right side has a denominator of sin⁴a, so we can make everything else have that denominator.

sin²a+cos²a-cos²a/sin²a + cos⁴a/sin⁴a

= sin²a + (cos²(a)sin⁴(a) - cos²(a)sin²(a) + cos⁴a)/sin⁴a

We can then factor cos²a out of the numerator

sin²a + (cos²(a)sin⁴(a) - cos²(a)sin²(a) + cos⁴a)/sin⁴a

= sin²a + cos²a (sin⁴a-sin²a+cos²a)/sin⁴a

Then, in the expanded right side, we can notice that the fraction has a numerator with only cos in it. We can therefore write sin⁴a in terms of cos (we don't want to write the sin²a term in terms of cos because it can easily add with cos²a to become 1, so we can hold that off for later) , so

sin²a = (1-cos²a)

sin⁴a = (1-cos²a)² = cos⁴a - 2cos²a + 1

sin²a + cos²a (sin⁴a-sin²a+cos²a)/sin⁴a

= sin²a + cos²a (cos⁴a-2cos²a+1-sin²a+cos²a)/sin⁴a

= sin²a + cos²a (cos⁴a-cos²a+1-sin²a)/sin⁴a

factor our the -cos²a-sin²a as -1(cos²a+sin²a) = -1(1) = -1

sin²a + cos²a (cos⁴a-cos²a+1-sin²a)/sin⁴a

=  sin²a + cos²a (cos⁴a-1 + 1)/sin⁴a

= sin²a + cos⁶a/sin⁴a

= sin²a(1+cos⁶a/sin⁶a)

= sin²a(1+cot⁶a)

8 0
3 years ago
Round to nearest hundredth 0.3238​
Alekssandra [29.7K]

Answer:

0.32

Have a nice day!

6 0
2 years ago
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