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Natalka [10]
2 years ago
6

Using the given table of an exponential function, determine the growth/decay factor as well as the rate of increase/decrease. Ty

pe your answers in as numbers. Example: Growth Factor: 3.5, Rate of Increase: 2.5. (Do not input your rate of increase as a percent.)

Mathematics
1 answer:
Westkost [7]2 years ago
7 0

Answer:

The growth rate is 0.6

The growth factor is 1.6

Step-by-step explanation:

Generally, the exponential equation can be written as;

y = a(1 + r)^x

Where a is the initial value of 100

r is the growth rate

So let us form equations;

160 = 100(1 + r)^1 •••••(i)

Also;

256 = 100(1 + r)^2 •••••••(ii)

Divide equation ii by i

256/160 = 1 + r

1.6 = 1 + r

r = 1.6-1

r = 0.6

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Evaluate the double integral.
Fynjy0 [20]

Answer:

\iint_D 8y^2 \ dA = \dfrac{88}{3}

Step-by-step explanation:

The equation of the line through the point (x_o,y_o) & (x_1,y_1) can be represented by:

y-y_o = m(x - x_o)

Making m the subject;

m = \dfrac{y_1 - y_0}{x_1-x_0}

∴

we need to carry out the equation of the line through (0,1) and (1,2)

i.e

y - 1 = m(x - 0)

y - 1 = mx

where;

m= \dfrac{2-1}{1-0}

m = 1

Thus;

y - 1 = (1)x

y - 1 = x ---- (1)

The equation of the line through (1,2) & (4,1) is:

y -2 = m (x - 1)

where;

m = \dfrac{1-2}{4-1}

m = \dfrac{-1}{3}

∴

y-2 = -\dfrac{1}{3}(x-1)

-3(y-2) = x - 1

-3y + 6 = x - 1

x = -3y + 7

Thus: for equation of two lines

x = y - 1

x = -3y + 7

i.e.

y - 1 = -3y + 7

y + 3y = 1 + 7

4y = 8

y = 2

Now, y ranges from 1 → 2 & x ranges from y - 1 to -3y + 7

∴

\iint_D 8y^2 \ dA = \int^2_1 \int ^{-3y+7}_{y-1} \ 8y^2 \ dxdy

\iint_D 8y^2 \ dA =8 \int^2_1 \int ^{-3y+7}_{y-1} \ y^2 \ dxdy

\iint_D 8y^2 \ dA =8 \int^2_1  \bigg ( \int^{-3y+7}_{y-1} \ dx \bigg)   dy

\iint_D 8y^2 \ dA =8 \int^2_1  \bigg ( [xy^2]^{-3y+7}_{y-1} \bigg ) \ dy

\iint_D 8y^2 \ dA =8 \int^2_1  \bigg ( [y^2(-3y+7-y+1)]\bigg ) \ dy

\iint_D 8y^2 \ dA =8 \int^2_1  \bigg ([y^2(-4y+8)] \bigg ) \ dy

\iint_D 8y^2 \ dA =8 \int^2_1  \bigg ( -4y^3+8y^2 \bigg ) \ dy

\iint_D 8y^2 \ dA =8 \bigg [\dfrac{ -4y^4}{4}+\dfrac{8y^3}{3} \bigg ]^2_1

\iint_D 8y^2 \ dA =8 \bigg [ -y^4+\dfrac{8y^3}{3} \bigg ]^2_1

\iint_D 8y^2 \ dA =8 \bigg [ -2^4+\dfrac{8(2)^3}{3} + 1^4- \dfrac{8\times (1)^3}{3}\bigg]

\iint_D 8y^2 \ dA =8 \bigg [ -16+\dfrac{64}{3} + 1- \dfrac{8}{3}\bigg]

\iint_D 8y^2 \ dA =8 \bigg [ -15+ \dfrac{64-8}{3}\bigg]

\iint_D 8y^2 \ dA =8 \bigg [ -15+ \dfrac{56}{3}\bigg]

\iint_D 8y^2 \ dA =8 \bigg [  \dfrac{-45+56}{3}\bigg]

\iint_D 8y^2 \ dA =8 \bigg [  \dfrac{11}{3}\bigg]

\iint_D 8y^2 \ dA = \dfrac{88}{3}

4 0
2 years ago
PLEASE HELP ME!!!!!!!!
juin [17]
That would be 242 divided by the total number of people surveyed, or 242/1028.

4 0
3 years ago
Does 1/7 divided by what equals 14
Paul [167]
2*7=14 7/14=2 do yes
5 0
3 years ago
PLEASE HELP ME AND SHOW ALL WORK THANK YOU!!!!!!
Natalija [7]
M>5=m>4
m>5 and m>7 are on line then m>5+m>7=180
by substituting
m>4+37=180
m>4=143
m>2=m>7=37
m>4=m>8
m<3+m>5=180
7 0
3 years ago
H(x) = -4x – 4; Find h(3x)<br><br> please help, I’m confused if it’s -4•3•3 or not
inna [77]

Answer:

Im really sprry i do not know tgis but good luck!!

4 0
2 years ago
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