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3 years ago

For what value of a does 9 equal 1/27 a + 3

Mathematics

2 answers:

Yakvenalex [24]3 years ago

6 0

Answer:

a = 162

Step-by-step explanation:

9 = 1/27 a +3

Subtract 3 from each side

9-3 = 1/27 a +3-3

6 = 1/27 a

Multiply each side by 27

6*27 = 1/27 a *27

162 = a

expeople1 [14]3 years ago

5 0

Answer:

a = 162.

Step-by-step explanation:

1/27 a + 3 = 9

1/27 a = 9 - 3 = 6

Multiply both sides by 27:

a = 6 * 27

a = 162.

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2. Describe the transformation of the graph of f(x) = 6* that yields the graph of f(x) = 6*+1

avanturin [10]

Answer: Translation

The difference between the first equation and the second is that the second is moved one space higher. All of its points will be a unit higher than the other. Because of this change in height, the name of the transformation is a translation

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3 0

3 years ago

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

Find 3√27 +3√64 - 3√125

serg [7]

Answer:

the answer is approximately 6.047

4 0

3 years ago

The table below shows the average yearly balance in a savings account where interest is compounded annually. no money is deposit

djverab [1.8K]

A. Linear with a negative rate of change

6 0

3 years ago

Raoul saves $14.50 each month for 3 months. He puts his money in the bank and earns $0.33 interest . Hector puts \$3.60 e each w

Norma-Jean [14]

Answer:

$0.63

Step-by-step explanation:

Raoul saves $14.50 each month for 3 months. Total amount saved is 14.50 * 3 = $43.50

He earned an interest of $0.33.

The total amount of money he has is thus: 43.50 + 0.33 = $43.83

Hector puts $3.60 each week for 12 weeks. Total amount save by Hector is thus 3.60 * 12 = $43.2

Now we are asked to get how much money Raoul has. To get this, we simply subtract the amount of money Hector has from that of Raoul

That would be 43.83 - 43.2 = $0.63

4 0

3 years ago

For what value of a does 9 equal 1/27 a + 3​

For what value of a does 9 equal 1/27 a + 3