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xenn [34]
3 years ago
11

28 squared minus 39x minus 80 equals 0​

Mathematics
1 answer:
Rudik [331]3 years ago
3 0

Answer:

x = 704/39 ( <----Fraction)

Step-by-step explanation:

28 squared minus 39x minus 80 equals 0​  

784-39x-80=0  evaluate the power

704-39x=0  Subtract the numbers

-39x=-704  move the constant to the right hand side and change it's sign

Divide on both side to get x = 704/39

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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
3 years ago
Can someone help me I’ll Mark brainliest I been struggling with this one :(
mariarad [96]

Answer:

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Step-by-step explanation:

5 0
3 years ago
A baker sells a dozen for &amp;3.99 the cost of to make three donuts is $0.45 how much is the total profit on 5 dozen donuts?
Ber [7]
19.95 is the correct answer I believe
5 0
3 years ago
A boy bought 9 new cards to add to his card collection of 25 cards. The next day his dog ate half of his collection. How many ca
lora16 [44]
This would equal (9+25)/2

Which could be simplified into 34/2

Which means the answer is 17
4 0
3 years ago
Suppose a square garden hose represented by 16x^2 ft^2. what is the length of each side?
Triss [41]

Answer:

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Step-by-step explanation:

As we know area of a square is side*side

side*side = 16^2

side=√16ft^2

side=4ft

7 0
3 years ago
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