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3 years ago

Which number is a solution of a² +4= 6a – 1? 07 6 7 8

Mathematics

1 answer:

LuckyWell [14K]3 years ago

6 0

Answer:

a=1 or a=5

Step-by-step explanation:

$a^{2}$ +4=6a−1

Step 1: Subtract 6a-1 from both sides.

$a^{2}$ +4−(6a−1)=6a−1−(6a−1)

$a^{2}$ −6a+5=0

Step 2: Factor left side of equation.

(a−1)(a−5)=0

Step 3: Set factors equal to 0.

a−1=0 or a−5=0

a=1 or a=5

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Two angles are complementary.The larger angle is 30 degrees more than 4 times the smaller

gregori [183]

Answer:

Step-by-step explanation:

Remark

2 angles are complementary means that 2 angles add up to 90°. That is the guiding princeple behind the problem.

Givens

Let the smallest angle = x

Let the largest angle = 4x + 30

Equation

4x + 30° + x = 90° Combine like terms on the left

Solution

5x + 30 = 90 Subtract 30 from both sides.

5x + 30 - 30 = 90 - 30 Combine

5x = 60 Divide both sides by 5

5x/5 = 60/5

x = 12

Answer

smallest angle = 12
Larger angle = 4*12 + 30 = 78

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2 years ago

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AlladinOne [14]

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3 years ago

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A tank has the shape of a surface generated by revolving the parabolic segment y = x2 for 0 ≤ x ≤ 3 about the y-axis (measuremen

Darina [25.2K]

Answer:

$100\pi\int\limits^9_0 {(\sqrt y)^2(14-y)} \, dy$ ft-lbs.

Step-by-step explanation:

Given:

The shape of the tank is obtained by revolving $y=x^2$ about y axis in the interval $0\leq x\leq 3$ .

Density of the fluid in the tank, $D=100\ lbs/ft^3$

Let the initial height of the fluid be 'y' feet from the bottom.

The bottom of the tank is, $y(0)=0^2=0$

Now, the height has to be raised to a height 5 feet above the top of the tank.

The height of top of the tank is obtained by plugging in $x=3$ in the parabolic equation . This gives,

$H=3^2=9\ ft$

So, the height of top of tank is, $y(3)=H=9\ ft$

Now, 5 ft above 'H' means $H+5=9+5=14$

Therefore, the increase in height of the top surface of the fluid in the tank is given as:

$\Delta y=(14-y)$ ft

Now, area of cross section of the tank is given as:

$A(y)=\pi r^2\\r\to radius\ of\ the\ cross\ section$

Radius is the distance of a point on the parabola from the y axis. This is nothing but the x-coordinate of the point.

We have, $y=x^2$

So, $x=\sqrt y$

Therefore, radius, $r=\sqrt y$

Now, area of cross section is, $A(y)=\pi (\sqrt y)^2$

Work done in pumping the contents to 5 feet above is given as:

$W=D\int\limits^{y(3)}_{y(0)} {A(y)(\Delta y)} \, dy$

Plug in all the values. This gives,

$W=100\int\limits^9_0 {\pi (\sqrt y)^2(14-y)} \, dy\\\\W=100\pi\int\limits^9_0 { (\sqrt y)^2(14-y)} \, dy\textrm{ ft-lbs}$

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