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2 years ago

The residents of a city voted on whether to raise property taxes. The ratio of yes votes to no votes was 5 to 3 . If there were

2829 no votes, what was the total number of votes?

Mathematics

1 answer:

dlinn [17]2 years ago

5 0

$\bf \begin{array}{ccll} yes&no\\ \text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\ 5&3\\ y&2829 \end{array}\implies \cfrac{5}{y}=\cfrac{3}{2829}\implies \cfrac{5\cdot 2829}{3}=y\implies 4715=y \\\\\\ \stackrel{\textit{total number of votes}}{yes+no}\implies 4715+2829\implies 7544$

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Please help me with this.

inna [77]

Answer:

Plot and connect (5,4), (3,0) and (7,0).

Step-by-step explanation:

This is in factored form and can be graphed directly from the equation. The equation shows the x-intercepts.

The x-intercepts are found by solving for x using the zero product rule.

(x-3)=0 so x = 3

(x-7) = 0 so x = 7

The intercepts are (3,0) and (7,0). Plot the points. The vertex will occur halfway between these points. 7-3 / 2 = 4/2 = 2. This means the axis of symmetry is at x = 5 and this is the x-coordinate of the vertex too.

Substitute x = 5 into the equation and solve for y.

-(5-3)(5-7) = -(2)(-2) = 4

The vertex is (5,4). Plot it and connect the points.

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2 years ago

If the area of the region bounded by the curve y^2 =4ax and the line x= 4a is 256/3 Sq units, using integration find the value o

almond37 [142]

If the area of the region bounded by the curve $y^2 =4ax$ and the line $x= 4a$ is $\frac{256}{3}$ Sq units, then the value of $a$ will be $2$ .

<h3>What is area of the region bounded by the curve ?</h3>

An area bounded by two curves is the area under the smaller curve subtracted from the area under the larger curve. This will get you the difference, or the area between the two curves.

Area bounded by the curve $=\int\limits^a_b {x} \, dx$

We have,

$y^2 =4ax$

⇒ $y=\sqrt{4ax}$

$x= 4a$ ,

Area of the region $=\frac{256}{3}$ Sq units

Now comparing both given equation to get the intersection between points;

$y^2=16a^2$

$y=4a$

So,

Area bounded by the curve $= \[ \int_{0}^{4a} y \,dx \]$

$\frac{256}{3} =\[ \int_{0}^{4a} \sqrt{4ax} \,dx \]$

$\frac{256}{3}= \[\sqrt{4a} \int_{0}^{4a} \sqrt{x} \,dx \]$

$\frac{256}{3}= \[2\sqrt{a} \int_{0}^{4a} x^{\frac{1}{2} } \,dx \]$

$\frac{256}{3}= 2\sqrt{a} \left[\begin{array}{ccc}\frac{(x)^{\frac{1}{2}+1 } }{\frac{1}{2}+1 }\end{array}\right] _{0}^{4a}$

$\frac{256}{3}= 2\sqrt{a} \left[\begin{array}{ccc}\frac{(x)^{\frac{3}{2} } }{\frac{3}{2} }\end{array}\right] _{0}^{4a}$

$\frac{256}{3}= 2\sqrt{a} *\frac{2}{3} \left[\begin{array}{ccc}(x)^{\frac{3}{2}\end{array}\right] _{0}^{4a}$

On applying the limits we get;

$\frac{256}{3}= \frac{4}{3} \sqrt{a} \left[\begin{array}{ccc}(4a)^{\frac{3}{2} \end{array}\right]$

$\frac{256}{3}= \frac{4}{3} \sqrt{a} *\sqrt{(4a)^{3} }$

$\frac{256}{3}= \frac{4}{3} \sqrt{a} * 8 *a^{2} \sqrt{a}$

$\frac{256}{3}= \frac{4}{3} * 8 *a^{3}$

⇒ $a^{3} =8$

$a=2$

Hence, we can say that if the area of the region bounded by the curve $y^2 =4ax$ and the line $x= 4a$ is $\frac{256}{3}$ Sq units, then the value of $a$ will be $2$ .

To know more about Area bounded by the curve click here

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1 year ago

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I need help to figure this out I don't understand it

vladimir1956 [14]

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