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Lady bird [3.3K]

3 years ago

10

Tisha and her academic team are working to go to state finals. They must have a certain number of points to advance. They have h

ad three local matches and will attend a district match. District match points count for three times the number of points than local matches do. Choose the equation that would help Tisha find how many points they need to earn in the district match, b, to advance.

1 answer:

kolbaska11 [484]3 years ago

4 0

Answer:

$b=\frac{T-a-c-d}{3}$

b = (T - a - c - d) / 3

Step-by-step explanation:

Let T be the total number of points required to advance.

a, c and d are points scored in the local matches, and b is the number of points scored in the district match. If b is worth 3 times as much as the other matches, the total number of points is given by:

$T=a+c+d+3b$

Isolate b in order to find out how many points they need in the district match:

$T=a+c+d+3b\\3b=T-a-c-d\\b=\frac{T-a-c-d}{3}$

They need to score (T - a - c - d)/3, in the district match in order to win.

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Condense the following logarithm: 5log_4(a) - 6log_4(b)

grin007 [14]

Hi there! We are given the expression:

$\large \boxed{5 log_{4}(a) - 6 log_{4}(b) }$

To condense or simplify the following logarithm. You have to remember these properties:

Properties - Logarithm

$\large{ log_{b}(a)^{n} = n log_{b}(a) } \\ \large{ log_{b}(a) - log_{b}(c) = log_{b}( \frac{a}{c} ) }$

These two properties are what we need for our problem. Therefore,

$\large{ log_{4}(a)^{5} - log_{4}(b)^{6} }$

We use the log_b(a)^n = nlog_b(a) property to convert in the form above. Next, we use the second property.

$\large{ log_{4}( \frac{ {a}^{5} }{ {b}^{6} } ) }$

Answer

log base 4 of (a^5/b^6) is our simplifed form.

Any questions can be asked through comment as I may reply soon. Thanks for using Brainly! Have a good day and happy learning! :)

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3 years ago

The portion of the parabola y²=4ax above the x-axis, where is form 0 to h is revolved about the x-axis. Show that the surface ar

castortr0y [4]

Answer:

See below for Part A.

Part B)

$\displaystyle h=\Big(\frac{125}{\pi}+27\Big)^\frac{2}{3}-9\approx7.4614$

Step-by-step explanation:

Part A)

The parabola given by the equation:

$y^2=4ax$

From 0 to <em>h</em> is revolved about the x-axis.

We can take the principal square root of both sides to acquire our function:

$y=f(x)=\sqrt{4ax}$

Please refer to the attachment below for the sketch.

The area of a surface of revolution is given by:

$\displaystyle S=2\pi\int_{a}^{b}r(x)\sqrt{1+\big[f^\prime(x)]^2} \,dx$

Where <em>r(x)</em> is the distance between <em>f</em> and the axis of revolution.

From the sketch, we can see that the distance between <em>f</em> and the AoR is simply our equation <em>y</em>. Hence:

$r(x)=y(x)=\sqrt{4ax}$

Now, we will need to find f’(x). We know that:

$f(x)=\sqrt{4ax}$

Then by the chain rule, f’(x) is:

$\displaystyle f^\prime(x)=\frac{1}{2\sqrt{4ax}}\cdot4a=\frac{2a}{\sqrt{4ax}}$

For our limits of integration, we are going from 0 to <em>h</em>.

Hence, our integral becomes:

$\displaystyle S=2\pi\int_{0}^{h}(\sqrt{4ax})\sqrt{1+\Big(\frac{2a}{\sqrt{4ax}}\Big)^2}\, dx$

Simplify:

$\displaystyle S=2\pi\int_{0}^{h}\sqrt{4ax}\Big(\sqrt{1+\frac{4a^2}{4ax}}\Big)\,dx$

Combine roots;

$\displaystyle S=2\pi\int_{0}^{h}\sqrt{4ax\Big(1+\frac{4a^2}{4ax}\Big)}\,dx$

Simplify:

$\displaystyle S=2\pi\int_{0}^{h}\sqrt{4ax+4a^2}\, dx$

Integrate. We can consider using u-substitution. We will let:

$u=4ax+4a^2\text{ then } du=4a\, dx$

We also need to change our limits of integration. So:

$u=4a(0)+4a^2=4a^2\text{ and } \\ u=4a(h)+4a^2=4ah+4a^2$

Hence, our new integral is:

$\displaystyle S=2\pi\int_{4a^2}^{4ah+4a^2}\sqrt{u}\, \Big(\frac{1}{4a}\Big)du$

Simplify and integrate:

$\displaystyle S=\frac{\pi}{2a}\Big[\,\frac{2}{3}u^{\frac{3}{2}}\Big|^{4ah+4a^2}_{4a^2}\Big]$

Simplify:

$\displaystyle S=\frac{\pi}{3a}\Big[\, u^\frac{3}{2}\Big|^{4ah+4a^2}_{4a^2}\Big]$

FTC:

$\displaystyle S=\frac{\pi}{3a}\Big[(4ah+4a^2)^\frac{3}{2}-(4a^2)^\frac{3}{2}\Big]$

Simplify each term. For the first term, we have:

$\displaystyle (4ah+4a^2)^\frac{3}{2}$

We can factor out the 4a:

$\displaystyle =(4a)^\frac{3}{2}(h+a)^\frac{3}{2}$

Simplify:

$\displaystyle =8a^\frac{3}{2}(h+a)^\frac{3}{2}$

For the second term, we have:

$\displaystyle (4a^2)^\frac{3}{2}$

Simplify:

$\displaystyle =(2a)^3$

Hence:

$\displaystyle =8a^3$

Thus, our equation becomes:

$\displaystyle S=\frac{\pi}{3a}\Big[8a^\frac{3}{2}(h+a)^\frac{3}{2}-8a^3\Big]$

We can factor out an 8a^(3/2). Hence:

$\displaystyle S=\frac{\pi}{3a}(8a^\frac{3}{2})\Big[(h+a)^\frac{3}{2}-a^\frac{3}{2}\Big]$

Simplify:

$\displaystyle S=\frac{8\pi}{3}\sqrt{a}\Big[(h+a)^\frac{3}{2}-a^\frac{3}{2}\Big]$

Hence, we have verified the surface area generated by the function.

Part B)

We have:

$y^2=36x$

We can rewrite this as:

$y^2=4(9)x$

Hence, a=9.

The surface area is 1000. So, S=1000.

Therefore, with our equation:

$\displaystyle S=\frac{8\pi}{3}\sqrt{a}\Big[(h+a)^\frac{3}{2}-a^\frac{3}{2}\Big]$

We can write:

$\displaystyle 1000=\frac{8\pi}{3}\sqrt{9}\Big[(h+9)^\frac{3}{2}-9^\frac{3}{2}\Big]$

Solve for h. Simplify:

$\displaystyle 1000=8\pi\Big[(h+9)^\frac{3}{2}-27\Big]$

Divide both sides by 8π:

$\displaystyle \frac{125}{\pi}=(h+9)^\frac{3}{2}-27$

Isolate term:

$\displaystyle \frac{125}{\pi}+27=(h+9)^\frac{3}{2}$

Raise both sides to 2/3:

$\displaystyle \Big(\frac{125}{\pi}+27\Big)^\frac{2}{3}=h+9$

Hence, the value of h is:

$\displaystyle h=\Big(\frac{125}{\pi}+27\Big)^\frac{2}{3}-9\approx7.4614$

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