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3 years ago

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A parabola has a line of symmetry x = -5. The minimum value of the quadratic function that it represents is -7. Find a possible

equation of this parabola and explain how you found it

1 answer:

Ilya [14]3 years ago

6 0

hello :<span>
<span>the parabola's equation is : f(x) = a(x-h)²+k
the verex is (h,k)
</span></span><span>line of symmetry x = h
</span><span>The minimum or maximum value is : k
</span>a possible equation of this parabola is : f(x) = a(x+5)²-7

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Write the equation of the line through (6,4) that is parallel to the line whose equation is y= 1/2 x+7. Show all work below and

soldier1979 [14.2K]

Answer:

$\displaystyle y = \frac{1}{2}x + 1$

Step-by-step explanation:

4 = ½[6] + b

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$\displaystyle 1 = b \\ \\ y = \frac{1}{2}x + 1$

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3 years ago

Demand for Tablet Computers The quantity demanded per month, x, of a certain make of tablet computer is related to the average u

soldier1979 [14.2K]

$x = f ( p ) = \frac { 100 } { 9 } \sqrt { 810,000 - p ^ { 2 } } } \\\\ \qquad { p ( t ) = \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { t } } + 200 \quad ( 0 \leq t \leq 60 ) }$

Answer:

12.0 tablet computers/month

Step-by-step explanation:

The average price of the tablet 25 months from now will be:

$p ( 25) = \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { 25 } } + 200 \\= \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \times 5 } + 200\\\\=\dfrac { 400 } { 1 + \dfrac { 5 } { 8 } } + 200\\p(25)=\dfrac { 5800 } {13}$

Next, we determine the rate at which the quantity demanded changes with respect to time.

Using Chain Rule (and a calculator)

$\dfrac{dx}{dt}= \dfrac{dx}{dp}\dfrac{dp}{dt}$

$\dfrac{dx}{dp}= \dfrac{d}{dp}\left[{ \dfrac { 100 } { 9 } \sqrt { 810,000 - p ^ { 2 } } }\right] =-\dfrac{100}{9}p(810,000-p^2)^{-1/2}$

$\dfrac{dp}{dt}=\dfrac{d}{dt}\left[\dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { t } } + 200 \right]=-25\left[1 + \dfrac { 1 } { 8 } \sqrt { t } \right]^{-2}t^{-1/2}$

Therefore:

$\dfrac{dx}{dt}= \left[-\dfrac{100}{9}p(810,000-p^2)^{-1/2}\right]\left[-25\left[1 + \dfrac { 1 } { 8 } \sqrt { t } \right]^{-2}t^{-1/2}\right]$

Recall that at t=25, $p(25)=\dfrac { 5800 } {13} \approx 446.15$

Therefore:

$\dfrac{dx}{dt}(25)= \left[-\dfrac{100}{9}\times 446.15(810,000-446.15^2)^{-1/2}\right]\left[-25\left[1 + \dfrac { 1 } { 8 } \sqrt {25} \right]^{-2}25^{-1/2}\right]\\=12.009$

The quantity demanded per month of the tablet computers will be changing at a rate of 12 tablet computers/month correct to 1 decimal place.

8 0

3 years ago

Holly wants to save money for an emergency. Holly invests 1,500 in an account that pays an interest rate of 6.75%. How many year

lina2011 [118]

The formula is
A=p (1+r)^t
A future value 7300
P present value 1500
R interest rate 0.0675
T time?
7300=1500 (1+0.0675)^t
Solve for t
Divide both sides by 1500
7300/1500=1.0675^t
Take the log for both sides
Log (7300/1500)=t×log (1.0675)
Divide both sides by log (1.0675)
T=log(7,300÷1,500)÷log(1.0675)
T=24.2 years round your answer to get 24 years

Hope it helps!

4 0

3 years ago

Read 2 more answers

Which equation shows the quadratic formula used correctly to solve 5x2 + 3x – 4 = 0 for x? x = StartFraction negative 3 plus-or-

Brrunno [24]

Answer: FIRST OPTION

Step-by-step explanation:

<h3> The missing picture is attached.</h3>

By definition, given a Quadratic equation in the form:

$ax^2+bx+c=0$

Where "a", "b" and "c" are numerical coefficients and "x" is the unknown variable, you caN use the Quadratic Formula to solve it.

The Quadratic Formula is the following:

$x=\frac{-b \±\sqrt{b^2-4ac} }{2a}$

In this case, the exercise gives you this Quadratic equation:

$5x^2 + 3x - 4 = 0$

You can identify that the numerical coefficients are:

$a=5\\\\b=3\\\\c= - 4$

Therefore, you can substitute values into the Quadratic formula shown above:

$x=\frac{-b \±\sqrt{b^2-4ac} }{2a}\\\\x=\frac{-3 \±\sqrt{(3)^2-4(5)(-4)} }{2(5)}$

You can identify that the equation that shows the Quadratic formula used correctly to solve the Quadratic equation given in the exercise for "x", is the one shown in the First option.

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3 years ago

Read 2 more answers

Evaluate the integral. (sec2(t) i t(t2 1)8 j t7 ln(t) k) dt

polet [3.4K]

If you're just integrating a vector-valued function, you just integrate each component:

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$=\displaystyle\left(\int\sec^2t\,\mathrm dt\right)\hat\imath+\left(\int t(t^2-1)^8\,\mathrm dt\right)\hat\jmath+\left(\int t^7\ln t\,\mathrm dt\right)\hat k$

The first integral is trivial since $(\tan t)'=\sec^2t$ .

The second can be done by substituting $u=t^2-1$ :

$u=t^2-1\implies\mathrm du=2t\,\mathrm dt\implies\displaystyle\frac12\int u^8\,\mathrm du=\frac1{18}(t^2-1)^9+C$

The third can be found by integrating by parts:

$u=\ln t\implies\mathrm du=\dfrac{\mathrm dt}t$

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3 years ago