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2 years ago

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Three equivalent forms of a quadratic function are given. Select the form which reveals the zeros of f(x) without needing to alt

er the equation. Then, select all values for x from the given list when f(x) = 0.

1 answer:

Vanyuwa [196]2 years ago

4 0

Answer:

Read explanation and apply it

Step-by-step explanation:

The vertex form of a quadratic equation is $f(x)=a(x-h)^2+k$

The general form of a quadratic equation is $f(x)=ax^2+bx+c$

The factored form of a quadratic equation is $f(x)=a(x-p)(x-q)$

The one which reveals the zeros of f(x) without needing to alter the equation is the factored form $f(x)=a(x-p)(x-q)$ .

When f(x)=0, we have $a(x-p)(x-q)=0$

The values of x are $x=p$ and $x=q$ .

I hope this guides you to choose the correct answer.

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Find the exact value of cos(sin^-1(-5/13))

son4ous [18]

bearing in mind that the hypotenuse is never negative, since it's just a distance unit, so if an angle has a sine ratio of -(5/13) the negative must be the numerator, namely -5/13.

$\bf cos\left[ sin^{-1}\left( -\cfrac{5}{13} \right) \right] \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{then we can say that}~\hfill }{sin^{-1}\left( -\cfrac{5}{13} \right)\implies \theta }\qquad \qquad \stackrel{\textit{therefore then}~\hfill }{sin(\theta )=\cfrac{\stackrel{opposite}{-5}}{\stackrel{hypotenuse}{13}}}\impliedby \textit{let's find the \underline{adjacent}}$

$\bf \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{13^2-(-5)^2}=a\implies \pm\sqrt{144}=a\implies \pm 12=a \\\\[-0.35em] ~\dotfill\\\\ cos\left[ sin^{-1}\left( -\cfrac{5}{13} \right) \right]\implies cos(\theta )=\cfrac{\stackrel{adjacent}{\pm 12}}{13}$

le's bear in mind that the sine is negative on both the III and IV Quadrants, so both angles are feasible for this sine and therefore, for the III Quadrant we'd have a negative cosine, and for the IV Quadrant we'd have a positive cosine.

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2 years ago

A continuous probability distribution having a rectangular shape, where the probability is evenly distributed over an interval o

aleksklad [387]

Answer:

A continuous probability distribution having a rectangular shape, where the probability is evenly distributed over an interval of numbers is a(n) __uniform__________ distribution

Step-by-step explanation:

Given that there is a continuous probability distribution having a rectangular shape, where the probability is evenly distributed over an interval of numbers

Since the pdf is rectangular in shape and total probability is one we can say all values in the interval would be equally likely

Say if the interval is (a,b) P(X) = p the same for all places

Since total probability is 1,

we get integral of P(X)=p(b-a) =1

Or p= $\frac{1}{b-a}$

this is nothing but a uniform distribution continuous defined in the interval

A continuous probability distribution having a rectangular shape, where the probability is evenly distributed over an interval of numbers is a(n) __uniform__________ distribution

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2 years ago

What is the measure of EAB in circle F?<br> 72°<br> 92°<br> 148°<br> 200°

Arada [10]

We have been given a diagram. We are asked find the measure of arc EAB.

First of all, we will find the measure of arcs ED and CB using our given information.

We know that measure of an inscribed angle is half the measure of intercepted arc.

We can see that angle EBC is inscribed angle of arc EDC, so measure of arc EDC will be twice the measure of angle EBC.

$\widehat{EDC}=2\times m\angle EBC$

$\widehat{EDC}=2\times 80^{\circ}$

$\widehat{EDC}=160^{\circ}$

Similarly, we will find the measure of arc DCB.

$\widehat{DCB}=2\times m\angle DEB$

$\widehat{DCB}=2\times 70^{\circ}$

$\widehat{DCB}=140^{\circ}$

$\widehat{ED}=\widehat{EDC}-\wdiehat{DC}$

$\widehat{ED}=160^{\circ}-88^{\circ}$

$\widehat{ED}=72^{\circ}$

$\widehat{ED}+\widehat{DCB}+\widehat{EAB}=360^{\circ}$

$72^{\circ}+140^{\circ}+\widehat{EAB}=360^{\circ}$

$212^{\circ}+\widehat{EAB}=360^{\circ}$

$\widehat{EAB}=360^{\circ}-212^{\circ}$

$\widehat{EAB}=148^{\circ}$

Therefore, the measure of arc EAB is 148 degrees and option C is the correct choice.

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2 years ago

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6. a semicircle has as its diameter the hypotenuse of a right triangle shown below. determine the area of the semicircle to the

jeyben [28]

Answer:

$A = 137.3cm^2$

Step-by-step explanation:

Given

See attachment

Required

The area of the semicircle

First, we calculate the hypotenuse (h) of the triangle

Considering only the triangle, we have:

$\cos(68) = \frac{7}{h}$ --- cosine formula

Make h the subject

$h = \frac{7}{\cos(68)}$

$h = \frac{7}{0.3746}$

$h = 18.7$

The area of the semicircle is then calculated as:

$A = \frac{\pi h^2}{8}$

This gives:

$A = \frac{3.14 * 18.7^2}{8}$

$A = \frac{1098.03}{8}$

$A = 137.3cm^2$

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2 years ago

What does 4.57x0.08

jeka57 [31]

Answer:

0.3656 :)

Step-by-step explanation:

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2 years ago

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