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3 years ago

Suppose f(x) = x2 and g(x)=(6)

Mathematics

1 answer:

hodyreva [135]3 years ago

4 0

Answer:

In Exercises 1-9, describe the transformation of f(x) = x² represented by g. ... 14. f(x) = 4x2 + 5; horizontal stretch by a factor of 2 and a translation 2 units up, ... graph of g be a horizontal shrink by a factor of the graph of f(x) = x2 x 2x ... 4(x-3)?–4.

Step-by-step explanation:

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Y =28547, 11504 is obese, what is the %

Damm [24]

Huh? What does that even mean lol

3 0

3 years ago

Find the area of the triangle ABC with the coordinates of A(10, 15) B(15, 15) C(30, 9).

lions [1.4K]

Check the picture below. so, that'd be the triangle's sides hmmm so let's use Heron's Area formula for it.

$~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ (\stackrel{x_1}{10}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{15}~,~\stackrel{y_2}{15}) ~\hfill a=\sqrt{[ 15- 10]^2 + [ 15- 5]^2} \\\\\\ ~\hfill \boxed{a=\sqrt{125}} \\\\\\ (\stackrel{x_1}{15}~,~\stackrel{y_1}{15})\qquad (\stackrel{x_2}{30}~,~\stackrel{y_2}{9}) ~\hfill b=\sqrt{[ 30- 15]^2 + [ 9- 15]^2} \\\\\\ ~\hfill \boxed{b=\sqrt{261}}$

$(\stackrel{x_1}{30}~,~\stackrel{y_1}{9})\qquad (\stackrel{x_2}{10}~,~\stackrel{y_2}{5}) ~\hfill c=\sqrt{[ 10- 30]^2 + [ 5- 9]^2} \\\\\\ ~\hfill \boxed{c=\sqrt{416}} \\\\[-0.35em] ~\dotfill$

$\qquad \textit{Heron's area formula} \\\\ A=\sqrt{s(s-a)(s-b)(s-c)}\qquad \begin{cases} s=\frac{a+b+c}{2}\\[-0.5em] \hrulefill\\ a=\sqrt{125}\\ b=\sqrt{261}\\ c=\sqrt{416}\\ s\approx 23.87 \end{cases} \\\\\\ A\approx\sqrt{23.87(23.87-\sqrt{125})(23.87-\sqrt{261})(23.87-\sqrt{416})}\implies \boxed{A\approx 90}$

6 0

2 years ago

Solve 3x^2 + 6x+15=0

Elden [556K]

Answer:

The solution of given equation is ( - 1 + 2 i ) , ( - 1 - 2 i )

Step-by-step explanation:

Given equation as :

3 x² + 6 x +15 = 0

The value of x fro the quadratic equation a x² + b x + c = 0 is obtained as

x = $\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$

So , from given eq , the value of x is now obtain as

x = $\frac{-b\pm \sqrt{b^{2}-4\times a\times c}}{2\times a}$

Or, x = $\frac{-6\pm \sqrt{6^{2}-4\times 3\times 15}}{2\times 3}$

Or, x = $\frac{\sqrt{-144} }{6}$

∴ x = ( - 1 + 2 i ) , ( - 1 - 2 i )

Hence The solution of given equation is ( - 1 + 2 i ) , ( - 1 - 2 i ) Answer

6 0

4 years ago

In Triangle XYZ, measure of angle X = 49° , XY = 18°, and

marissa [1.9K]

Answer:

There are two choices for angle Y: $Y \approx 54.987^{\circ}$ for $XZ \approx 15.193$ , $Y \approx 27.008^{\circ}$ for $XZ \approx 8.424$ .

Step-by-step explanation:

There are mistakes in the statement, correct form is now described:

<em>In triangle XYZ, measure of angle X = 49°, XY = 18 and YZ = 14. Find the measure of angle Y:</em>

The line segment XY is opposite to angle Z and the line segment YZ is opposite to angle X. We can determine the length of the line segment XZ by the Law of Cosine:

$YZ^{2} = XZ^{2} + XY^{2} -2\cdot XY\cdot XZ \cdot \cos X$ (1)