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2 years ago

The graphs below have the same shape. What is the equation of the red graph? G(x)=___

Mathematics

1 answer:

aliya0001 [1]2 years ago

5 0

Answer:

g(x)= 4- $x^{4}$

Step-by-step explanation:

From the blue parabola, we would have to move up one unit to make the red parabola. This means that $g(x)= 4-x^{4}$ .

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PLSSSSSSSSSSS HELPPP>> Show work solve the sysytem- y=1/2x^2+4x+4 y=-4x+12 1/2

Mrac [35]

Answer:

(1, 8.5 ) and ( - 17, 80.5 )

Step-by-step explanation:

Given the 2 equations

y = $\frac{1}{2}$ x² + 4x + 4 → (1)

y = - 4x + 12 $\frac{1}{2}$ → (2)

Substitute (1) into (2), that is

$\frac{1}{2}$ x² + 4x + 4 = - 4x + 12 $\frac{1}{2}$ ( multiply through by 2 to clear the fractions )

x² + 8x + 8 = - 8x + 25 ( add 8x to both sides )

x² + 16x + 8 = 25 ( subtract 25 from both sides )

x² + 16x - 17 = 0 ← in standard form

(x + 17)(x - 1) = 0 ← in factored form

Equate each factor to zero and solve for x

x + 17 = 0 ⇒ x = - 17

x - 1 = 0 ⇒ x = 1

Substitute these values into either of the 2 equations and evaluate for y

Substituting into (2 )

x = 1 : y = - 4(1) + 12.5 = - 4 + 12.5 = 8.5 ⇒ (1, 8.5 )

x = - 17 : y = - 4(- 17) + 12.5 = 68 + 12.5 = 80.5 ⇒ (- 17, 80.5 )

5 0

3 years ago

A cone has a diameter of 9 inches and a height of 12 inches. Find the volume of the cone to the nearest tenth.

Vanyuwa [196]

1)
The formula for calculating the cone volume is:
$V = \frac{ \pi *r^2*h}{3}$

Data:
V (volume) = ?
h (height) → h = 12 in
d (diagonal) = 9 in , if: d = 2*r , soon: 9 = 2*r → 2r = 9 → r = 9/2 → r = 4.5 in
Use: $\pi \approx 3.14$

Solving:
$V = \frac{ \pi *r^2*h}{3}$
$V = \frac{ 3.14 *4.5^2*12}{3}$
$V = \frac{ 3.14 *20.25*12}{3}$
$V = \frac{763.02}{3}$
$\boxed{\boxed{V = 254.34\:in^3}}\end{array}}\qquad\quad\checkmark$

2)
The circle area formula is: $A = \pi * r^2$

Data:
A (area) = ?
d (diagonal) = 19 Km , if: d = 2*r , soon: 19 = 2*r → 2r = 19 → r = 19/2 → r = 9.5 Km
Use: $\pi \approx 3.14$

Solving:
$A = \pi * r^2$
$A = 3.14*9.5^2$
$A = 3.14*90.25$
$\boxed{\boxed{A = 283.385\:Km^2}}\end{array}}\qquad\quad\checkmark$

3) False, because it's the volume, the number of cubic units needed to fill a space.

6 0

3 years ago

Solve |x| + 7 < 4.

Kobotan [32]

Answer: ur inbox in y=my+b form

Step-by-step explanation:

5 0

3 years ago

Find the laplace transform by intergration<br> f(t)=tcosh(3t)

Shkiper50 [21]

$\mathcal L_s\{t\cosh3t\}=\displaystyle\int_0^\infty t\cosh3t e^{-st}\,\mathrm dt$

Integrate by parts, setting

$u_1=t\implies\mathrm du_1=\mathrm dt$
$\mathrm dv_1=\cosh3t e^{-st}\,\mathrm dt\implies v_1=\displaystyle\int\cosh3t e^{-st}\,\mathrm dt$

To evaluate $v_1$ , integrate by parts again, this time setting

$u_2=\cosh3t\implies\mathrm du_2=3\sinh3t\,\mathrm dt$
$\mathrm dv_2=\displaystyle\int e^{-st}\,\mathrm dt\implies v_2=-\frac1se^{-st}$

$\implies\displaystyle\int\cosh3te^{-st}\,\mathrm dt=-\frac1s\cosh3te^{-st}+\frac3s\int \sinh3te^{-st}$

Integrate by parts yet again, with

$u_3=\sinh3t\implies\mathrm du_3=3\cosh3t\,\mathrm dt$
$\mathrm dv_3=e^{-st}\,\mathrm dt\implies v_3=-\dfrac1se^{-st}$

$\implies\displaystyle\int\cosh3te^{-st}\,\mathrm dt=-\frac1s\cosh3te^{-st}+\frac3s\left(-\frac1s\sinh3te^{-st}+\frac3s\int\cosh3te^{-st}\,\mathrm dt\right)$
$\displaystyle\int\cosh3te^{-st}\,\mathrm dt=-\frac1s\cosh3te^{-st}-\frac3{s^2}\sinh3te^{-st}+\frac9{s^2}\int\cosh3te^{-st}\,\mathrm dt$
$\displaystyle\frac{s^2-9}{s^2}\int\cosh3te^{-st}\,\mathrm dt=-\frac1s\cosh3te^{-st}-\frac3{s^2}\sinh3te^{-st}$
$\implies\displaystyle\underbrace{\int\cosh3te^{-st}\,\mathrm dt}_{v_1}=-\frac{(s\cosh3t+3\sinh3t)e^{-st}}{s^2-9}$

So we have

$\displaystyle\int_0^\infty t\cosh3t e^{-st}\,\mathrm dt=u_1v_1\big|_{t=0}^{t\to\infty}-\int_0^\infty v_1\,\mathrm du_1$
$=\displaystyle-\frac{t(s\cosh3t+3\sinh3t)e^{-st}}{s^2-9}\bigg|_{t=0}^{t\to\infty}-\int_0^\infty \left(-\frac{(s\cosh3t+3\sinh3t)e^{-st}}{s^2-9}\right)\,\mathrm dt$
$=\displaystyle\frac1{s^2-9}\int_0^\infty(s\cosh3t+3\sinh3t)e^{-st}\,\mathrm dt$

We already have the antiderivative for the first term:

$\displaystyle\frac s{s^2-9}\int_0^\infty \cosh3te^{-st}\,\mathrm dt=\frac s{s^2-9}\left(-\frac{(s\cosh3t+3\sinh3t)e^{-st}}{s^2-9}\right)\bigg|_{t=0}^{t\to\infty}$
$=\dfrac{s^2}{(s^2-9)^2}$

And we can easily find the remaining term's antiderivative by integrating by parts (for the last time!), or by simply exchanging $\cosh$ with $\sinh$ in the derivation of $v_1$ , so that we have

$\displaystyle\frac3{s^2-9}\int_0^\infty\sinh3te^{-st}\,\mathrm dt=\frac3{s^2-9}\left(-\frac{(s\sinh3t+3\cosh3t)e^{-st}}{s^2-9}\right)\bigg|_{t=0}^{t\to\infty}$
$=\dfrac9{(s^2-9)^2}$

(The exchanging is permissible because $(\sinh x)'=\cosh x$ and $(\cosh x)'=\sinh x$ ; there are no alternating signs to account for.)

And so we conclude that

$\mathcal L_s\{t\cosh3t\}=\dfrac{s^2+9}{(s^2-9)^2}$

8 0

3 years ago

Please help I need help ASAP

Bogdan [553]

Answer:

rotational

Step-by-step explanation:

or revolutionary nakalimutan ko sagot ko basta dyan sa dalawang yan

3 0

3 years ago