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il63 [147K]
2 years ago
11

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
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