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2 years ago

11

Marisol is building a frame for a sandbox but the boards she has is too short she must join 2 boards together to build as shown

at what angle did she cut the first board

1 answer:

ira [324]2 years ago

7 0

The anwser would be a 90 degree angle because t batter what the frame would be 360 degrees cuz each angle being 90 degrees

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The function f is such that f(x) = x^2 – 2x + 3

aev [14]

Answer:

x = - 4, x = 6

Step-by-step explanation:

Given

f(x) = x² - 2x + 3 and f(x) = 27, then equating the 2 gives

x² - 2x + 3 = 27 ( subtract 27 from both sides )

x² - 2x - 24 = 0 ← in standard form

Consider the factors of the constant term (- 24) which sum to give the coefficient of the x- term (- 2)

The factors are - 6 and + 4, since

- 6 × 4 = - 24 and - 6 + 4 = - 2, thus

(x - 6)(x + 4) = 0

Equate each factor to zero and solve for x

x - 6 = 0 ⇒ x = 6

x + 4 = 0 ⇒ x = - 4

4 0

2 years ago

Read 2 more answers

Example of propertional relationship

Leokris [45]

Answer:

1/2 and 6/12 have a proportional relationship, which means they are the same.

Step-by-step explanation:

3 0

2 years ago

Add fractions with unlike denominators

DIA [1.3K]

3/10 + 2/6
use their lowest common multiple — which is 30 so:
3/10 -> 9/30 (multiply by 3)
2/6 -> 10/30 (multiply by 5)
then we can get the new equation of:
9/30 + 10/30 = 19/30

6 0

2 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

2 years ago

Please Solve and explain how to find the surface area of these triangular prisms

BARSIC [14]

<span>I think these are the right answers.
</span>
<span>---------------------------------------------------------------------------

For the first one:

The volume of the triangular prism is: 1030.25cm
The surface area of the triangular prism is:</span><span> 696.95cm

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For the second one:
</span>
T<span><span>he volume of the triangular prism is: 42cm
</span>The surface area of the triangular prism is: 96cm</span>
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As for explaining how to find the volume of triangular prisms, use your G00GLE search engine and type "find the volume of triangular prism".

Enter in the data/numbers and it will calculate the answer along with step by step instructions on how it got the answer.

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Have a good day!!!

4 0

2 years ago