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2 years ago

8

5

1 answer:

nignag [31]2 years ago

3 0

Answer:60hours

Step-by-step explanation:

465/155×4

=60 hours

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Using the Law of Cosines, in triangle DEF, if e=18yd, d=10yd, f=22yd, find measurement of angle D

Ivahew [28]

Check the picture below.

$\bf \textit{Law of Cosines}\\ \quad \\
c^2 = {{ a}}^2+{{ b}}^2-(2{{ a}}{{ b}})cos(C)\implies 
c = \sqrt{{{ a}}^2+{{ b}}^2-(2{{ a}}{{ b}})cos(C)}
\\\\\\
\cfrac{{{ a}}^2+{{ b}}^2-c^2}{2{{ a}}{{ b}}}=cos(C)\implies cos^{-1}\left(\cfrac{{{ a}}^2+{{ b}}^2-c^2}{2{{ a}}{{ b}}}\right)=\measuredangle C\\\\
-------------------------------\\\\
\begin{cases}
d=10\\
e=18\\
f=22
\end{cases}\implies cos^{-1}\left(\cfrac{{{ 18}}^2+{{ 22}}^2-10^2}{2(18)(22)}\right)=\measuredangle D$

$\bf \implies cos^{-1}\left(0.893\overline{93}\right)=\measuredangle D\implies 26.63^o\approx \measuredangle D$

7 0

3 years ago

Helppppppppppppppppppp me Write this rate as a unit rate. 7/5 miles 2/3 hour

malfutka [58]

Answer: Your answer is 3.5 miles per hour.

Step-by-step explanation:

7 0

3 years ago

John paid $300 for a new xbox with a discount of 40% what was the original price before discount

irakobra [83]

Answer:

$500 is the original price

Step-by-step explanation:

0.6x = 300 Divide 0.6 on both sides

x = $500

3 0

2 years ago

Read 2 more answers

Choose the missing step in the given solution to the inequality −6x − 10 > 14 + 2x. −6x − 10 > 14 + 2x _______________ −8x

Vanyuwa [196]

-6x - 10 > 14 + 2x

<u>-2x </u> <u> -2x </u>

-8x - 10 > 14 THIS IS THE MISSING STEP

<u> +10</u> <u>+10 </u>

-8x > 24

$\frac{-8x}{-8}$ $\frac{24}{-8}$

x < -3

Answer: x < -3

8 0

2 years ago

Find the solution of the initial value problem<br><br> dy/dx=(-2x+y)^2-7 ,y(0)=0

Leokris [45]

Substitute $v(x)=-2x+y(x)$ , so that $\dfrac{\mathrm dv}{\mathrm dx}=-2+\dfrac{\mathrm dy}{\mathrm dx}$ . Then the ODE is equivalent to

$\dfrac{\mathrm dv}{\mathrm dx}+2=v^2-7$

which is separable as

$\dfrac{\mathrm dv}{v^2-9}=\mathrm dx$

Split the left side into partial fractions,

$\dfrac1{v^2-9}=\dfrac16\left(\dfrac1{v-3}-\dfrac1{v+3}\right)$

so that integrating both sides is trivial and we get

$\dfrac{\ln|v-3|-\ln|v+3|}6=x+C$

$\ln\left|\dfrac{v-3}{v+3}\right|=6x+C$

$\dfrac{v-3}{v+3}=Ce^{6x}$

$\dfrac{v+3-6}{v+3}=1-\dfrac6{v+3}=Ce^{6x}$

$\dfrac6{v+3}=1-Ce^{6x}$

$v=\dfrac6{1-Ce^{6x}}-3$

$-2x+y=\dfrac6{1-Ce^{6x}}-3$

$y=2x+\dfrac6{1-Ce^{6x}}-3$

Given the initial condition $y(0)=0$ , we find

$0=\dfrac6{1-C}-3\implies C=-1$

so that the ODE has the particular solution,

$\boxed{y=2x+\dfrac6{1+e^{6x}}-3}$

5 0

3 years ago