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3 years ago

Find the measure of Angle 5. measure of angle 7 = 2x+15 measure of angle 8 = 3x

Mathematics

1 answer:

belka [17]3 years ago

5 0

*<span> 7 = 2x+15
-2X = 15 - 7
X = 8/-2
x = -8/2

*8 = 3x
3x = 8
x = 8/3 </span><span> </span><span>

</span>

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If 3 to the power y = x then y = ?

Oksanka [162]

Answer:

$y=\log_3(x)$

Step-by-step explanation:

$3^y=x$

$\ln(3^y)=\ln(x)$

$y\ln(3)=\ln(x)$

$y=\frac{\displaystyle\ln(x)}{\displaystyle\ln(3)}$

but if

$f(x)=\frac{\displaystyle\ln(x)}{\displaystyle\ln(3)}$

then since

$f(3)=\frac{\displaystyle\ln(3)}{\displaystyle\ln(3)}=1$

$f(x)=\log_3(x)$

$y=\log_3(x)$

$3^y=x$

$\log_3(3^y)=\log_3(x)$

$y\log_3(3)=\log_3(x)$

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6 0

3 years ago

The management of an electronics firm has determined that the profit the company earns is a function of thenumber of employees i

ruslelena [56]

we have the function

$P(x)=\ln\left|4x+1\right|+3x-x^2$

For x=2

substitute

$\begin{gathered} P(x)=\operatorname{\ln}\lvert4(2)+1\rvert+3(2)-(2)^2 \\ P(x)=\operatorname{\ln}\lvert9\rvert+2 \\ P(x)=4.20\text{ millions of dollars} \end{gathered}$

therefore

<h2>the answer is 4.20 million dollars</h2>

6 0

1 year ago

What isthe answer of: 2/3 - 1/2

Kaylis [27]

Answer:

1/6

Step-by-step explanation:

first of all find the LCD of 3 and 2. The LCD would be 6.

2/3(2/2) - 1/2(3/3)

4/6 and 3/6

then you will subtract 4 and 3 and you end up with 1/6

5 0

3 years ago

Let F⃗ =2(x+y)i⃗ +8sin(y)j⃗ .

Alik [6]

Answer:

-42

Step-by-step explanation:

The objective is to find the line integral of $F$ around the perimeter of the rectangle with corners (4,0), (4,3), (−3,3), (−3,0), traversed in that order.

We will use <em>the Green's Theorem </em>to evaluate this integral. The rectangle is presented below.

We have that

$F(x,y) = 2(x+y)i + 8j \sin y = \langle 2(x+y), 8\sin y \rangle$

Therefore,

$P(x,y) = 2(x+y) \quad \wedge \quad Q(x,y) = 8\sin y$

Let's calculate the needed partial derivatives.

$P_y = \frac{\partial P}{\partial y} (x,y) = (2(x+y))'_y = 2\\Q_x =\frac{\partial Q}{\partial x} (x,y) = (8\sin y)'_x = 0$

Thus,

$Q_x -P_y = 0 -2 = - 2$

Now, by the Green's theorem, we have

$\oint_C F \,dr = \iint_D (Q_x-P_y)\,dA = \int \limits_{-3}^{4} \int \limits_{0}^{3} (-2)\,dy\, dx \\ \\\phantom{\oint_C F \,dr = \iint_D (Q_x-P_y)\,dA}= \int \limits_{-3}^{4} (-2y) \Big|_{0}^{3} \; dx\\ \phantom{\oint_C F \,dr = \iint_D (Q_x-P_y)\,dA}= \int \limits_{-3}^{4} (-6)\; dx = -6x \Big|_{-3}^{4} = -42$

4 0

3 years ago

Is it A. B. C. or D.

podryga [215]

Your answer would be b i'm pretty sure

5 0

3 years ago