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3 years ago

Solve for x in the diagram in the pic

Mathematics

1 answer:

julia-pushkina [17]3 years ago

8 0

Answer:

x=34

Step-by-step explanation:

The angle measure of a straight line is 180 degrees.

So if you add up all of the angles to make an equation, you get 2x + 112=180.

Then solve.

2x + 112=180

-112

2x=68

x=34

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Si 15 obreros construyen un muro en 45 minutos, ¿en cuánto tiempo lo construirán 3 obreros con similar habilidad?

Alexxx [7]

Answer:

3 obreros tardaran 225 minutos en completar el trabajo.

Step-by-step explanation:

Sea R la velocidad con la que un obrero puede construir 1 muro.

Sabemos que 15 obreros construyen un muro en 45 minutos, entonces:

(15*R)*45min = 1 muro

Con esta ecuación podemos encontrar el valor R

R = (1 muro)/(15*45min)

Ahora queremos saber cuando tiempo tardan 3 obreros en construir un muro, entonces tenemos que resolver:

(3*R)*T = 1 muro

Reemplazando el valor de R obtenemos:

(3*(1 muro)/(15*45min))*T = 1 muro

T = (1 muro)*(15*45 min)/(3*1 muro)

T = 5*45min = 225 min

3 obreros tardaran 225 minutos en completar el trabajo.

7 0

3 years ago

Sally decides to buy quarts of ice cream at $5 each and boxes of pizza at $10 each for her softball team.

MakcuM [25]

Answer:

5x + 10y

5*4 + 10*5

20+50 = $70

Step-by-step explanation:

8 0

3 years ago

Y = 12 <br>2x -y = 4<br><br>What is the solution of the system?

Keith_Richards [23]

2x-12=4
put the 12 in place of y since they equal each other
2x=16
add 12 to both sides
x=8
divide both sides by 2

6 0

4 years ago

Read 2 more answers

Relative extrema of f(x)=(x+3)/(x-2)

Salsk061 [2.6K]

Answer:

$\displaystyle f(x) = \frac{x + 3}{x - 2}$ has no relative extrema when the domain is $\mathbb{R} \backslash \lbrace 2 \rbrace$ (the set of all real numbers other than $2$ .)

Step-by-step explanation:

Assume that the domain of $\displaystyle f(x) = \frac{x + 3}{x - 2}$ is $\mathbb{R} \backslash \lbrace 2 \rbrace$ (the set of all real numbers other than $2$ .)

Let $f^{\prime}(x)$ and $f^{\prime\prime}(x)$ denote the first and second derivative of this function at $x$ .

Since this domain is an open interval, $x = a$ is a relative extremum of this function if and only if $f^{\prime}(a) = 0$ and $f^{\prime\prime}(a) \ne 0$ .

Hence, if it could be shown that $f^{\prime}(x) \ne 0$ for all $x \in \mathbb{R} \backslash \lbrace 2 \rbrace$ , one could conclude that it is impossible for $\displaystyle f(x) = \frac{x + 3}{x - 2}$ to have any relative extrema over this domain- regardless of the value of $f^{\prime\prime}(x)$ .

$\displaystyle f(x) = \frac{x + 3}{x - 2} = (x + 3) \, (x - 2)^{-1}$ .

Apply the product rule and the power rule to find $f^{\prime}(x)$ .

$\begin{aligned}f^{\prime}(x) &= \frac{d}{dx} \left[ (x + 3) \, (x - 2)^{-1}\right] \\ &= \left(\frac{d}{dx}\, [(x + 3)]\right)\, (x - 2)^{-1} \\ &\quad\quad (x + 3)\, \left(\frac{d}{dx}\, [(x - 2)^{-1}]\right) \\ &= (x - 2)^{-1} \\ &\quad\quad+ (x + 3) \, \left[(-1)\, (x - 2)^{-2}\, \left(\frac{d}{dx}\, [(x - 2)]\right) \right] \\ &= \frac{1}{x - 2} + \frac{-(x+ 3)}{(x - 2)^{2}} \\ &= \frac{(x - 2) - (x + 3)}{(x - 2)^{2}} = \frac{-5}{(x - 2)^{2}}\end{aligned}$ .

In other words, $\displaystyle f^{\prime}(x) = \frac{-5}{(x - 2)^{2}}$ for all $x \in \mathbb{R} \backslash \lbrace 2 \rbrace$ .

Since the numerator of this fraction is a non-zero constant, $f^{\prime}(x) \ne 0$ for all $x \in \mathbb{R} \backslash \lbrace 2 \rbrace$ . (To be precise, $f^{\prime}(x) < 0$ for all $x \in \mathbb{R} \backslash \lbrace 2 \rbrace\!$ .)

Hence, regardless of the value of $f^{\prime\prime}(x)$ , the function $f(x)$ would have no relative extrema over the domain $x \in \mathbb{R} \backslash \lbrace 2 \rbrace$ .

7 0

3 years ago

PLEASE HELP!!!!!!!!!!!!!

Alenkasestr [34]

Answer:

Tasha made her mistake in step 4.

7 0

3 years ago

Read 2 more answers