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3 years ago

12

Can someone tell me the answer. i don’t understand this.

1 answer:

dangina [55]3 years ago

3 0

The angle that is included is <m because both lines share the M point.

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Find the slope of a line parallel to this given line

dybincka [34]

bearing in mind that parallel lines have the same exact slope, then any parallel line to the one above will have the same slope as that one.

$\bf \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array} \qquad \implies y = \stackrel{\stackrel{m}{\downarrow }}{-\cfrac{7}{3}}x-4$

3 0

3 years ago

Help plz ASAP !!!!!!!!!!!!!!!!!!!!!!!!!!

Yakvenalex [24]

Answer:

-2

Step-by-step explanation:

Let's solve your equation step-by-step.

−(3v+1)+7(6v+6)=−37

Step 1: Simplify both sides of the equation.

−(3v+1)+7(6v+6)=−37

−3v+−1+(7)(6v)+(7)(6)=−37(Distribute)

−3v+−1+42v+42=−37

(−3v+42v)+(−1+42)=−37(Combine Like Terms)

39v+41=−37

39v+41=−37

Step 2: Subtract 41 from both sides.

39v+41−41=−37−41

39v=−78

Step 3: Divide both sides by 39.

39v

39

=

−78

39

v=−2

Answer:

v=−2

8 0

3 years ago

Read 2 more answers

Need help w this please

rusak2 [61]

Solved using proportions.

8 0

3 years ago

Does anyone have a coursehero account that I can borrow

vagabundo [1.1K]

Answer:

No but an easy way is to just make up random email names and passwords and on each acc u get 5 free questions,

Step-by-step explanation:

Can u pls mark me brainliest

5 0

3 years ago

Haya la integral de xarcsenx/((1-x^2)^1/2). Utiliza la sustitución de t= arcsenx

ikadub [295]

Answer:

<em>Explicación más abajo</em>

Step-by-step explanation:

Integración Indefinida

La integral

$I=\int \dfrac{x .arcsen\left(x\right)}{\sqrt{1-x^2}}$

Se resuelve con el cambio de variables:

t=arcsen(x)

Una vez hechos los cambios, la integral se resuelve en función de t:

$I=sen(t)-t.cos(t)+C$

Hay que devolver los cambios para mostrarla en función de x.

El cambio de variables también se puede escribir:

x=sen(t)

Recordando que

$cos(t)=\sqrt{1-sen^2(t)}$

Entonces:

$cos(t)=\sqrt{1-x^2}$

Devolviendo los cambios:

$I=sen(t)-t.cos(t)+C=x-arcsen(x)\sqrt{1-x^2}+C$

Es la respuesta correcta

7 0

3 years ago