All categories

3 years ago

Lines x and y are parallel.

Mathematics

2 answers:

densk [106]3 years ago

7 0

On the bottom line, the 106 and number 1 make a straight line and needs to equal 180 degrees.

This means number 1 = 180-106 = 74 degrees.

Because x and y are parallel, the top outside angle is the same as number 1.

6x +8 = 74

Subtract 8 from both sides:

6x = 66

divide both sides by 6:

x = 66 / 6

x = 11

Now you have x, replace x in the equation for 7x-2 to find that angle:

7(11) -2 = 77-2 = 75 degrees.

The three inside angles need to equal 180 degrees.

Angle 2 = 180 - 74 - 75 = 31 degrees.

fomenos3 years ago

4 0

Answer:

all work is shown and pictured

You might be interested in

Which equation best matches the scenario below?

siniylev [52]

Answer:

Definetly t = 4n

4 0

3 years ago

two companies offer different charges for text messaging company 1 charges 0.06 per text message while company 2 charges rates a

yanalaym [24]

Answer:

Step-by-step explanation:

I need help too

8 0

3 years ago

Olivia bowled 5 games. Her scores were 80, 95, 100, 85, and 95. What is the mean of her scores for the 5 games? * A. 20 B. 72 C.

egoroff_w [7]

The answer is gonna be C,which is 91.Hope this helps

5 0

4 years ago

Read 2 more answers

1.) Find the length of the arc of the graph x^4 = y^6 from x = 1 to x = 8.

xxTIMURxx [149]

First, rewrite the equation so that y is a function of x :

$x^4 = y^6 \implies \left(x^4\right)^{1/6} = \left(y^6\right)^{1/6} \implies x^{4/6} = y^{6/6} \implies y = x^{2/3}$

(If you were to plot the actual curve, you would have both $y=x^{2/3}$ and $y=-x^{2/3}$ , but one curve is a reflection of the other, so the arc length for 1 ≤ x ≤ 8 would be the same on both curves. It doesn't matter which "half-curve" you choose to work with.)

The arc length is then given by the definite integral,

$\displaystyle \int_1^8 \sqrt{1 + \left(\frac{\mathrm dy}{\mathrm dx}\right)^2}\,\mathrm dx$

We have

$y = x^{2/3} \implies \dfrac{\mathrm dy}{\mathrm dx} = \dfrac23x^{-1/3} \implies \left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2 = \dfrac49x^{-2/3}$

Then in the integral,

$\displaystyle \int_1^8 \sqrt{1 + \frac49x^{-2/3}}\,\mathrm dx = \int_1^8 \sqrt{\frac49x^{-2/3}}\sqrt{\frac94x^{2/3}+1}\,\mathrm dx = \int_1^8 \frac23x^{-1/3} \sqrt{\frac94x^{2/3}+1}\,\mathrm dx$

Substitute

$u = \dfrac94x^{2/3}+1 \text{ and } \mathrm du = \dfrac{18}{12}x^{-1/3}\,\mathrm dx = \dfrac32x^{-1/3}\,\mathrm dx$

This transforms the integral to

$\displaystyle \frac49 \int_{13/4}^{10} \sqrt{u}\,\mathrm du$

and computing it is trivial:

$\displaystyle \frac49 \int_{13/4}^{10} u^{1/2} \,\mathrm du = \frac49\cdot\frac23 u^{3/2}\bigg|_{13/4}^{10} = \frac8{27} \left(10^{3/2} - \left(\frac{13}4\right)^{3/2}\right)$

We can simplify this further to

$\displaystyle \frac8{27} \left(10\sqrt{10} - \frac{13\sqrt{13}}8\right) = \boxed{\frac{80\sqrt{10}-13\sqrt{13}}{27}}$

7 0

3 years ago

How many cm make 1m? Can some pls tell me from mm to km

Alborosie

Answer:

100cm= 1m

100000mm= 1km

5 0

3 years ago