All categories

2 years ago

Hey! please help i’ll give brainliest

Mathematics

2 answers:

Kitty [74]2 years ago

8 0

Answer:

Anxeity (Option 4)

Step-by-step explanation:

They would most likley be anxious and nervous about what they scored on their test while waiting for the results

MariettaO [177]2 years ago

3 0

Answer:

I think its C

hope this helps

have a good day :)

Step-by-step explanation:

You might be interested in

Write an equation for the line passing through the points (-5,15) and (20,25). Show how you found your answer.

olasank [31]

The point-slope form:

$y-y_1=m(x-x_1)\\\\m=\dfrac{y_2-y_1}{x_2-x_1}$

We have the points (-5, 15) and (20, 25). substitute:

$m=\dfrac{25-15}{20-(-5)}=\dfrac{10}{25}=\dfrac{2}{5}\\\\y-15=\dfrac{2}{5}(x-(-5))\\\\y-15=\dfrac{2}{5}(x+5)\qquad|\text{use distributive property}\\\\y-15=\dfrac{2}{5}x+2\qquad|\text{add 15 from both sides}\\\\y=\dfrac{2}{5}x+17\qquad|\text{multiply both sides by 5}\\\\5y=2x+85\qquad|\text{subtract 2x from both sides}\\\\-2x+5y=85\qquad|\text{change the signs}\\\\2x-5y=-85$

Answer:

point-slope form: $y-15=\dfrac{2}{5}(x+5)$

slope-intercept form: $y=\dfrac{2}{5}x+17$

standard form: $2x-5y=-85$

7 0

3 years ago

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:

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