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3 years ago

What is the most difficult in equations??

1 answer:

3 0

I'll go with graphing cause when you try graphing a picewise function for example it's much harder to graph because it's to many numbers and you can't figure out what to graph especially for me.) mark me brainliest please

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How many pounds of coffee worth $9 a pound should be added to 10 pounds of coffee worth $3 a pound to get a mixture worth $7 a p

Dima020 [189]

Here is your answer......

6 0

3 years ago

Which sum will be irrational

ivolga24 [154]

<h2>Hey there! </h2>

<h3>The answer is: </h3>

<h3>Option 'B' </h3>

$\sqrt{19} + \frac{7}{2}$

<h2>BECAUSE root 19 is irrational. </h2>

5 0

2 years ago

PLZ ANSWER: A sandwich store charges a $10 delivery fee and $4.50 for each sandwich which is the total cost language and deliver

Irina-Kira [14]

Answer

$37

Explanation

it costs $4.50 per sandwich so you multiply that by 6 which gets you $27. then you add the delivery fee ($10) to that and you get $37

hope this helps!!

6 0

2 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 <em>y</em> is a function of <em>x</em> :

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

(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 ≤ <em>x</em> ≤ 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

The lines 3y + 1 = 6x + 4 and 2y + 1 = x - 9 are ?

harkovskaia [24]

Answer:

lines are not perpendicular or parallel

6 0

2 years ago