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crimeas [40]
3 years ago
12

Find the diffrence between temperature -30 degrees c and 70 degrees c?

Mathematics
2 answers:
kati45 [8]3 years ago
6 0
One is very cold below freezing and one is just cold hope this helps
andreyandreev [35.5K]3 years ago
4 0
One is colder than the other. 
     Hope this helps :)
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6 0
3 years ago
WILL GIVE A BRAINLIEST IF UR CORRECT PLEASE HELP!!
Otrada [13]
A pair of numbers is a solution of y = -3x if the second number, y, is -3 times the first number, x.

What about the pair (-1, 4)?  If you replace x with -1, you get y = -3(-1) = 3, not 4.  So (-1, 4) is NOT a solution.

For each answer choice, you have to check all three of the given pairs.

(0, 0) is a solution because if x = 0, then y = -3(0) = 0.
(3, -9) is a solution because if x = 3, then y = 3(-3) = -9.

Can you finish it?
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 <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
Will someone please help me solve this !!
zimovet [89]
  1. 90 degrees
  2. 27 degrees
  3. 63 degrees
  4. 63 degrees
Rhombus
8 0
3 years ago
What is the the comparison number for 30and 60
Anestetic [448]

Answer:

2 i think

Step-by-step explanation:

8 0
3 years ago
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