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2 years ago

Answer the following questions.

Mathematics

2 answers:

krok68 [10]2 years ago

6 0

Answer:

B, x=83

Step-by-step explanation:

35+62+x=180

97+x=180

x=180-97

x=83

kakasveta [241]2 years ago

3 0

Answer:

x=83

Step-by-step explanation:

The sum of interior angles in a triangle will always be equal to 180 degrees. Knowing this, we can set up the following equation:

35+62+x=180

97+x=180

Subtract both sides by 97

97+x-97=180-97

x=83

I hope this helps!

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Solve the system and enter the smallest x-coordinate first

ankoles [38]

Answer:
(2,-3)(6,5)
Step-by-step explanation:
We can use substitution to get the equation: x^2-6x+5 = 2x-7
Solve:
x^2-6x+5 = 2x-7
x^2 = 8x-12
x^2-8x+12 = 0 (we now have a polynomial)
(x-6)(x-2) = 0 (set each equation to equal 0 and solve)
x-6 = 0 --> x=6
x-2 = 0 --> x=2
To get the Y coordinates:
y=2(2)-7 --> y = -3
y = 2(6)-7 --> y = 5

Check work:
5 = 6^2-6(6)+5 --> 5 = 36-36 +5 --> 5=5
-3 = 2^2-6(2)+5 --> -3 = 4-12+5 --> -3=-3

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1 year ago

How to do math easily

katrin [286]

Step-by-step explanation:

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2 years ago

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defon

I've attached a photo of the solution below. Hope it helps!

3 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 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

2 years ago

Find a 4 digit odd number using each of the digit 1,2,3,4 and 5 only once such than when the first and the last digit are in int

Vesna [10]

Answer:

The possible numbers could be : 4521, 4125, and 2415

3 0

2 years ago