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borishaifa [10]
2 years ago
14

What is the value of x in the equation X=-45? O-75 O -27 O 27 O 75

Mathematics
2 answers:
swat322 years ago
6 0

Answer:b

Step-by-step explanation:

miskamm [114]2 years ago
5 0

Answer:

b

Step-by-step explanation:

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B. triangle ABC is isosceles. The angles add up 180 degrees.
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Miguel's teacher instructed him to graph the linear equation y = (3/5)x + 1. He
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Laura sees a horse pulling a buggy. She wonders how it can accelerate if the action of the horse pulling the cart would causes a
KatRina [158]

Answer:

The net forces exerted on the horse and cart are not the same, so they are not balanced forces.

Step-by-step explanation:

Please see the Newton's 2nd Law which states  that an object accelerates if there is a net or unbalanced force on it. In this scenario there is just one force exerted on the wagon i.e: the force that the horse exerts on it. The wagon accelerates because the horse pulls on it. And the amount of acceleration equals the net force on the wagon divided by its mass.

As there are two forces the push and pull the horse; the wagon pulls the horse backwards, and the ground pushes the horse forward. The net force is determined by the relative sizes of these two forces.

If the ground pushes harder on the horse than the wagon pulls, there is a net force in the forward direction, and the horse accelerates forward, and if the wagon pulls harder on the horse than the ground pushes, there is a net force in the backward direction, and the horse accelerates backward.

If the force that the wagon exerts on the horse is the same size as the force that the ground exerts, the net force on the horse is zero, and the horse does not accelerate.

7 0
3 years ago
Read 2 more answers
I need help on this question
tatiyna
It is a right agn have a good day
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
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