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2 years ago

I really don’t understand

Mathematics

1 answer:

Korolek [52]2 years ago

8 0

Answer:

x=5

Step-by-step explanation:

The exterior angle of a triangle theorem states the angles V and U (in the diagram) will equal angle UTK.

40°+12x=21x-5

Now we just solve like a regular equation

40+5=21x-12x

45°=9x

45°÷9=9x÷9

5=×

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A stunt driver wants to make his car jump over 8 cars parked side by side below a horizontal ramp. The vertical height of the ra

mr Goodwill [35]

Answer:

v=26.41 m/s

Step-by-step explanation:

From the Newtons laws of motions, we sew that x= (2v₁²sin∅os∅)/g where x is the horizontal distance v is the initial speed and ∅ is the launch angle.

From trigonometry we see that 2 sin∅cos∅=sin 2∅

Therefore, x=(v²sin2∅)/g

x=22m

∅=9°

g=9.8m/s²

22m=v²×sin(2×9)/(9.8m/s²)

v²=(22×9.8)/(sin 18)

v²=697.696

v=√697.696

v=26.41 m/s

3 0

3 years ago

a group of 4 people share 1/2 pound of almonds. how many pounds of almonds does each person get choose an equation you can use t

Natasha_Volkova [10]

4•1\2=1/16
I don’t know the equation but you can just change it around a little

8 0

3 years ago

Read 2 more answers

Help with 30 please. thanks.

Svet_ta [14]

Answer:

See Below.

Step-by-step explanation:

We have the equation:

$\displaystyle y = \left(3e^{2x}-4x+1\right)^{{}^1\! / \! {}_2}$

And we want to show that:

$\displaystyle y \frac{d^2y }{dx^2} + \left(\frac{dy}{dx}\right) ^2 = 6e^{2x}$

Instead of differentiating directly, we can first square both sides:

$\displaystyle y^2 = 3e^{2x} -4x + 1$

We can find the first derivative through implicit differentiation:

$\displaystyle 2y \frac{dy}{dx} = 6e^{2x} -4$

Hence:

$\displaystyle \frac{dy}{dx} = \frac{3e^{2x} -2}{y}$

And we can find the second derivative by using the quotient rule:

$\displaystyle \begin{aligned}\frac{d^2y}{dx^2} & = \frac{(3e^{2x}-2)'(y)-(3e^{2x}-2)(y)'}{(y)^2}\\ \\ &= \frac{6ye^{2x}-\left(3e^{2x}-2\right)\left(\dfrac{dy}{dx}\right)}{y^2} \\ \\ &=\frac{6ye^{2x} -\left(3e^{2x} -2\right)\left(\dfrac{3e^{2x}-2}{y}\right)}{y^2}\\ \\ &=\frac{6y^2e^{2x}-\left(3e^{2x}-2\right)^2}{y^3}\end{aligned}$