On the last history quiz, Knesha answered 0.75 of the 20 question correctly How many questions did Kanesha answer correctly?

1 answer:

5 0

.75 is 75% of the test, and 75 of 20 is equal to 1/4. So 20/4 is equal to 5, meaning 5 questions were wrong, and 15 were answered correctly.

I hope this was not too confusing for you.

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2. An articulated truck is loaded with cylindrical containers filled with oil. The height of each container is 80cm and the diam

RoseWind [281]

The volume of each container is: 25,133 $cm^{3}$

Given:

height of container (h) = 80 cm

diameter of container (d) = 20 cm

radius = half of diameter = $\frac{1}{2}$ of $20 = 10 cm$

Recall:

Volume of a cylinder = $\pi r^{2} h$

<em>Substitute the values into the formula:</em>

$Volume = \pi \times 10^{2} \times 80 = 25132.74123$

Therefore, volume of each cylinder is approximately 25,133 $cm^{3}$

Learn more about volume of cylinder here:

brainly.com/question/971817

8 0

3 years ago

If g(x)= x=2 and h(x) = 4 - X, what is the value of (gºh)(-3)?

navik [9.2K]

The required value of (g°h)(-3) is 5.

Step-by-step explanation:

Given,

g(x)= x-2 and h(x) = 4 - x

To find (g°h)(-3)

Now,

(g°h)(x) = g(h(x))

= g(4-x)

= (4-x)-2 = 2-x

So,

(g°h)(-3) = 2-(-3) = 5 [ putting x=-3]

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3 years ago

Write down a differential equation of the form dy/dt = ay + b whose solutions have the required behavior as t → ∞. all other sol

nordsb [41]

The ODE has an equilibrium point at

$\dfrac{\mathrm dy}{\mathrm dt}=ay+b=0\implies y=-\dfrac ba$

We want $y=2$ to be an unstable equilibrium solution - in other words, we want all solutions with initial condition $y(t_0)=c$ for $c$ near 2 to diverge from $y=2$ - so $-dfrac ba=2$ .

Divergence from $y=2$ would require that for $y>2$ , any solution is increasing, and for $y$ , and solution is decreasing. In order for that to happen, we need