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AleksAgata [21]
3 years ago
8

What is the distance between the points? Round to the nearest tenth if necessary.

Mathematics
1 answer:
Yuri [45]3 years ago
8 0

Answer:

10.7 units

Step-by-step explanation:

distance formula

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Find the volume of a rectangular prism that is 8 inches high, 12 inches long, and 5 inches wide.
icang [17]

Answer:

B  480 in.³

Step-by-step explanation:

Volume of a rectangular prism is given by

V = l*w*h

V = 8*12*5

V =480 in ^3

3 0
3 years ago
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Solve for x. 4(5x - 20) = -20 Enter your answer in the box. PLEASE HELP!
Colt1911 [192]
20x - 80 = -20
20x = 60
x = 3
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3 years ago
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⦁ What are the key aspects of any parabola? What do the key aspects tell you about the graph?
olya-2409 [2.1K]

One form of the equation of a vertical parabola is y = x^2, which is the same as y-0 = a(x-0)^2.

If the coefficient a is positive, the parabola opens up.   If a is - the parabola opens down.

The vertex of this parabola is (0,0).  

More generally, y - k = a(x - h)^2 represents a vertical parabola that opens up if a is + and opens down if a is - and has its vertex at (h,k).

Often a = 1.  If a is greater than 1, the graph of the parabola is stretched vertically; if less than 1, the graph is compressed vertically (and thus appears to be flatter).

y - k = a(x - h)^2 is called the 'general vertex form' of a vertical parabola.

This is a quadratic equation.  With some algebra, we could rewrite

y - k = a(x - h)^2 in the form y = ax^2 + bx + c.

x-intercepts of this parabola, if any, can be found using the quadratic formula, involving the constant coefficients a, b and c.



6 0
3 years ago
F(prime)(t) = t^2 (1+f(t))<br><br> f(0) = 3<br><br> f(t) = ?
Andreyy89

Answer: F(t) = 4 e ^ {\frac{t^{3} }{3}}-1

Step-by-step explanation:

F'(t) = t^{2} (1+ F(t))

\frac{dF}{dt} = t^{2} (1+F)

First, we separated the variables:

\frac{dF}{1+F} = t^{2} dt

We integrate:

\int\limits^{F(t)}_{F(0)} {\frac{1}{1+F'} } \, dF' = \int\limits^t_0 {t'^{3}} \, dt

We change the variable for make more easy the integral:

u = 1+F', du = dF'

\int\limits^{}_{} {\frac{1}{u} } \, du = \int\limits^0_t {t'^{3}} \, dt

ln(u)= \frac{t^{3}}{3}

Now replace with u = 1+F' and evaluate the limit:

ln(1+F(t)) - ln (1+F(0)) = \frac{t^{3}}{3}

Using properties of natural logarithm

ln(1+F(t))= \frac{t^{3}}{3} + ln(4)

Taking exponential of both sides

e^{ln(1+F(t))} = e ^ {\frac{t^{3} }{3} + ln4}

e^{ln(1+F(t))} = e ^ {\frac{t^{3} }{3}}e^{ln4} =e ^ {\frac{t^{3} }{3}} * 4

1+F(t) = 4 e ^ {\frac{t^{3} }{3}}

F(t) = 4 e ^ {\frac{t^{3} }{3}}-1

7 0
3 years ago
Through (4,-4), slope=-2
liq [111]
There is not anything else to the question. What are you asking? Tell more info
7 0
4 years ago
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