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3 years ago

6x^3 +8x^2 – 4x I need help now ‼️‼️‼️‼️‼️‼️

Mathematics

1 answer:

Vinvika [58]3 years ago

5 0

Answer:

Are there any answers??

Step-by-step explanation:

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Find the number c that satisfies the conclusion of the Mean Value Theorem on the given interval. (Enter your answers as a comma-

Shtirlitz [24]

Answer:

The number c is 2.

Step-by-step explanation:

Mean Value Theorem:

If f is a continuous function in a bounded interval [0,4], there is at least one value of c in (a,b) for which:

$f(c) = \frac{1}{b-a}\int\limits^a_b {f(x)} \, dx$

In this problem, we have that:

$f(x) = x, a = 0, b = 4$

So $f(c) = c$

----------

$f(c) = \frac{1}{b-a}\int\limits^a_b {f(x)} \, dx$

$c = \frac{1}{4-0}\int\limits^0_4 {x} \, dx$

$c = \frac{1}{4-0}*8$

$c = 2$

The number c is 2.

3 0

2 years ago

PLS HELP ASAP! LAST DAY OF SCHOOL!

lianna [129]

Area of triangles: 1/2 x b x h
area of square: l x w

Triangle
1/2 x 7 x 4= 14, but there are 3 of them so 42
Square: 7 x7 = 49

49 + 42 = 91

6 0

2 years ago

Two-step equations. Solve X 3x -14 = 46

DedPeter [7]

Answer:

x is 20

Step-by-step explanation:

3x-14=46

3x=60

x=20

5 0

2 years ago

Read 2 more answers

Will give brainliest!

Gala2k [10]

Check the picture below, it hits the ground when y = 0.

$\bf ~~~~~~\textit{initial velocity}\\\\
\begin{array}{llll}
~~~~~~\textit{in feet}\\\\
h(t) = -16t^2+v_ot+h_o \\\\
\end{array} 
\quad 
\begin{cases}
v_o=\stackrel{0}{\textit{initial velocity of the object}}\\\\
h_o=\stackrel{162}{\textit{initial height of the object}}\\\\
h=\stackrel{}{\textit{height of the object at "t" seconds}}
\end{cases}
\\\\\\
h(t)=-16t^2+0t+162\implies 0=-16t^2+162\implies 16t^2=162
\\\\\\
t=\cfrac{162}{16}\implies t=\cfrac{81}{8}\implies t=10\frac{1}{8}$

6 0

2 years ago

aliyah tapi brianna and greg always split two pizzas for lunch. each pizza is cut into 8 slices. if aliyah eats 3 pieces tapi ea

NemiM [27]

Answer: 5

Step-by-step explanation: There 2 pizza boxes cut into 8 so that make it 16 slices 16-3-4-1-3= 5.

7 0

3 years ago

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