Evaluate the function below for x=2. F(x)=x^3+5x^2+1

Mathematics

1 answer:

olga2289 [7]4 years ago

7 0

2^3=8. 2^2=4 5*4= 20. Your answer is 8+20+1= 29

You might be interested in

If QRST is a rectangle with QT=50 and RS=4x+8, find x

Alenkinab [10]

If QRST is a rectangle with QT = 50 and RS = 4x + 8, find x.
.
QT = RS (opposite sides of a rectangle are equal)

50 = 4x + 8
42 = 4x
10.5 = x

4 0

4 years ago

Read 2 more answers

12. (05.02 LC) Look at the figure below: an image of a right triangle is shown with an angle labeled y If sin y° = 7 over q and

meriva

Answer:

$sec~y=\frac{q}{r}$

Step-by-step explanation:

$tan ~y=\frac{7}{r} \\\\\frac{sin~y}{cos~y} =\frac{7}{r} \\\\sin~y=\frac{7}{r} cos~y\\sin ~y=\frac{7}{q} \\\frac{7}{q} =\frac{7}{r} cos~y\\sec~y=\frac{7}{r} \times \frac{q}{7} =\frac{q}{r}$

6 0

3 years ago

Can anyone help me?

frozen [14]

2x + x - 1 x
= ------------------ * ---------
x(x - 1) 8

3x - 1
= --------------
8x - 8

6 0

3 years ago

Checking the Mean Value Theorem: Number 3

mrs_skeptik [129]

$\bf f(x)=x+\cfrac{1}{x}\qquad \left[\frac{1}{2},2 \right]\\\\
-----------------------------\\\\
\cfrac{df}{dx}=1+\left(-1x^{-2} \right)\implies \cfrac{df}{dx}=1-\cfrac{1}{x^2}
\\\\\\
f'(c)=1-\cfrac{1}{c^2}\quad \quad 1-\cfrac{1}{c^2}=\cfrac{f(2)-f\left( \frac{1}{2} \right)}{2-\frac{1}{2}}
\\\\\\
1-\cfrac{1}{c^2}=\cfrac{\frac{5}{2}-\frac{5}{2}}{\frac{3}{2}}\implies 1-\cfrac{1}{c^2}=\cfrac{0}{\frac{3}{2}}\implies 1-\cfrac{1}{c^2}=0
\\\\\\
1=\cfrac{1}{c^2}\implies c^2=1\implies c=\pm\sqrt{1}\implies c=\pm 1$

there's a quick graph below of the bounds and the tangent at "c"

not happening -2 or 2 will have a tangent parallel to a,b, needless to say -2 is out of the range [a,b] anyway, so the only value is really 1, on the positive 1st quadrant

6 0

3 years ago

5 4/7 + 3 13/21 Enter your answer as a mixed number in simplest form

Rama09 [41]

Answer:

9 1/7

Take 5 4/7 + 3 13/21 and get 9 1/7

8 0

3 years ago

Read 2 more answers