The volume, v, of a solid is v = bh, where b is the area of the base and h is the height.solve v = bh for h.

The graph of the continuous function g, the derivative of the function f, is shown above. The function g is piecewise linear for

4vir4ik [10]

A) $g=f'$ is continuous, so $f$ is also continuous. This means if we were to integrate $g$ , the same constant of integration would apply across its entire domain. Over $0$ , we have $g(x)=2x$ . This means that

$f_{0$

For $f$ to be continuous, we need the limit as $x\to1^-$ to match $f(1)=3$ . This means we must have

$\displaystyle\lim_{x\to1}x^2+C=1+C=3\implies C=2$

Now, over $x$ , we have $g(x)=-3$ , so $f_{x$ , which means $f(-5)=17$ .

b) Integrating over [1, 3] is easy; it's just the area of a 2x2 square. So,

$\displaystyle\int_1^6g(x)=4+\int_3^62(x-4)^2\,\mathrm dx=4+6=10$

c) $f$ is increasing when $f'=g>0$ , and concave upward when $f''=g'>0$ , i.e. when $g$ is also increasing.

We have $g>0$ over the intervals $0$ and $x>4$ . We can additionally see that $g'>0$ only on $0$ and $x>4$ .

d) Inflection points occur when $f''=g'=0$ , and at such a point, to either side the sign of the second derivative $f''=g'$ changes. We see this happening at $x=4$ , for which $g'=0$ , and to the left of $x=4$ we have $g$ decreasing, then increasing along the other side.

We also have $g'=0$ along the interval $-1$ , but even if we were to allow an entire interval as a "site of inflection", we can see that $g'>0$ to either side, so concavity would not change.

5 0

2 years ago

1.

Luba_88 [7]

Hey there!

a = 5 ; b = 2

Option A.

a + b

= 5 + 2

= 7

Option B.

a - b

= 5 - 2

= 3

Option C.

2a + 3b

= 2(5) + 3(2)

= 10 + 6

= 16

Option D.

5a - b

= 5(5) - 2

= 25 - 2

= 23

Good luck on your assignment and enjoy your day!

~Amphitrite1040:)

5 0

2 years ago

Please help me thank you if u do

astra-53 [7]

Answer:

Step-by-step explanation:

1/2 * ( 9 * 6 * 10)

1/2 * 540

270 $in^{2}$

7 0

3 years ago

The value 5 is an upper bound for the zeros of the function shown below.

Mice21 [21]

Answer:

The given statement that value 5 is an upper bound for the zeros of the function f(x) = x⁴ + x³ - 11x² - 9x + 18 will be true.

Step-by-step explanation:

Given

$f\left(x\right)\:=\:x^2\:+\:x^3\:-\:11x^2\:-\:9x\:+\:18$

We know the rational zeros theorem such as:

if $x=c$ is a zero of the function $f(x)$ ,

then $f(c) = 0$ .

As the $f\left(x\right)\:=\:x^2\:+\:x^3\:-\:11x^2\:-\:9x\:+\:18$ is a polynomial of degree $4$ , hence it can not have more than $4$ real zeros.

Let us put certain values in the function,

$f(5) = 448$ , $f(4) = 126$ , $f(3) = 0$ , $f(2) = -20$ ,

$f(1) = 0$ , $f(0) = 18$ , $f(-1) = 16$ , $f(-2) = 0$ , $f(-3) = 0$

From the above calculation results, we determined that $4$ zeros as

$x = -3, -2, 1,$ and $3$ .

Hence, we can check that

$f(x) = (x+3)(x+2)(x-1)(x-3)$

Observe that,

$for x > 3$ , $f(x)$ increases rapidly, so there will be no zeros for $x>3$ .

Therefore, the given statement that value 5 is an upper bound for the zeros of the function f(x) = x⁴ + x³ - 11x² - 9x + 18 will be true.

5 0

3 years ago

A bicycle was originally sold for $400 was on sale for $320. what percent was the bike marked down?

Eva8 [605]

100% of the price = $400
x % of the price = $320
we need to find x

using a proportion
100 : 400 = x : 320
100 * 320 = 400 * x
32000 = 400x
x = 32000 / 400
x = 80

That means the new price is 80% of the previous price.
100% - 80% = 20%
and based on this we can make a conclusion that the discount was 20%

I hope it is clear to you, feel free to ask anything about it :)

6 0

3 years ago

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