Parker marks sixths on a number line. He writes

I need to know if 93 is a multiple of 9 ×

iris [78.8K]

93 is not a multiple of 9. Multiple of 9 would be 90 or 99. These are multiples of 9 in a range from 0 up to 100:

9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99

5 0

3 years ago

How many times bigger is 1.8x10^27 than 1x10^25

victus00 [196]

To find the number of times it is bigger

$= \dfrac{1.8 \times 10^{27} }{1 \times 10^{25}}$

$= 1.8 \times 10^{27-25}$

$= 1.8 \times 10^{2}$

$= 180 \text { times}$

3 0

3 years ago

Read 2 more answers

X – y = 2 3x + y = 10 all of yalls answer r wrong the answer is x= 3 and y= 1

Y_Kistochka [10]

Answer:

Step-by-step explanation:

People always get it wrong just for points

3 0

3 years ago

Find the value of x, rounded to the nearest tenth. Correct answers only please..

Ratling [72]

Cos28 = 11/x
x = 11/cos28
x = 11 / 0.88294759
x = 12.46

Answer
C. 12.5 (nearest tenth)

8 0

3 years ago

Read 2 more answers

Need help on both of these Calculus AB mock exams.

babymother [125]

AB1

(a) $g$ is continuous at $x=3$ if the two-sided limit exists. By definition of $g$ , we have $g(3)=f(3^2-5)=f(4)$ . We need to have $f(4)=1$ , since continuity means

$\displaystyle\lim_{x\to3^-}g(x)=\lim_{x\to3}f(x^2-5)=f(4)$

$\displaystyle\lim_{x\to3^+}g(x)=\lim_{x\to3}5-2x=-1$

By the fundamental theorem of calculus (FTC), we have

$f(6)=f(4)+\displaystyle\int_4^6f'(x)\,\mathrm dx$

We know $f(6)=-4$ , and the graph tells us the *signed* area under the curve from 4 to 6 is -3. So we have

$-4=f(4)-3\implies f(4)=-1$

and hence $g$ is continuous at $x=3$ .

(b) Judging by the graph of $f'$ , we know $f$ has local extrema when $x=0,4,6$ . In particular, there is a local maximum when $x=4$ , and from part (a) we know $f(4)=-1$ .

For $x>6$ , we see that $f'\ge0$ , meaning $f$ is strictly non-decreasing on (6, 10). By the FTC, we find

$f(10)=f(6)+\displaystyle\int_6^{10}f'(x)\,\mathrm dx=7$

which is larger than -1, so $f$ attains an absolute maximum at the point (10, 7).

(c) Directly substituting $x=3$ into $k(x)$ yields

$k(3)=\dfrac{\displaystyle3\int_4^9f'(t)\,\mathrm dt-12}{3e^{2f(3)+5}-3}$

From the graph of $f'$ , we know the value of the integral is -3 + 7 = 4, so the numerator reduces to 0. In order to apply L'Hopital's rule, we need to have the denominator also reduce to 0. This happens if

$3e^{2f(3)+5}-3=0\implies e^{2f(3)+5}=1\implies 2f(3)+5=\ln1=0\implies f(3)=-\dfrac52$

Next, applying L'Hopital's rule to the limit gives

$\displaystyle\lim_{x\to3}k(x)=\lim_{x\to3}\frac{9f'(3x)-4}{6f'(x)e^{2f(x)+5}-1}=\frac{9f'(9)-4}{6f'(3)e^{2f(3)+5}-1}=-\dfrac4{21e^{2f(3)+5}-1}$

which follows from the fact that $f'(9)=0$ and $f'(3)=3.5$ .

Then using the value of $f(3)$ we found earlier, we end up with

$\displaystyle\lim_{x\to3}k(x)=-\dfrac4{20}=-\dfrac15$

(d) Differentiate both sides of the given differential equation to get

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=2f''(2x)(3y-1)+f'(2x)\left(3\dfrac{\mathrm dy}{\mathrm dx}\right)=[2f''(2x)+3f'(2x)^2](3y-1)$

The concavity of the graph of $y=h(x)$ at the point (1, 2) is determined by the sign of the second derivative. For $y>\frac13$ , we have $3y-1>0$ , so the sign is entirely determined by $2f''(2x)+3f'(2x)^2$ .

When $x=1$ , this is equal to

$2f''(2)+3f'(2)^2$

We know $f''(2)=0$ , since $f'$ has a horizontal tangent at $x=2$ , and from the graph of $f'$ we also know $f'(2)=5$ . So we find $h''(1)=75>0$ , which means $h$ is concave upward at the point (1, 2).

(e) $h$ is increasing when $h'>0$ . The sign of $h'$ is determined by $f'(2x)$ . We're interested in the interval $-1$ , and the behavior of $h$ on this interval depends on the behavior of $f'$ on $-2$ .