Ask question

Ask question

All categories

2 years ago

13

which explains the correct way to move the decimal point to find the quotient in the problem 12.46 divided by 100? My choices ar

e; (A) 2 places to the left (B) 2 places to the right (C) 3 places to the left (D) 3 places to the right?

1 answer:

jarptica [38.1K]2 years ago

6 0

Well, since it has 2 zeros, you move the point 2 places to the right.

You might be interested in

The solution of a certain differential equation is of the form y(t)=aexp(7t)+bexp(11t), where a and b are constants. The solutio

Solnce55 [7]

Answer:

Step-by-step explanation:

Given that the solution of a certain differential equation is of the form

$y(t) = ae^{7t} +be^{11t}$

Use the initial conditions

i) y(0) =1

$1=a(1)+b(1)\\a+b=1$ ... I

ii) y'(0) = 4

Find derivative of y first and then substitute

$y'(t) = 7ae^{7t} +11be^{11t}\\y'(0) =7a+11b \\7a+11b =4 ...II$

Now using I and II we solve for a and b

Substitute b = 1-a in II

$7a+11(1-a) = 4\\-4a+11 =4\\-4a =-7\\a = 1.75 \\b = -0.75$

Hence solution is

$y(t) = 1.75e^{7t} -0.75e^{11t}$

8 0

2 years ago

Read 2 more answers

Help please. Screenshot included.

Alla [95]

The following information will help us with this problem:

$\sqrt[m]{x^n} = x^{\frac{n}{m}$

When we use that information in the context of this problem, we can find:

$\sqrt[4]{15^7} = 15^{\frac{7}{4}}$

Thus, a = 15, b = 7, and c = 4.

4 0

3 years ago

Please help me solve this

Sever21 [200]

The answer is C) all real numbers greater than or equal to one.

Hope this helps!

4 0

2 years ago

1) Let f(x)=6x+6/x. Find the open intervals on which f is increasing (decreasing). Then determine the x-coordinates of all relat

brilliants [131]

Answer:

1) increasing on (-∞,-1] ∪ [1,∞), decreasing on [-1,0) ∪ (0,1]

$x = -1$ is local maximum, $x = 1$ is local minimum

2) increasing on [1,∞), decreasing on (-∞,0) ∪ (0,1]

$x = 1$ is absolute minimum

3) increasing on (-∞,0] ∪ [8,∞), decreasing on [0,4) ∪ (4,8]

$x = 0$ is local maximum, $x = 8$ is local minimum

4) increasing on [2,∞), decreasing on (-∞,2]

$x = 2$ is absolute minimum

5) increasing on the interval (0,4/9], decreasing on the interval [4/9,∞)

$x = 0$ is local minimum, $x = 4/9$ is absolute maximum

Step-by-step explanation:

To find minima and maxima the of the function, we must take the derivative and equalize it to zero to find the roots.

1) $f(x) = 6x + 6/x$

$f\prime(x) = 6 - 6/x^2 = 0$ and $x \neq 0$

So, the roots are $x = -1$ and $x = 1$

The function is increasing on the interval (-∞,-1] ∪ [1,∞)

The function is decreasing on the interval [-1,0) ∪ (0,1]

$x = -1$ is local maximum, $x = 1$ is local minimum.

2) $f(x)=6-4/x+2/x^2$

$f\prime(x)=4/x^2-4/x^3=0$ and $x \neq 0$

So the root is $x = 1$

The function is increasing on the interval [1,∞)

The function is decreasing on the interval (-∞,0) ∪ (0,1]

$x = 1$ is absolute minimum.

3) $f(x) = 8x^2/(x-4)$

$f\prime(x) = (8x^2-64x)/(x-4)^2=0$ and $x \neq 4$

So the roots are $x = 0$ and $x = 8$

The function is increasing on the interval (-∞,0] ∪ [8,∞)

The function is decreasing on the interval [0,4) ∪ (4,8]

$x = 0$ is local maximum, $x = 8$ is local minimum.

4) $f(x)=6(x-2)^{2/3} +4=0$

$f\prime(x) = 4/(x-2)^{1/3}$ has no solution and $x = 2$ is crtitical point.

The function is increasing on the interval [2,∞)

The function is decreasing on the interval (-∞,2]

$x = 2$ is absolute minimum.

5) $f(x)=8\sqrt x - 6x$ for $x>0$

$f\prime(x) = (4/\sqrt x)-6 = 0$

So the root is $x = 4/9$

The function is increasing on the interval (0,4/9]

The function is decreasing on the interval [4/9,∞)

$x = 0$ is local minimum, $x = 4/9$ is absolute maximum.

5 0

2 years ago

(a) One ball is selected at random from the bag.

lana [24]

Answer:

a) The probability is 1/5

bi) The probability each ball is even will be 2/5

bii) The probability the sum of the numbers on the ball is not more than 10 is 16/25

(that's what I think it is)

3 0

3 years ago