Ask question

Ask question

All categories

3 years ago

13

Find a formula in n, a, m, and d for the sum (a md) (a (m 1)d) (a (m 2)d) · · · (a (m n)d), where m and n are integers, n ≥ 0, a

nd a and d are real numbers. justify your answer.

1 answer:

Anettt [7]3 years ago

3 0

Write this as the sum of:
a+md
a+md+d
a+md+d+d
...
a+md+d+d+...+d

Total=n(a+md)+d(1+2+3+...+n-1)=n(a+md)+dn(n-1)/2

You might be interested in

Han divided 3 10 by 6. His work is shown below. Is han's answer correct? 3 10 ÷ 6 = 6 10 ÷ 6 = 1 10

Marat540 [252]

Han’s answer is incorrect

4 0

3 years ago

What is the solution to the system of equations y = -x - 5 and y = 2x + 4

Delvig [45]

Answer:

(-3,-2)

Step-by-step explanation:

y = -x - 5

y = 2x + 4

plug in one of the y equations

(2x+4)= -x - 5

2x+4=-x-5

3x=-9

x= -3

plug in x to one of the y equations

y= -(-3) -5

y=3-5

y= -2

x= -3, y= -2

4 0

3 years ago

In triangle ABC, (angle A) is a right angle and (m angle B) = 45deg.

Ratling [72]

We can use the ratios for special triangles (see the attachment below). We'll be using the 45-45-90 triangle. We notice that side BC is equal to $x \sqrt{2}$ , and side AC (x using the special right triangle) is equal to 16. We can therefore say that side BC is equal to $16 \sqrt{2}$ ft

:)

4 0

3 years ago

Read 2 more answers

A set of data has a normal distribution with a mean of 46 and a standard deviation of 7. Find the percent of data within the fol

kvasek [131]

Answer:

50%

the mean (46) marks the halfway point, so over the mean is 50% and under the mean is 50%

7 0

1 year ago

A baseball diamond is a square with side 90 ft. A batter hits the ball and runs toward first base with a speed of 31 ft/s. (a) A

julsineya [31]

Answer:

a) -13.9 ft/s

b) 13.9 ft/s

Step-by-step explanation:

a) The rate of his distance from the second base when he is halfway to first base can be found by differentiating the following Pythagorean theorem equation respect t:

$D^{2} = (90 - x)^{2} + 90^{2}$ (1)

$\frac{d(D^{2})}{dt} = \frac{d(90 - x)^{2} + 90^{2})}{dt}$

$2D\frac{d(D)}{dt} = \frac{d((90 - x)^{2})}{dt}$

$D\frac{d(D)}{dt} = -(90 - x) \frac{dx}{dt}$ (2)

Since:

$D = \sqrt{(90 -x)^{2} + 90^{2}}$

When x = 45 (the batter is halfway to first base), D is:

$D = \sqrt{(90 - 45)^{2} + 90^{2}} = 100. 62$

Now, by introducing D = 100.62, x = 45 and dx/dt = 31 into equation (2) we have:

$100.62 \frac{d(D)}{dt} = -(90 - 45)*31$

$\frac{d(D)}{dt} = -\frac{(90 - 45)*31}{100.62} = -13.9 ft/s$

Hence, the rate of his distance from second base decreasing when he is halfway to first base is -13.9 ft/s.

b) The rate of his distance from third base increasing at the same moment is given by differentiating the folowing Pythagorean theorem equation respect t:

$D^{2} = 90^{2} + x^{2}$

$\frac{d(D^{2})}{dt} = \frac{d(90^{2} + x^{2})}{dt}$

$D\frac{dD}{dt} = x\frac{dx}{dt}$ (3)

We have that D is:

$D = \sqrt{x^{2} + 90^{2}} = \sqrt{(45)^{2} + 90^{2}} = 100.63$

By entering x = 45, dx/dt = 31 and D = 100.63 into equation (3) we have:

$\frac{dD}{dt} = \frac{45*31}{100.63} = 13.9 ft/s$

Therefore, the rate of the batter when he is from third base increasing at the same moment is 13.9 ft/s.

I hope it helps you!

4 0

2 years ago

Read 2 more answers