All categories

3 years ago

What is 2 times 5 to the 3rd power divided by (9-4) raised to the power of 4

1 answer:

8 0

Heyo Kiddo. How are you doing today?

Let's figure this out.

So in number form, the equation will be looking a little like this:

$2(5) ^{3} / (9-4)^4$

To solve.

$2(5) ^{3} / (9-4)^4 \ \textless \ Parenthesis first\ \textgreater \ 

2(5)^{3} / 5^4 \ \textless \ Exponents\ \textgreater \ 

2(125) / 625 (Multiply)

250 / 625 (Divide)

0.4 (Answer)$

So I hope this answers your question and that you have a great day. :)

You might be interested in

LAM 1 Mod 6 Question 2.

Artyom0805 [142]

Answer:

As a triangle because it's a quadratic shape

6 0

2 years ago

A force of 60 N is used to stretch two springs that are initially the same length. Spring A has a spring constant of 4 N/m, and

goblinko [34]

Answer:

D:Spring A is 3 m longer than spring B because 15 – 12 = 3.

Step-by-step explanation:

In this question, you should remember the Hooke's Law in physics.

The Hooke's Law simply explains that the extension that occurs on a spring is directly proportional to the load applied on it.

The mathematical expression for this law is

$F=-kx$

where;

F= force applied on the spring

x = the extension on the spring

k= the spring constant which varies in spring.

The question will need you to calculate the extension on the springs A and B then compare the values obtained.

In spring A

Force, F=60N and spring constant ,k=4 N/m

To find the extension x apply the expression;

$F=-kx\\\\60=-4*x\\\\60=-4x\\\\\frac{60}{-4} =\frac{-4x}{-4} \\\\\\-15=x$

Here the spring extension is 15 m

In spring B

Force, F=60N and spring constant , k=5N/m

To find the extension x apply the same expression

$F=-kx\\\\60=-5*x\\\\60=-5x\\\\\\\frac{60}{-5} =\frac{-5x}{-5} \\\\\\-12=x$

Here the extension on the spring is 12 m

Compare

The extension on spring A is 3 m longer than that in spring B because when you subtract the value of spring B from that in spring A you get 3m

$=15m-12m=3m$

5 0

2 years ago

What is the distance between the points (5, 6) and (5, 2)? a. 1 b. 3 c. 4 d. 8

postnew [5]

$The\ distance\ between\ A(x_A;\ y_A)\ and\ B(x_B;\ y_B):\\\\d=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}$

$(5;\ 6);\ (5;\ 2),\ subtitute\\\\d=\sqrt{(2-6)^2+(5-5)^2}=\sqrt{(-4)^2+0^2}=\sqrt{16}=4$

4 0

3 years ago

A 9 B 18 C 4.5 D 19 Thanks

irina1246 [14]

Answer:

The answer is C. 4.5. Hope that helps.

Step-by-step explanation:

7 0

3 years ago

A sample of 10 adult elephants had an average weight of 12,556 pounds. The standard deviation

kirill [66]

Answer:

The 99% confidence interval of the population mean for the weights of adult elephants is between 12,475 pounds and 12,637 pounds.

Step-by-step explanation:

We have the standard deviation for the sample, so we use the t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 10 - 1 = 9

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 9 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.99}{2} = 0.995$ . So we have T = 3.25

The margin of error is:

M = T*s = 3.25*25 = 81

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 12,556 - 81 = 12,475 pounds

The upper end of the interval is the sample mean added to M. So it is 12,556 + 81 = 12,637 pounds.

The 99% confidence interval of the population mean for the weights of adult elephants is between 12,475 pounds and 12,637 pounds.

6 0

2 years ago