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3 years ago

[Ignore the 2 problems] I need help on #9

Mathematics

1 answer:

Lena [83]3 years ago

5 0

Comment.
9A
Let's use (a) as an example. What number is in front of the x inside be brackets? If I gave you the choice of 1,2,3,4,5 which number is it out of those 5?

If you chose 5, you have the second best answer. You are reading outside the brackets. Not what we need.

If you choose 1, you are well on your way. There is a 1 beside the x.

Now go to the outside of the brackets. What's in front of the x? 1,2,3,4,5.. There is only one number you can choose. It's a 5. So the result outside the brackets is 5x. How did 1x become 5x? You had to put a 5 on the blank to the left of the brackets..

5(5 + x) = 5x + ___ What goes on the line to the right of 5x? Rule whatever you do to one side of a plus sign inside the brackets, you must do to the other side of the plus sign.

It becomes 5*5

So the blank is filled with 25. And here's the answer.
5(5 + x) = 5x + 25 <<<< answer..
Stop don't do anything more. 5x and 25 are not like terms. They don't mix

9B
Let's make this one brief so that you have something to do. What did you multiply 5 by to get 10? You should answer 2.
2(x + 5) = ___ x + 10

What you do on one side of the plus sign, you must do on the other. So what's in front of the x on the right?

2(x + 5) = 2x + 10 <<<< answer

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If the 1500-lblb boom ABAB, the 220-lblb cage BCDBCD, and the 169-lblb man have centers of gravity located at points G1G1, G2G2

ANEK [815]

Answer:

(M)_a = -8171 lb-ft

Step-by-step explanation:

Step 1:

- We will first mark each weight from left most to right most.

Point: Weight: Moment arm r_a

G_3 W_g3 = 169 30*Cos(75) + 4.25

G_2 W_g2 = 220 30*Cos(75) + 2.5

G_1 W_g1 = 1500 10*cos(75)

Step 2:

- Set up a sum of moments about pivot point A, the expression would be as follows:

(M)_a = -W_g3*(30cos(75) + 4.25) - W_g2*(30*Cos(75) + 2.5) - W_g1*10*cos(75)

Step 3:

- Plug in the values and solve for (M)_b, as follows:

(M)_a = -169*(30cos(75) + 4.25) - 220*(30*Cos(75) + 2.5) - 1500*10*cos(75)

(M)_a = -2030.462559 -2258.205698 - 3882.285677

(M)_a = -8171 lb-ft

4 0

2 years ago

JJack, Jill, and Bill each carried a 48-fluid ounce bucket full of water down the hill. By the time they reached the bottom, Jac

Advocard [28]

Answer:

Jill lost 1/3

Bill lost 1/6

and Jack lost 1/4

Make a common denominator - let's use 12

Jill lost 4/12

Bill lost 2/12

Jack lost 3/12

Add them together = 9/12

reduce by dividing by 3.

total of 3/4 spilt

48 divided into 4 parts then multiplied by 3 of those parts

48 / 4 * 3 = 36

Step-by-step explanation:

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2 years ago

The equation (x-6)^2/16 + (y+7)^2/4 = 1 represents an ellipse. What are the vertices of the ellipse?

topjm [15]

Answer:

The correct option is C.

Step-by-step explanation:

The given equation is

$\frac{(x-6)^2}{16}+\frac{(y+7)^2}{4}= 1$

It can be rewritten as

$\frac{(x-6)^2}{4^2}+\frac{(y+7)^2}{2^2}= 1$ .....(1)

The standard form of an ellipse is

$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}= 1$ ....(2)

Where (h,k) is center of the ellipse.

If a>b, then the vertices of the ellipse are $(h\pm a, k)$ .

From (1) and (2) we get

$h=6,k=-7,a=4,b=2$

Since a>b, therefore the vertices of the ellipse are

$(h+a, k)=(6+4,-7)\Rightarrow (10,-7)$

$(h-a, k)=(6-4,-7)\Rightarrow (2,-7)$

The vertices of the given ellipse are (10, –7) and (2, –7). Therefore the correct option is C.

5 0

2 years ago

Read 2 more answers

The weight of football players is normally distributed with a mean of 200 pounds and a standard deviation of 25 pounds. the prob

Anni [7]

Given:
μ = 200 lb, the mean
σ = 25, the standard deviation

For the random variable x = 250 lb, the z-score is
z = (x-μ)/σ =(250 - 200)/25 = 2

From standard tables for the normal distribution, obtain
P(x < 250) = 0.977

Answer: 0.977

7 0

3 years ago

The number of errors in a textbook follow a poisson distribution with a mean of 0.01 errors per page. what is the probability th

neonofarm [45]

Mean number of errors in each page = 0.01
Mean number of errors in 100 pages = 0.01*100=1

It is possible to use the cumulative distribution function (CMF), but the math is a little more complex, involving the gamma-function. Tables and software are available for that purpose.

Thus it is easier to evaluate with a calculator for the individual cases of k=0,1,2 and 3.

The Poisson distribution has a PMF (probability mass function)

$P(k):=\frac{\lambda^ke^{-\lambda}}{k!}$
with λ = 1
=>
$P(0):=\frac{1^0e^{-1}}{0!}=0.3678794$
$P(1):=\frac{1^1e^{-1}}{1!}=0.3678794$
$P(2):=\frac{1^2e^{-1}}{2!}=0.1839397$
$P(3):=\frac{1^3e^{-1}}{3!}=0.0613132$
=>
$P(k$
or
P(k<=3)=0.9810 (to four decimal places)

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2 years ago