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a_sh-v [17]
4 years ago
5

What is the following simplified product? Assume x greater-than-or-equal-to 0 2 StartRoot 8 x cubed EndRoot (3 StartRoot 10 x Su

perscript 4 Baseline EndRoot minus x StartRoot 5 x squared EndRoot) 24 x cubed StartRoot 5 x EndRoot minus 4 x squared StartRoot 10 x EndRoot 24 x cubed StartRoot 5 x EndRoot minus 4 x cubed StartRoot 10 x EndRoot 24 x cubed StartRoot 5 EndRoot minus 4 x cubed StartRoot 10 x EndRoot 24 x cubed StartRoot 5 x EndRoot + 4 x cubed StartRoot 10 x EndRoot
Mathematics
2 answers:
Lady bird [3.3K]4 years ago
5 0
Wow that’s a lot let me try to help
Sedbober [7]4 years ago
4 0

Answer:

The answer to this should be D on ed2020

Step-by-step explanation:

There's another post for this question where Cheerchar13 had answered.

You might be interested in
Macy estimated a product to be 800. The actual product is 768. Her estimate was an
Vladimir79 [104]

Answer:

Overestimate

Step-by-step explanation:

it would be an underestimate if the number was lower than the actual number but since its larger than it is an overestimation.

hope this helps

6 0
4 years ago
The graph shows the equation x2 + y2 = 5. Use the slider for a to move the vertical line on the graph. Based on the vertical
lidiya [134]

Answer: This equation is not a function.

Step-by-step explanation:

Ok, a function is something like a "machine".

It takes an input value, x, and transforms it into an output value, y.

Such that for every input x, the function can transform it into only one value of y.

for example, if we have a "function" f(x)

such that f(2) = 3 and f(2) = 4

(for the same input, x = 2, we have two different outputs)

Then this is not a function.

Now, the given equation is:

x^2 + y^2 = 5

This is actually the equation for a circle, centered at the point (0, 0) and with a radius equal to √5.

Then, if we want to transform it into y(x), we can see a problem

If x = 0, we have:

0^2 + y^2 = 5

y^2 = 5

then we have two possible values for y:

y = +√5

y = -√5

Then for one value of x, we have two values of y.

This means that this is not a function.

4 0
4 years ago
I'm giving brainliest to the most helpful answer. For 50 points:
Alenkinab [10]

Answer:

1) \displaystyle\frac{5}{18}\approx27.78\%

2) \displaystyle\frac{4}{7}\approx57.14\%

3) \displaystyle  \frac{13}{16}=81.25\%

Step-by-step explanation:

We are given a two-way frequency table. Using the table, we will determine the probability for each case.

Question 1)

P(A Student With A Part Time Job Without A Car)

Using the first column, the total number of students that have a part time job is 78+30=108.

Likewise, using the first column, we can see that out of those 108 students, 30 do not have a car.

Hence, the probability that a student with a part time job without a car is:

\displaystyle P=\frac{30}{108}=\frac{5}{18}\approx27.78\%

Question 2)

P(No Car | Does Not Have A Part Time Job)

Remember that the vertical line means conditional probability.

So, we want the probability of a student having no car given that they do not have a part time job.

Using the second column, we can see that the total number of students that do not have a part time job is 18+24=42.

Likewise, using the second column, 24 do not have a car.

Hence, the probability that a student with a part time job without a car is:

\displaystyle P=\frac{24}{42}=\frac{4}{7}\approx57.14\%

Question 3)

P(Part Time Job | Car)

So, we want to probability of a student having a part time job given that they have a car.

Using the first row, the total students that have a car is 78+18=96.

And of those 96 students, 78 have a part time job.

Hence, the probability that a student with a car has a part time job is given by:

\displaystyle P=\frac{78}{96}=\frac{13}{16}=81.25\%

4 0
3 years ago
Using the figure below, find the area of the circle.
Pachacha [2.7K]
Diameter= 10 cm
radius = 5 cm
area of circle = πr2
= (3.142)(5)^2
=78.55 cm^2
5 0
3 years ago
Read 2 more answers
Suppose that Adam rolls a fair six-sided die and a fair four-sided die simultaneously. Let AAA be the event that the six-sided d
myrzilka [38]

The question is incomplete. Here is the complete question.

Suppose that Adam rolls a fair six-sided die and a fair four-sided die simultaneously. Let A be the event that the six-sided die is an even number and B be the event that the four-sided die is an odd number. Using the sample space of possible outcomes below, answer each of the following questions.

What is P(A), the probabillity that the six-sided die is an even number?

What is P(B), the probability of the four-sided die is an odd number?

What is P(A and B), the probability that the six-sided die is an even number  <em>and</em> the four-sided die is an odd number?

Are events A and B independent?

a) Yes, events A and B are independent events.

b) No, events A and B are not independent events.

Answer: P(A) = 1/2

P(B) = 1/2

P(A and B) = 1/4

a) Yes, events A and B are independent events.

Step-by-step explanation: The event A is related to a six-sided die, so total possibilities is 6.

For a six-sided die to show a even number, there are 3 possibilities: (2,4,6)

so, P(A) = 3/6 = 1/2

The event B is for a 4-sided die, i.e. total possibilities is 4.

To show an odd number, there are 2 possibilities: (1,3).

Then, P(B) = 2/4 = 1/2

Now, the probability of occuring A and B is:

P(A and B) = P(A).P(B)

P(A and B) = 1/2*1/2

P(A and B) = 1/4

The events are <u>independent</u> events because the probability of A happening does not influence the occuring of event B.

3 0
4 years ago
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