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3 years ago

What are the factors in 20x-11x+3 ?

Mathematics

2 answers:

sergiy2304 [10]3 years ago

8 0

The factor(s) of this problem is→ 3(3x + 1)

sergey [27]3 years ago

7 0

There are no factors , the answer would essentially be prime. since there are no numbers that add up to -11 and multiply to be positive 3.

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Explain why the exponents cannot be added in the product 12^3 x<br> 11^3.

sergij07 [2.7K]

Answer:

see explanation

Step-by-step explanation:

The exponents can only be added if the bases are the same

Here one base is 12 and the other is 11

So the rule of addition is not applicable.

4 0

2 years ago

Read 2 more answers

Evaluate if a = 28 and b = -3

Nata [24]

Answer:

The answer is 4

Step-by-step explanation:

hope this helps

8 0

3 years ago

Read 2 more answers

Type the correct answer in each box. A circle is centered at the point (5, -4) and passes through the point (-3, 2). The equatio

Galina-37 [17]

Answer:

$(x+ \boxed{-5})^2+(y+\boxed4)^2=\boxed{100}$

Step-by-step explanation:

Given:

Center of circle is at (5, -4).

A point on the circle is $(x_1,y_1)=(-3, 2)$

Equation of a circle with center $(h,k)$ and radius 'r' is given as:

$(x-h)^2+(y-k)^2=r^2$

Here, $(h,k)=(5,-4)$

Radius of a circle is equal to the distance of point on the circle from the center of the circle and is given using the distance formula for square of the distance as:

$r^2=(h-x_1)^2+(k-y_1)^2$

Using distance formula for the points (5, -4) and (-3, 2), we get

$r^2=(5-(-3))^2+(-4-2)^2\\r^2=(5+3)^2+(-6)^2\\r^2=8^2+6^2\\r^2=64+36=100$

Therefore, the equation of the circle is:

$(x-5)^2+(y-(-4))^2=100\\(x-5)^2+(y+4)^2=100$

Now, rewriting it in the form asked in the question, we get

$(x+ \boxed{-5})^2+(y+\boxed4)^2=\boxed{100}$

4 0

3 years ago

3(x+y)=y if (x y) is a solution to the equation above and y=0 what is the ratio x/y

Alona [7]

The equation can be rewritten as x = -2/3 y, so x/y = -2/3
However, for y=0, the ratio is undefined as you would be dividing by 0??

3 0

3 years ago

A random sample of size 15 taken from a normally distributed population revealed a sample mean of 75 and a sample variance of 25

GREYUIT [131]

Answer:

The upper limit of a 95% confidence interval for the population mean would equal 83.805.

Step-by-step explanation:

The standard deviation is the square root of the variance. Since the variance is 25, the sample's standard deviation is 5.

We have the sample standard deviation, not the population, so we use the t-distribution to solve this question.

T interval:

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 = 15 - 1 = 14

Now, we have to find a value of T, which is found looking at the t table, with 14 degrees of freedom(y-axis) and a confidence level of 0.95( $t_{95}$ ). So we have T = 1.761

The margin of error is:

M = T*s = 1.761*5 = 8.805.

The upper end of the interval is the sample mean added to M. So it is 75 + 8.805 = 83.805.

The upper limit of a 95% confidence interval for the population mean would equal 83.805.

4 0

3 years ago