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Kamila [148]
3 years ago
13

(x+1)^2 (x-4)(-2x+3) degree? Ld. Coef? y-int?

Mathematics
1 answer:
8_murik_8 [283]3 years ago
6 0
(x + 1)²(x - 4)(-2x + 3)
(x + 1)(x + 1)(x - 4)(-2x + 3)
(x² + x + x + 1)(-8x² + 3x + 8x - 12)
(x² + 2x + 1)(-8x² + 11x - 12)
(-8x^4 + 11x³ - 12x² - 16x³ + 11x² - 12x - 8x² + 11x - 12)
(8x^4 + 11x³ - 16x³ - 12x² + 11x² - 8x² - 12x - 11x - 12)
8x^4 - 5x³ + 15x² - 23x - 12
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Write an equivalent expression for the following by combining like terms: 4y − y + 7y
Assoli18 [71]
4y - y + 7y

4y - y = 3y

3y + 7y = 10y

10y is your answer

hope this helps
4 0
3 years ago
Suppose each of the following data sets is a simple random sample from some population. For each dataset, make a normal QQ plot.
adell [148]

Answer:

a) For this case the histogram is not too skewed and we can say that is approximately symmetrical so then we can conclude that this dataset is similar to a normal distribution

b) For this case the data is skewed to the left and we can't assume that we have the normality assumption.

c) This last case the histogram is not symmetrical and the data seems to be skewed.

Step-by-step explanation:

For this case we have the following data:

(a)data = c(7,13.2,8.1,8.2,6,9.5,9.4,8.7,9.8,10.9,8.4,7.4,8.4,10,9.7,8.6,12.4,10.7,11,9.4)

We can use the following R code to get the histogram

> x1<-c(7,13.2,8.1,8.2,6,9.5,9.4,8.7,9.8,10.9,8.4,7.4,8.4,10,9.7,8.6,12.4,10.7,11,9.4)

> hist(x1,main="Histogram a)")

The result is on the first figure attached.

For this case the histogram is not too skewed and we can say that is approximately symmetrical so then we can conclude that this dataset is similar to a normal distribution

(b)data = c(2.5,1.8,2.6,-1.9,1.6,2.6,1.4,0.9,1.2,2.3,-1.5,1.5,2.5,2.9,-0.1)

> x2<- c(2.5,1.8,2.6,-1.9,1.6,2.6,1.4,0.9,1.2,2.3,-1.5,1.5,2.5,2.9,-0.1)

> hist(x2,main="Histogram b)")

The result is on the first figure attached.

For this case the data is skewed to the left and we can't assume that we have the normality assumption.

(c)data = c(3.3,1.7,3.3,3.3,2.4,0.5,1.1,1.7,12,14.4,12.8,11.2,10.9,11.7,11.7,11.6)

> x3<-c(3.3,1.7,3.3,3.3,2.4,0.5,1.1,1.7,12,14.4,12.8,11.2,10.9,11.7,11.7,11.6)

> hist(x3,main="Histogram c)")

The result is on the first figure attached.

This last case the histogram is not symmetrical and the data seems to be skewed.

7 0
3 years ago
Are these ratios equivalent?
zhenek [66]
No because 5/2= 2,5 however 63/19=3,66...
7 0
2 years ago
PLEASE SOLVE DUE RIGHT NOW!!!!!!!!!!!!
erma4kov [3.2K]

Solution:

<u>We know that:</u>

Area \ of \ triangle = \dfrac{1}{2} \times Base \times Height

<em>Since we are already given the base and the height, we can substitute them into the formula to find the area of the triangle.</em>

Area \ of \ triangle = \dfrac{1}{2} \times 18 \times24

<u>Cancel out the "2" by dividing it by 18.</u>

Area \ of \ triangle =  9 \times24

<u>Simplify the multiplication.</u>

\bold{Area \ of \ triangle =  216 \ ft^{2}}

6 0
2 years ago
What is the completely factored form of this expression?
Eddi Din [679]

Answer:

B.

(3x + 4)(x − 7)

Step-by-step explanation:

If you multiply.

3x2-21x+4x-28

3X2-17x-28

8 0
3 years ago
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