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Mariulka [41]
3 years ago
7

Name the property illustrated be each statement. (3x+2)-5=(3x+2)-5

Mathematics
1 answer:
Evgesh-ka [11]3 years ago
8 0
The associative property states that we can add numbers in any way we want to still get the same sum, so the parenthesis here can essentially be tossed out to get 3x+2-5=3x-3. However, make sure, for example, that you don't add x and 1 together in 3(x)+1 as you still have to multiply the x with 3 before adding the 1 according to PEMDAS
You might be interested in
I need help from question 11- 16! Please help!
insens350 [35]
<h2>11. Find discriminant.</h2>

Answer: D) 0, one real solution

A quadratic function is given of the form:

ax^2+bx+c=

We can find the roots of this equation using the quadratic formula:

x_{12}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}

Where \Delta=b^2-4ac is named the discriminant. This gives us information about the roots without computing them. So, arranging our equation we have:

4a^2-4a-6=-7 \\ \\ Adding \ 7 \ to \ both \ sides \ of \ the \ equation: \\ \\ 4a^2-4a-6+7=-7+7 \\ \\ 4a^2-4a+1=0 \\ \\ Then \ the \ discriminant: \\ \\ \Delta=(-4)^2-4(4)(1) \\ \\ \Delta=16-16 \\ \\ \boxed{Delta=0}

<em>Since the discriminant equals zero, then we just have one real solution.</em>

<h2>12. Find discriminant.</h2>

Answer: D) -220, no real solution

In this exercise, we have the following equation:

-r^2-2r+14=-8r^2+6

So we need to arrange this equation in the form:

ax^2+bx+c=

Thus:

-r^2-2r+14=-8r^2+6 \\ \\ Adding \ 8r^2 \ to \ both \ sides \ of \ the \ equation: \\ \\ -r^2-2r+14+8r^2=-8r^2+6+8r^2 \\ \\ Associative \ Property: \\ \\ (-r^2+8r^2)-2r+14=(-8r^2+8r^2)+6 \\ \\ 7r^2-2r+14=6 \\ \\ Subtracting \ 6 \ from \ both \ sides: \\ \\ 7r^2-2r+14-6=6-6 \\ \\ 7r^2-2r+8=0

So the discriminant is:

\Delta=(-2)^2-4(7)(8) \\ \\ \Delta=4-224 \\ \\ \boxed{\Delta=-220}

<em>Since the discriminant is less than one, then there is no any real solution</em>

<h2>13. Value that completes the squares</h2>

Answer: C) 144

What we need to find is the value of c such that:

x^2+24x+c=0

is a perfect square trinomial, that are given of the form:

a^2x^2\pm 2axb+b^2

and can be expressed in squared-binomial form as:

(ax\pm b)^2

So we can write our quadratic equation as follows:

x^2+2(12)x+c \\ \\ So: \\ \\ a=1 \\ \\ b=12 \\ \\ c=b^2 \therefore c=12^2 \therefore \boxed{c=144}

Finally, the value of c that completes the square is 144 because:

x^2+24x+144=(x+12)^2

<h2>14. Value that completes the square.</h2>

Answer: C) \frac{121}{4}

What we need to find is the value of c such that:

z^2+11z+c=0

So we can write our quadratic equation as follows:

z^2+2\frac{11}{2}z+c \\ \\ So: \\ \\ a=1 \\ \\ b=\frac{11}{2} \\ \\ c=b^2 \therefore c=\left(\frac{11}{2}\left)^2 \therefore \boxed{c=\frac{121}{4}}

Finally, the value of c that completes the square is \frac{121}{4} because:

z^2+11z+\frac{121}{4}=(x+\frac{11}{2})^2

<h2 /><h2>15. Rectangle.</h2>

In this problem, we need to find the length and width of a rectangle. We are given the area of the rectangle, which is 45 square inches. We know that the formula of the area of a rectangle is:

A=L\times W

From the statement we know that the length of the rectangle is is one inch less than twice the width, this can be written as:

L=2W-1

So we can introduce this into the equation of the area, hence:

A=L\times W \\ \\ \\ Where: \\ \\ W:Width \\ \\ L:Length

A=(2W-1)(W) \\ \\ But \ A=45: \\ \\ 45=(2W-1)(W) \\ \\ Distributive \ Property:\\ \\ 45=2W^2-W \\ \\ 2W^2-W-45=0 \\ \\ Quadratic \ Formula: \\ \\ x_{12}=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ \\ W_{1}=\frac{-(-1)+ \sqrt{(-1)^2-4(2)(-45)}}{2(2)} \\ \\ W_{1}=\frac{1+ \sqrt{1+360}}{4} \therefore W_{1}=5 \\ \\ W_{2}=\frac{-(-1)- \sqrt{(-1)^2-4(2)(-45)}}{2(2)} \\ \\ W_{2}=\frac{1- \sqrt{1+360}}{4} \therefore W_{2}=-\frac{9}{2}

The only valid option is W_{1} because is greater than zero. Recall that we can't have a negative value of the width. For the length we have:

L=2(5)-1 \\ \\ L=9

Finally:

The \ length \ is \ 9 \ inches \\ \\ The \ width \ is \ 5 \ inches

<h2>16. Satellite</h2>

The distance in miles between mars and a satellite is given by the equation:

d=-9t^2+776

where t is the number of hours it has fallen. So we need to find when the satellite will be 452 miles away from mars, that is, d=452:

d=-9t^2+776 \\ \\ 452=-9t^2+776 \\ \\ 9t^2=776-452 \\ \\ 9t^2=324 \\ \\ t^2=\frac{324}{9} \\ \\ t^2=36 \\ \\ t=\sqrt{36} \\ \\ \boxed{t=6h}

Finally, <em>the satellite will be 452 miles away from mars in 6 hours.</em>

3 0
3 years ago
I really need help for this geometry question, I would appreciate any help thank you.
irinina [24]

Answer:

75=x

Step-by-step explanation:

They are the same triangle indicated by the dashes and the angle marks

so x would be the same as the angle A

Triangles add up to 180 degrees

so

180-45-60=75

5 0
3 years ago
An arrangement of two dimensionals figures that can be folded to make a three dimensionals figure
kirill115 [55]

wht are you trying to say


7 0
3 years ago
The box plots show the weights, in pounds, of the dogs in two different animal shelters. Weights of Dogs in Shelter A 2 box plot
katovenus [111]

Answer:

1) The median weight for shelter A is greater than that for shelter B.

True, the median weight for shelter A is 21 pounds as compared to the 18 pounds median weight for shelter B.

4) The data for shelter B are a symmetric data set.

True, the median is exactly 2 pounds away from each quartile which means it is symmetric.

5) The interquartile range of shelter A is greater than the interquartile range of shelter B.

True, the interquartile range of shelter A is 11 pounds as compared to the 4 pounds interquartile range of shelter B.

Step-by-step explanation:

We are given two box-plots, a box plot basically shows 5 statistical characteristics of a data set and they are  

1. minimum value  

2. Lower quartile  

3. Median value  

4. Upper quartile  

5. Maximum value  

Box-plot of dogs in shelter A:

The whiskers range from 8 to 30

Which means that the range is = 30 - 8 = 22

The box ranges from 17 to 28

Which means that the interquartile range is = 28 - 17 = 11

21 - 17 = 4

28 - 21 = 9

The median is not equal distance away from each quartile which means that the box-plot is not symmetric.

A line divides the box at 21

Which means that the median is = 21

Box-plot of dogs in shelter B:

The whiskers range from 10 to 28

Which means that the range is = 28 - 10 = 20

The box ranges from 16 to 20

Which means that the interquartile range is = 20 - 16 = 4

18 - 16 = 2

20 - 18 = 2

The median is exactly 2 pounds away from each quartile which means it is symmetric.

A line divides the box at 18

Which means that the median is = 18

Which is true of the data in the box plots?

1) The median weight for shelter A is greater than that for shelter B.

True, the median weight for shelter A is 21 pounds as compared to the 18 pounds median weight for shelter B.

2) The median weight for shelter B is greater than that for shelter A.

False

3) The data for shelter A are a symmetric data set.

False

4) The data for shelter B are a symmetric data set.

True

5) The interquartile range of shelter A is greater than the interquartile range of shelter B.

True, the interquartile range of shelter A is 11 pounds as compared to the 4 pounds interquartile range of shelter B.

5 0
3 years ago
50% of 180 dancers said they prefer hip hop class over tap class. How many dancers prefer hip hop class?
Simora [160]

Answer:

90

Step-by-step explanation:

50% of 180 = 90%

90% is the answer.

3 0
2 years ago
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