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earnstyle [38]
3 years ago
12

100 POINTS!!!! Please be quick and make sure you have a correct answer, no plagiarizing, and no silly answers. Will give BRAINLI

EST!
Part A: Create a fifth-degree polynomial with four terms in standard form. How do you know it is in standard form? (5 points)

Part B: Explain the closure property as it relates to subtraction of polynomials. Give an example. (5 points)
Mathematics
2 answers:
NikAS [45]3 years ago
6 0

Part A

5x 3y^3 2z

I know it is in standard form because there are no more like terms.

Part B: Polynomials are always closed under multiplication. Unlike with addition and subtraction, both the coefficients and exponents can change. The variables and coefficients will automatically fit in a polynomial. When there are exponents in a multiplication problem, they are added, so they will also fit in a polynomial.

ivanzaharov [21]3 years ago
3 0

Answer:

A: 3x^2 + 4y^5 - 3. I know it's in standard form because I cannot simplify it anymore.

B:  When subtracting polynomials, the variables and their exponents do not change. Only their coefficients will possibly change.  the variables and exponents will only change with multiplication and division.

Step-by-step explanation:

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two objects are travelling toward each other on a collision course. One object is doing 800 km / hour. The other is doing 1000 k
sammy [17]

Answer:

Step-by-step explanation:

The relative speed of the two trains is the sum of the speeds they are traveling. (If you're on either of the trains, this is the speed you appear to be moving when you see the other train.) In our problem, the relative speed of the two trains is 70 mph + 60 mph = 130 mph. What if the trains were traveling in the same direction? Then we'd need to subtract the speed of the slower train from the speed of the faster train, and their relative speed would be 10 mph.

6 0
3 years ago
It costs $164.34 to rent a standard-size car for 5 days. Find the price per day to rent this car.
g100num [7]
164.34 divided by the amount of days, in this case 5.
164.34 divided by 5 =32.868.

That answer can be rounded to $33 or $32.87. I would go with the first one if asked to round.
7 0
2 years ago
If A(4 -6) B(3 -2) and C (5 2) are the vertices of a triangle ABC fine the length of the median AD from A to BC. Also verify tha
Gnoma [55]

Answer:

a) The median AD from A to BC has a length of 6.

b) Areas of triangles ABD and ACD are the same.

Step-by-step explanation:

a) A median is a line that begin in a vertix and end at a midpoint of a side opposite to vertix. As first step the location of the point is determined:

D (x,y) = \left(\frac{x_{B}+x_{C}}{2},\frac{y_{B}+y_{C}}{2}  \right)

D(x,y) = \left(\frac{3 + 5}{2},\frac{-2 + 2}{2}  \right)

D(x,y) = (4,0)

The length of the median AD is calculated by the Pythagorean Theorem:

AD = \sqrt{(x_{D}-x_{A})^{2}+ (y_{D}-y_{A})^{2}}

AD = \sqrt{(4-4)^{2}+[0-(-6)]^{2}}

AD = 6

The median AD from A to BC has a length of 6.

b) In order to compare both areas, all lengths must be found with the help of Pythagorean Theorem:

AB = \sqrt{(x_{B}-x_{A})^{2}+ (y_{B}-y_{A})^{2}}

AB = \sqrt{(3-4)^{2}+[-2-(-6)]^{2}}

AB \approx 4.123

AC = \sqrt{(x_{C}-x_{A})^{2}+ (y_{C}-y_{A})^{2}}

AC = \sqrt{(5-4)^{2}+[2-(-6)]^{2}}

AC \approx 4.123

BC = \sqrt{(x_{C}-x_{B})^{2}+ (y_{C}-y_{B})^{2}}

BC = \sqrt{(5-3)^{2}+[2-(-2)]^{2}}

BC \approx 4.472

BD = CD = \frac{1}{2}\cdot BC (by the definition of median)

BD = CD = \frac{1}{2} \cdot (4.472)

BD = CD = 2.236

AD = 6

The area of any triangle can be calculated in terms of their side length. Now, equations to determine the areas of triangles ABD and ACD are described below:

A_{ABD} = \sqrt{s_{ABD}\cdot (s_{ABD}-AB)\cdot (s_{ABD}-BD)\cdot (s_{ABD}-AD)}, where s_{ABD} = \frac{AB+BD+AD}{2}

A_{ACD} = \sqrt{s_{ACD}\cdot (s_{ACD}-AC)\cdot (s_{ACD}-CD)\cdot (s_{ACD}-AD)}, where s_{ACD} = \frac{AC+CD+AD}{2}

Finally,

s_{ABD} = \frac{4.123+2.236+6}{2}

s_{ABD} = 6.180

A_{ABD} = \sqrt{(6.180)\cdot (6.180-4.123)\cdot (6.180-2.236)\cdot (6.180-6)}

A_{ABD} \approx 3.004

s_{ACD} = \frac{4.123+2.236+6}{2}

s_{ACD} = 6.180

A_{ACD} = \sqrt{(6.180)\cdot (6.180-4.123)\cdot (6.180-2.236)\cdot (6.180-6)}

A_{ACD} \approx 3.004

Therefore, areas of triangles ABD and ACD are the same.

4 0
4 years ago
Solve 75=125 (0.8)t^
valentinak56 [21]
Really night what's the problem
7 0
3 years ago
Read 2 more answers
I need help on this question as well.
Y_Kistochka [10]

8 is the answer, because 4 x 8 = 32

7 0
3 years ago
Read 2 more answers
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