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2 years ago

WILL MARK BRAINLIEST!!! 40 POINTS!! ACTUAL ANSWERS, PLZZZ

Mathematics

2 answers:

vitfil [10]2 years ago

4 0

Answer:

Step-by-step explanation:

PART A: 2x^5 + 3x^2-3

It is a fifth degree polynomial because, the highest degree is 5 and it is in the standard form since, the polynomial is written in descending order i.e, from highest degree to the least

part B:

Closure property is applicable to the subtraction of polynomials.

for example 2x^4+2x^2-5 is a polynomial and 2x^3+2x+2 is also a polynomial.

if we subtract these two polynomials, the outcome

2x^4-2x^3+2x^2-2x-7 is also a polynomial

AlexFokin [52]2 years ago

4 0

Step-by-step explanation:

Part A

The fifth degree polynomial

x^5 + x^4 +3x²+4x+9

the three terms 3x², 4x and 9 are forming a second degree polynomial written in standard form

since it has this form : ax²+bx+c

Part B

The closure property for real numbers means that the sum oe the substraction of two real numbers

The substraction of two polynomials is a polynom using the closure property

The nature remains the same

Here is an example :

3x²+5x+8 and x²+x+1

calculate the substraction

3x²+5x+8-(x²+x+1)
3x²+5x+8-x²-x-1
2x²+4x+7

The result is a polynomial

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2 years ago

What two formulas are necessary to find the area of a segment of a circle?

tresset_1 [31]

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Step-by-step explanation:

Area of segment=area of sector - area of triangle

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3 years ago

Please help!!! 30 minutes left

NeX [460]

Answer:

Option "D" is correct.

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2 years ago

Problem 4: Solve the initial value problem

pishuonlain [190]

Separate the variables:

$y' = \dfrac{dy}{dx} = (y+1)(y-2) \implies \dfrac1{(y+1)(y-2)} \, dy = dx$

Separate the left side into partial fractions. We want coefficients a and b such that

$\dfrac1{(y+1)(y-2)} = \dfrac a{y+1} + \dfrac b{y-2}$

$\implies \dfrac1{(y+1)(y-2)} = \dfrac{a(y-2)+b(y+1)}{(y+1)(y-2)}$

$\implies 1 = a(y-2)+b(y+1)$

$\implies 1 = (a+b)y - 2a+b$

$\implies \begin{cases}a+b=0\\-2a+b=1\end{cases} \implies a = -\dfrac13 \text{ and } b = \dfrac13$

So we have

$\dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = dx$

Integrating both sides yields

$\displaystyle \int \dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = \int dx$

$\dfrac13 \left(\ln|y-2| - \ln|y+1|\right) = x + C$

$\dfrac13 \ln\left|\dfrac{y-2}{y+1}\right| = x + C$

$\ln\left|\dfrac{y-2}{y+1}\right| = 3x + C$

$\dfrac{y-2}{y+1} = e^{3x + C}$

$\dfrac{y-2}{y+1} = Ce^{3x}$

With the initial condition y(0) = 1, we find

$\dfrac{1-2}{1+1} = Ce^{0} \implies C = -\dfrac12$

so that the particular solution is

$\boxed{\dfrac{y-2}{y+1} = -\dfrac12 e^{3x}}$

It's not too hard to solve explicitly for y; notice that

$\dfrac{y-2}{y+1} = \dfrac{(y+1)-3}{y+1} = 1-\dfrac3{y+1}$

Then

$1 - \dfrac3{y+1} = -\dfrac12 e^{3x}$

$\dfrac3{y+1} = 1 + \dfrac12 e^{3x}$

$\dfrac{y+1}3 = \dfrac1{1+\frac12 e^{3x}} = \dfrac2{2+e^{3x}}$

$y+1 = \dfrac6{2+e^{3x}}$

$y = \dfrac6{2+e^{3x}} - 1$

$\boxed{y = \dfrac{4-e^{3x}}{2+e^{3x}}}$

7 0

2 years ago

What is the value of z?

Licemer1 [7]

Answer:

z = 23 degrees

Step-by-step explanation:

Total degree inside a triangle = 180

We can add up the 2 angles we got.

62 + 95 = 157

Subtract 157 from 180.

180 - 157 = 23

Check = the angle missing is an acute angle (less than 90 degrees) and 23 is less than 90 degrees so it is possible.

I hope this helped and please mark me as brainliest!

6 0

2 years ago

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