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3 years ago

I need help getting the answer to this -71+blank=21

Mathematics

1 answer:

vagabundo [1.1K]3 years ago

6 0

Answer:

It is Positive 92

Step-by-step explanation:

-71 + 92 = 21

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Help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! waaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

yaroslaw [1]

Answer:

I believe it would be - (n+4) x 2

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

Find the constant $a$ such that\[(x^2 - 3x + 4)(2x^2 +ax + 7) = 2x^4 -11x^3 +30x^2 -41x +28.\]

Colt1911 [192]

The value of constant a is -5

Further explanation:

We will use the comparison of co-efficient method for finding the value of a

So,

Given

$(x^2 - 3x + 4)(2x^2 +ax + 7)\\= x^2((2x^2 +ax + 7)-3x((2x^2 +ax + 7)+4(2x^2 +ax + 7)\\= 2x^4+ax^3+7x^2-6x^3-3ax^2-21x+8x^2+4ax+28\\Combining\ alike\ terms\\=2x^4+ax^3-6x^3+7x^2-3ax^2+8x^2-21x+4ax+28\\= 2x^4 +(a-6)x^3+(15-3a)x^2-(21-4a)x+28\\$

As it is given that

(x^2 - 3x + 4)(2x^2 +ax + 7) = 2x^4 -11x^3 +30x^2 -41x +28

In this case, co-efficient of variables will be equal, so we can compare the coefficients of x^3, x^2 or x

Comparing coefficient of x^3

$a-6 = -11\\a = -11+6\\a = -5$

So the value of constant a is -5

Keywords: Polynomials, factorization

Learn more about factorization at:

#LearnwithBrainly

4 0

3 years ago

HELPPPP PLEASE OMGGG

vlabodo [156]

Answer:

positive

negetive

negative

positive

Step-by-step explanation:

multiplication of of negetive values is positive

multiplication of positive value and negetive value is negetive

4 0

2 years ago

Write an equation for this

Cerrena [4.2K]

Answer:

Y=(-x/2)-3

Step-by-step explanation:

You have two points:

(0,-3) and (-4,-1)

The formula of line trough two points is:

(Y-y1)/(y2-y1)=(x-x1)/(x2-x1)

So put these two point like (x1,y1) and (x2,y2) you will get

(Y+3)/(-1+3)=(x-0)/(-4-0)

Y+3=2x/(-4)

Y=-x/2-3

8 0

3 years ago

Consider the implicit differential equation <img src="https://tex.z-dn.net/?f=%2849%20y%5E%7B3%7D%20%2B%2045%20xy%29%20dx%20%2B%

BaLLatris [955]

We're looking for an integrating factor $\mu(x,y)=x^py^q$ such that

$\mu\underbrace{(49y^3+45xy)}_M\,\mathrm dx+\mu\underbrace{(98xy^2+50x^2)}_N\,\mathrm dy=0$

is exact, which would require that

$(\mu M)_y=(\mu N)_x$
$(49x^py^{q+3}+45x^{p+1}y^{q+1})_y=(98x^{p+1}y^{q+2}+50x^{p+2}y^q)_x$
$49(q+3)x^py^{q+2}+45(q+1)x^{p+1}y^q=98(p+1)x^py^{q+2}+50(p+2)x^{p+1}y^q$
$\implies\begin{cases}49(q+3)=98(p+1)\\45(q+1)=50(p+2)\end{cases}\implies p=\dfrac52,q=4$

You can verify that $(\mu M)_y=(\mu N)_x$ if you'd like. With the ODE now exact, we have a solution $F(x,y)=C$ such that

$F_x=\mu M$
$F=\displaystyle\int(49y^3+45xy)x^{5/2}y^4\,\mathrm dx$
$F=10x^{9/2}y^5+14x^{7/2}y^7+f(y)$

$F_y=\mu N$
$50x^{9/2}y^4+98x^{7/2}y^6+f'(y)=98x^{7/2}y^2+50x^{9/2}y^4$
$f'(y)=0$
$\implies f(y)=C$

and so the general solution is

$F(x,y)=10x^{9/2}y^5+14x^{7/2}y^7=C$

8 0

3 years ago