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3 years ago

1/2x+3y=4 for x when y=6

Mathematics

2 answers:

vredina [299]3 years ago

6 0

X
=
2
+
3
y
2 I got the answer right

goldfiish [28.3K]3 years ago

4 0

Answer:

are you looking for the substiution

Step-by-step explanation:

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Find the radius...<br><br> Help plz

Vladimir [108]

It is 20 for the radius

6 0

3 years ago

Need help on this question

Alja [10]

He uses 15 gallons. Because 270/6 = 45. Meaning he drives 45 miles per gallon. So 675/45=15. 15 gallons he will need to use to drive 675 miles.

7 0

3 years ago

Evaluate the line integral, where c is the given curve. (x + 9y) dx + x2 dy, c c consists of line segments from (0, 0) to (9, 1)

viktelen [127]

$\displaystyle\int_C(x+9y)\,\mathrm dx+x^2\,\mathrm dy=\int_C\langle x+9y,x^2\rangle\cdot\underbrace{\langle\mathrm dx,\mathrm dy\rangle}_{\mathrm d\mathbf r}$

The first line segment can be parameterized by $\mathbf r_1(t)=\langle0,0\rangle(1-t)+\langle9,1\rangle t=\langle9t,t\rangle$ with $0\le t\le1$ . Denote this first segment by $C_1$ . Then

$\displaystyle\int_{C_1}\langle x+9y,x^2\rangle\cdot\mathbf dr_1=\int_{t=0}^{t=1}\langle9t+9t,81t^2\rangle\cdot\langle9,1\rangle\,\mathrm dt$
$=\displaystyle\int_0^1(162t+81t^2)\,\mathrm dt$
$=108$

The second line segment ( $C_2$ ) can be described by $\mathbf r_2(t)=\langle9,1\rangle(1-t)+\langle10,0\rangle t=\langle9+t,1-t\rangle$ , again with $0\le t\le1$ . Then

$\displaystyle\int_{C_2}\langle x+9y,x^2\rangle\cdot\mathrm d\mathbf r_2=\int_{t=0}^{t=1}\langle9+t+9-9t,(9+t)^2\rangle\cdot\langle1,-1\rangle\,\mathrm dt$
$=\displaystyle\int_0^1(18-8t-(9+t)^2)\,\mathrm dt$
$=-\dfrac{229}3$

Finally,

$\displaystyle\int_C(x+9y)\,\mathrm dx+x^2\,\mathrm dy=108-\dfrac{229}3=\dfrac{95}3$

5 0

3 years ago

What is the constant term of the quotient when 2x^3-3x^2 + 4x-2 is divided by x^2-x+2?

AlladinOne [14]

Answer:

-1

Step-by-step explanation:

See the attachment for the polynomial long division. The constant in the quotient is -1.

_____

Here, there is a remainder of -x. If there were no remainder the constant in the quotient is the ratio of the constant in the dividend to the constant in the divisor: -2/2 = -1.

That could be a first guess in a "guess and check" solution approach.

<em>Guess</em>: first term of binomial quotient is (2x^3)/x^2 = 2x; last term of binomial quotient is -2/2 = -1. So, the quotient is guessed to be (2x -1).

<em>Check</em>: (2x -1)(x^2 -x +2) = 2x^3 -3x^2 +5x -2

Subtracting this from the actual dividend gives a remainder of -x. This has a lower degree than the divisor, so no further adjustment of the quotient is required.

6 0

3 years ago

What second degree polynomial function has a leading coefficient of -2 and root 4 with a multiplicity of 2

OverLord2011 [107]

The second degree polynomial with leading coefficient of -2 and root 4 with multiplicity of 2 is:

$p(x) = -2*(x - 4)*(x - 4) = -2*(x - 4)^2$

<h3>How to write the polynomial?</h3>

A polynomial of degree N, with the N roots {x₁, ..., xₙ} and a leading coefficient a is written as:

$p(x) = a*(x - x_1)*(x - x_2)*...*(x - x_n)$

Here we know that the degree is 2, the only root is 4 (with a multiplicity of 2, this is equivalent to say that we have two roots at x = 4) and a leading coefficient equal to -2.

Then this polynomial is equal to:

$p(x) = -2*(x - 4)*(x - 4) = -2*(x - 4)^2$

If you want to learn more about polynomials:

brainly.com/question/4142886

#SPJ1

8 0

2 years ago