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alexdok [17]
3 years ago
6

HELP MEEEEEEEEEE......?.?.?(.?:).).).).).).).)

Mathematics
2 answers:
stepladder [879]3 years ago
8 0

Answer:

The answer is C

Step-by-step explanation:

Let X and Y be two proportional quantities and let K be constant.

Y=KX so Y/X=K

So for Equation number 1 we have 4y=3x and we will fill it in for 4/3=K

solving to get 1.33...

Now repeat this for the rest to complete the chart

1-  1.33...

2.  1.6

3-  1.33...

4-  0.625

Now we can see that 1 and 3 are the same, so that's how you get your answer, hope this helps!

Naddik [55]3 years ago
4 0

Answer:

c i think

Step-by-step explanation:

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What is the quotient when 4x3 + 2x + 7 is divided by x + 3?
Arte-miy333 [17]

Answer:

The quotient of this division is (4x^2 -12x + 38). The remainder here would be -26.

Step-by-step explanation:

The numerator 4x^3 + 2x + 7 is a polynomial about x with degree 3.

The divisor x + 3 is a polynomial, also about x, but with degree 1.

By the division algorithm, the quotient should be of degree 3 - 1 = 2, while the remainder shall be of degree 1 - 1 = 0 (i.e., the remainder would be a constant.) Let the quotient be a\,x^2 + b\, x + c with coefficients a, b, and c.

4x^3 + 2x + 7 = \left(a\,x^2 + b\, x + c\right)(x + 3).

Start by finding the first coefficient of the quotient.

The degree-three term on the left-hand side is 4 x^3. On the right-hand side, that would be a\, x^3. Hence a = 4.

Now, given that a = 4, rewrite the right-hand side:

\begin{aligned}&\left(4\,x^2 + b\, x + c\right)(x + 3) \cr =& \left(4x^2 + (b\, x + c)\right)(x + 3) \cr =& 4x^2(x + 3) + (bx + c)(x + 3) \cr =& 4x^3 + 12x^2 + (bx + c)(x + 3)\end{aligned}.

Hence:

4x^3 + 2x + 7 = 4x^3 + 12x^2 + (b\,x + c)(x + 3)

Subtract \left(4x^3 + 12x^2\right from both sides of the equation:

-12x^2 + 2x + 7 = (b\,x + c)(x + 3).

The term with a degree of two on the left-hand side has coefficient (-12). Since the only term on the right hand side with degree two would have coefficient b, b = -12.

Again, rewrite the right-hand side:

\begin{aligned}&\left(-12 x + c\right)(x + 3) \cr =& \left(-12 x+ c\right)(x + 3) \cr =& (-12x)(x + 3) + c(x + 3) \cr =& -12x^2 -36x + (bx + c)(x + 3)\end{aligned}.

Subtract -12x^2 -36x from both sides of the equation:

38x + 7 = c(x + 3).

By the same logic, c = 38.

Hence the quotient would be (4x^2 - 12x + 38).

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