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3 years ago

7

Which is the product or 11/12 and 4? A 1 1/4 B 3 2/3 C 4 1/3 D 33

1 answer:

bazaltina [42]3 years ago

8 0

Answer:

Step-by-step explanation:

11/12 × 4 = 11/3 = 3⅔

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Raj fully simplified the polynomial and then put it in standard form with the last term being 6y3. 2x2y + 8x3 – xy2 – 2x3 + 3xy2

JulijaS [17]

Original polynomial:

2x^2 y + 8x^3 - xy^2 - 2x^3 + 3xy^2 + 6y^3

Order the polynomial in ascending order of y

8x^3 - 2x^3 + 2x^2y + 3xy^2 - xy^2 + 6y^3

Add up the similar terms:

6x^3 + 2x^2y + 2xy^2 + 6y^3

Thats is the polynomial Raj ended up with.

And the first term is 6x^3.

Answer: 6x^3

3 0

3 years ago

Intersection point of Y=logx and y=1/2log(x+1)

GalinKa [24]

Answer:

The intersection is $(\frac{1+\sqrt{5}}{2},\log(\frac{1+\sqrt{5}}{2})$ .

The Problem:

What is the intersection point of $y=\log(x)$ and $y=\frac{1}{2}\log(x+1)$ ?

Step-by-step explanation:

To find the intersection of $y=\log(x)$ and $y=\frac{1}{2}\log(x+1)$ , we will need to find when they have a common point; when their $x$ and $y$ are the same.

Let's start with setting the $y$ 's equal to find those $x$ 's for which the $y$ 's are the same.

$\log(x)=\frac{1}{2}\log(x+1)$

By power rule:

$\log(x)=\log((x+1)^\frac{1}{2})$

Since $\log(u)=\log(v)$ implies $u=v$ :

$x=(x+1)^\frac{1}{2}$

Squaring both sides to get rid of the fraction exponent:

$x^2=x+1$

This is a quadratic equation.

Subtract $(x+1)$ on both sides:

$x^2-(x+1)=0$

$x^2-x-1=0$

Comparing this to $ax^2+bx+c=0$ we see the following:

$a=1$

$b=-1$

$c=-1$

Let's plug them into the quadratic formula:

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{1 \pm \sqrt{(-1)^2-4(1)(-1)}}{2(1)}$

$x=\frac{1 \pm \sqrt{1+4}}{2}$

$x=\frac{1 \pm \sqrt{5}}{2}$

So we have the solutions to the quadratic equation are:

$x=\frac{1+\sqrt{5}}{2}$ or $x=\frac{1-\sqrt{5}}{2}$ .

The second solution definitely gives at least one of the logarithm equation problems.

Example: $\log(x)$ has problems when $x \le 0$ and so the second solution is a problem.

So the $x$ where the equations intersect is at $x=\frac{1+\sqrt{5}}{2}$ .

Let's find the $y$ -coordinate.

You may use either equation.

I choose $y=\log(x)$ .

$y=\log(\frac{1+\sqrt{5}}{2})$

The intersection is $(\frac{1+\sqrt{5}}{2},\log(\frac{1+\sqrt{5}}{2})$ .

6 0

3 years ago

Write the equation y - 4 = -2(x - 5) in slope-intercept form.

scoundrel [369]

The answer is y=-2x+14 .

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3 years ago

Consider a variable, , that follows the process For the first three years, and ; for the next three years, and . If the initial

bearhunter [10]

Answer:

N (20, 75)

Step-by-step explanation:

Explanation described in the attached document.

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3 years ago

Let x = 8.99999 . (a) Is x < 9 or is x = 9? x < 9 It is neither; x > 9 x = 9 x < 9 and x = 9 It cannot be determined

qaws [65]

Answer: idk

Step-by-step explanation:

idk

4 0

4 years ago