HELP MIDDLE SCHOOL QUESTION

Three students from Milton Middle School are running for class president. A preliminary poll was taken in three homeroom classes

xxTIMURxx [149]

Your answer is ian because he has the most votes

5 0

4 years ago

Read 2 more answers

n parallelogram LMNO, MP = 21 m, LP = (y + 3) m, NP = (3y – 1) m, and OP = (2x – 1) m. What are the values of x and y? x = 10, y

seropon [69]

Answer:

The value of x = 11 and y = 2

Step-by-step explanation:

Given : parallelogram LMNO, MP = 21 m, LP = (y + 3) m, NP = (3y – 1) m, and OP = (2x – 1) m.

We have to find values of x and y.

Let P be the point of intersection of diagonals OM and LN.

In a parallelogram diagonal bisects at right angles and point of intersection divide diagonal in equal parts.

Thus, OP = MP and LP = PN

OP = MP , substitute the values, we get,

(2x-1) = 21

⇒ 2x = 22

⇒ x = 11

LP = PN , substitute the values, we get,

y + 3 = 3y -1

⇒ 3y - y = 4

⇒ 2y = 4

⇒ y = 2

Thus, the value of x = 11 and y = 2

8 0

4 years ago

Write out the form of the partial fraction decomposition of the function. Do not determine the numerical values of the coefficie

Dvinal [7]

For part (a), you have

$\dfrac x{x^2+x-6}=\dfrac x{(x+3)(x-2)}=\dfrac a{x+3}+\dfrac b{x-2}$
$x=a(x-2)+b(x+3)$

If $x=2$ , then $2=b(2-3)\implies b=-2$ .

If $x=-3$ , then $-3=a(-3-2)\implies a=\dfrac35$ .

So,

$\dfrac x{x^2+x-6}=\dfrac 3{5(x+3)}-\dfrac 2{x-2}$

For part (b), since the degrees of the numerator and denominator are the same, you first need to find the quotient and remainder upon division.

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

In the remainder term, the denominator $x^2+x+2$ can't be factorized into linear components with real coefficients, since the discriminant is negative $(1-4\times1\times2=-7)$ . However, you can still factorized over the complex numbers, so a partial fraction decomposition in terms of complexes does exist.

$x^2+x+2=0\implies x=-\dfrac12\pm\dfrac{\sqrt7}2i$
$\implies x^2+x+2=\left(x-\left(-\dfrac12+\dfrac{\sqrt7}2i\right)\right)\left(x-\left(-\dfrac12-\dfrac{\sqrt7}2i\right)\right)$
$\implies x^2+x+2=\left(x+\dfrac12-\dfrac{\sqrt7}2i\right)\left(x+\dfrac12+\dfrac{\sqrt7}2i\right)$

Then you have

$\dfrac{x+2}{x^2+x+2}=\dfrac a{x+\dfrac12-\dfrac{\sqrt7}2i}+\dfrac b{x+\dfrac12+\dfrac{\sqrt7}2i}$
$x+2=a\left(x+\dfrac12+\dfrac{\sqrt7}2i\right)+b\left(x+\dfrac12-\dfrac{\sqrt7}2i\right)$

When $x=-\dfrac12-\dfrac{\sqrt7}2i$ , you have

$-\dfrac12-\dfrac{\sqrt7}2i+2=b\left(-\dfrac12-\dfrac{\sqrt7}2i+\dfrac12-\dfrac{\sqrt7}2i\right)$
$\dfrac32-\dfrac{\sqrt7}2i=-\sqrt7ib$
$b=\dfrac12+\dfrac3{2\sqrt7}i=\dfrac1{14}(7+3\sqrt7i)$

When $x=-\dfrac12+\dfrac{\sqrt7}2i$ , you have

$-\dfrac12+\dfrac{\sqrt7}2i+2=a\left(-\dfrac12+\dfrac{\sqrt7}2i+\dfrac12+\dfrac{\sqrt7}2i\right)$
$\dfrac32+\dfrac{\sqrt7}2i=\sqrt7ia$
$a=\dfrac12-\dfrac3{2\sqrt7}i=\dfrac1{14}(7-3\sqrt7i)$

So, you could write

$\dfrac{x^2}{x^2+x+2}=1-\dfrac{x+2}{x^2+x+2}=1-\dfrac {7-3\sqrt7i}{14\left(x+\dfrac12-\dfrac{\sqrt7}2i\right)}-\dfrac {7+3\sqrt7i}{14\left(x+\dfrac12+\dfrac{\sqrt7}2i\right)}$

but that may or may not be considered acceptable by that webpage.

5 0

3 years ago

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6-2x=2(15-x)<br> A. No solutions <br> B. -5<br> C. 5<br> D. All real numbers

andrezito [222]

6-2x=2(15-x)
distribute
6-2x=30-2x
add 2x to both sides
6=30
fasle

no solution
A is answer

5 0

3 years ago

X - 15 = -20 please show work

Natasha2012 [34]

Answer:

x = -5