All categories

3 years ago

PLS HELP I NEED HELP SO SOOON PLEASEEDSrewrite without absolute value for the given condition y=|x-5|+|x+5| if -5

Mathematics

1 answer:

polet [3.4K]3 years ago

5 0

answer: i think i failed this part in class

You might be interested in

Order the sides or angles of each triangle from least to greatest.

lubasha [3.4K]

Answer:

SU , UT , ST

Step-by-step explanation:

7 0

3 years ago

3 minus the product of 2 and a number X

almond37 [142]

Answer:

3 - 2x

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Given CS=2x+1, SB=6x, CR=7.5, and RA=18. What must the value of x be in order to prove SR || BA? Justify your answer

Anna [14]

The picture in the attached figure

we know that
In similar triangles. The ratio of the lengths of the sides CS and CB must be equal to the ratio of the lengths of sides CR and CA. CS / CB = CR / CA
which can also be written as,
CS / (CS + SB) = CR / (CR + RA)
CS*(CR+RA)=CR*(CS+RA)
CS=2x+1
SB=6x
CR=7.5
RA=18
(2x+1)*[7.5+18]=7.5*[2x+1+18]
(2x+1)*[25.5]=7.5*[2x+19]
(51x+25.5)=15x+142.5
51x-15x=142.5-25.5
36x=117
x=117/36
x=3.25

the answer is
x=3.25

5 0

3 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

Read 2 more answers

Write each height using scientific notation 1700000000 and 152000000

Gnesinka [82]

1.7 times 10 to the 9 power 1.52 times10 to the 8 power

5 0

3 years ago