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2 years ago

2y=12x-10 and y+=6x+12 Do the lines intersect, coincide, or are they parallel? 50 pts and ill give brainly to correct answer

Mathematics

2 answers:

astraxan [27]2 years ago

7 0

Answer:intersect

Step-by-step explanation:

NikAS [45]2 years ago

6 0

Answer:

parallel

Step-by-step explanation:

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Evaluate for x=4 and y=-3 write in simplest form<br> 2y/x2

lorasvet [3.4K]

Hello from MrBillDoesMath!

Answer:

I think your formula (2y/x2) is equivalent to

(2*y)/(x^2) where x^2 means x-squared.

If so, then

(2*y)/(x^2) = (2*3)/(4*4) = 6/16 = 3/8

Thank you,

MrB

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3 years ago

Which of the following is a result of the construction of a median of a triangle?

olga2289 [7]

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Answer:

A side of the triangle is bisected

Step-by-step explanation:

The median of a triangle connects a vertex to the midpoint of the opposite side. That is, it bisects the side it meets:

A side of the triangle is bisected

_____

<em>Comment on other choices</em>

A median will be an angle bisector only if the triangle is isosceles. Likewise, it will only be perpendicular to the side if the triangle is isosceles.

There are several points called "center" of a triangle, including the centroid (on the median), the incenter, the circumcenter, and the orthocenter.

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3 years ago

An airplane makes a 15° angle of elevation from the runway

Jet001 [13]

Answer: 536 feet tall

Step-by-step explanation:

2000 ( tan 15 ) = 536

6 0

3 years ago

Simplifying square root

oee [108]

Answer is A. 6
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3 0

3 years ago

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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