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3 years ago

Find an equation of the line that bisects the acute angles formed by the lines with equations 2x+y-5=0 and 3x-2y+6=0

1 answer:

5 0

The solving for the first one should be like this

$2x+y-5/sqrt 2^2 1^2=2x+y-5/sqrt5$
And the next part $3x-2y+6/sqrt 3^2 -2^2=3x-2y+6/sqrt13$

After these manipulations just multiply them
$
(2sqrt13-3sqrt5)x+(sqrt13+2sqrt5) y-5sqrt13-6sqrt5=0 
$
And finally you need to square and simplify for the completed answer

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A site is 90m long and 45m wide what is the area of the site

Andreyy89

Answer:4050m^2

Step-by-step explanation:

Assuming that the site is rectangular

Area= l x W

90 X 45

=4050

5 0

3 years ago

Read 2 more answers

A bag contain 3 black balls and 2 white balls.

Troyanec [42]

Answer:

Step-by-step explanation:

Total number of balls = 3 + 2 = 5

$Probability \ of \ taking \ 2 \ black \ ball \ with \ replacement\\\\ = \frac{3C_1}{5C_1} \times \frac{3C_1}{5C_1} =\frac{3}{5} \times \frac{3}{5} = \frac{9}{25}\\\\$

$Probability \ of \ one \ black \ and \ one\ white \ with \ replacement \\\\= \frac{3C_1}{5C_1} \times \frac{2C_1}{5C_1} = \frac{3}{5} \times \frac{2}{5} = \frac{6}{25}$

Probability of at least one black( means BB or BW or WB)

$=\frac{3}{5} \times \frac{3}{5} + \frac{3}{5} \times \frac{2}{5} + \frac{2}{5} \times \frac{3}{5} \\\\= \frac{9}{25} + \frac{6}{25} + \frac{6}{25}\\\\= \frac{21}{25}$

Probability of at most one black ( means WW or WB or BW)

$=\frac{2}{5} \times \frac{2}{5} + \frac{3}{5} \times \frac{2}{5} \times \frac{2}{5} + \frac{3}{5}\\\\= \frac{4}{25} + \frac{6}{25} + \frac{6}{25}\\\\=\frac{16}{25}$

a) Probability both black without replacement

$=\frac{3}{5} \times \frac{2}{4}\\\\=\frac{6}{20}\\\\=\frac{3}{10}$

b) Probability of one black and one white

$=\frac{3}{5} \times \frac{2}{4}\\\\=\frac{6}{20}\\\\=\frac{3}{10}$

c) Probability of at least one black ( BB or BW or WB)

$=\frac{3}{5} \times \frac{2}{4} + \frac{3}{5} \times \frac{2}{4} + \frac{2}{5} \times \frac{3}{4}\\\\=\frac{6}{20} + \frac{6}{20} + \frac{6}{20} \\\\=\frac{18}{20} \\\\=\frac{9}{10}$

d) Probability of at most one black ( BW or WW or WB)

$=\frac{3}{5} \times \frac{2}{4} + \frac{2}{5} \times \frac{1}{4} + \frac{2}{5} \times \frac{3}{4}\\\\=\frac{6}{20} + \frac{2}{20} + \frac{6}{20} \\\\=\frac{14}{20}\\\\=\frac{7}{10}$

6 0

3 years ago

Find the unknown number 5/12+()=1/2

xenn [34]

Hello from MrBillDoesMath!

Answer: 1/12

Discussion:

We solve 5/12 + x = 1/2 for x. Begin by subtracting 5/12 from each side

5/12 - 5/12 + x = 1/2 - 5/12

0 +x = 6/12 - 5/12 (as 1/2 = 6/12)

So x = 1/12

Regards, MrB

8 0

3 years ago

What does d equal in -24 + 12 (d - 3) + 22

zubka84 [21]

Answer:

-24+12(d-3)+22=-24+34(d-3)

10(d-3)

10d=-30

d=-30/10

d=3

4 0

3 years ago

A ball is thrown into the air with an upward velocity of 80ft/s. Its height H in feet after T seconds id given by the function H

pashok25 [27]

1. The function H= -16T^2+80T+5 is a parabola of the form $a x^{2} +bx+c$ , so to find the maximum height of the ball, we are going to find the y-coordinate of the vertex of the parabola. To find the y-coordinate of the vertex we are going to evaluate the function at the point $\frac{-b}{2a}$ .
From our function we can infer that $b=80$ and $a=-16$ , so the point \frac{-b}{2a} [/tex] will be $\frac{-80}{-2(16)} = \frac{5}{2}$ . Lets evaluate the function at that point:
$h=-16t^2+80t+5$
$h=-16( \frac{5}{2} )^2+80( \frac{5}{2} )+5$
$h=105$

We can conclude that the ball reaches a maximum height of 105 feet.

2. Since we now know that the maximum height the ball reaches is 105 feet, we are going to replace $h$ with 105 in our function, then we are going to solve for $t$ to find how long the ball takes to reach its maximum height:
$h=-16t^2+80t+5$
$105=-16t^{2}+80t+5$
$-16t^2+80t-100=0$
$-4(4t^{2}+20t-25)=0$
$4(2t-5)^2=0$
$2t-5=0$
$t= \frac{5}{2}$
$t=2.5$

We can conclude that the ball reaches its maximum height in 2.5 seconds.

3. Just like before, we are going to replace $h$ with 5 in our original function, then we are going to solve for $t$ to find how long will take for the ball to be caught 5 feet off the ground:
$h=-16t^2+80t+5$
$5=-16t^2+80t+5$
$-16t^{2}+80t=0$
$-16t(t-5)=0$
$t-5=0$
$t=5$

We can conclude that it takes 5 seconds for the ball to be caught 5 feet off the ground.

5 0

4 years ago