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3 years ago

Can someone please help me?????!!!!

Mathematics

1 answer:

Snowcat [4.5K]3 years ago

7 0

What changes might help "Happiness is a charming Charlie Brown at orlando rep" to take on the general purpose of "You're a good man, Charlie Brown?

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Urgent please help!!

wolverine [178]

Answer:

i think it would be B because if AC= 80 and thats like the distane between then abc would be 80 if im thinking right

Step-by-step explanation:

but hey if im wrong ill have my friend punch me in the arm to make it even

4 0

2 years ago

Find the values of x and y?

lakkis [162]

Answer:x=20 y=70

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

Find the midpoint of the segment having endpoints (0 ,1/8) and (-4/5 ,0)

kvasek [131]

Answer:

Mid point of the given end points = $(\frac{-2}{5},\frac{1}{16} )$

Step-by-step explanation:

Given the end points of the line segment are :

(0, $\frac{1}{8}$ ) & ( $\frac{-4}{5}$ ,0)

We will use the mid point formula when two points are: (x₁,y₁) & (x₂,y₂)

mid point = $(\frac{x_{1} +x_{2} }{2},\frac{y_{1} +y_{2} }{2} )$

now put the value of x₁ = 0

x₂ = $\frac{-4}{5}$

y₁ = $\frac{1}{8}$

y₂ = 0

mid point = $(\frac{0-\frac{4}{5} }{2},\frac{\frac{1}{8}+0 }{2})$

= $(\frac{-2}{5},\frac{1}{16} )$

That's the final answer.

3 0

2 years ago

Pls need due in 10 minutes

denpristay [2]

Answer:

392.00

Step-by-step explanation:

I'm pretty sure you just multiply the 46m x 2 and 150 x 2 both = the length of the whole track. and then add the two together

3 0

2 years ago

Read 2 more answers

Please solve the problem

jek_recluse [69]

Treat the matrices on the right side of each equation like you would a constant.

Let 2X + Y = A and 3X - 4Y = B.

Then you can eliminate Y by taking the sum

4A + B = 4 (2X + Y) + (3X - 4Y) = 11X

==> X = (4A + B)/11

Similarly, you can eliminate X by using

-3A + 2B = -3 (2X + Y) + 2 (3X - 4Y) = -11Y

==> Y = (3A - 2B)/11

It follows that

$X=\dfrac4{11}\begin{bmatrix}12&-3\\10&22\end{bmatrix}+\dfrac1{11}\begin{bmatrix}7&-10\\-7&11\end{bmatrix} \\\\ X=\dfrac1{11}\left(4\begin{bmatrix}12&-3\\10&22\end{bmatrix}+\begin{bmatrix}7&-10\\-7&11\end{bmatrix}\right) \\\\ X=\dfrac1{11}\left(\begin{bmatrix}48&-12\\40&88\end{bmatrix}+\begin{bmatrix}7&-10\\-7&11\end{bmatrix}\right) \\\\ X=\dfrac1{11}\begin{bmatrix}55&-22\\33&99\end{bmatrix} \\\\ X=\begin{bmatrix}5&-2\\3&9\end{bmatrix}$

Similarly, you would find

$Y=\begin{bmatrix}2&1\\4&4\end{bmatrix}$

You can solve the second system in the same fashion. You would end up with

$P=\begin{bmatrix}2&-3\\0&1\end{bmatrix} \text{ and } Q=\begin{bmatrix}1&2\\3&-1\end{bmatrix}$

3 0

2 years ago