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MaRussiya [10]
3 years ago
8

Write the equation of the line in slope-intercept form.

Mathematics
1 answer:
Butoxors [25]3 years ago
5 0
Equation of the perpendicular
y= 2x+6
slope of the perpendicular = 2

slope of the req. line = -1/2

equation of the req. line
y - y1 = m(x - x1)
y - 2 = -1/2(x + 4)
2(y - 2) = -1(x + 4)
2y - 4 = -x - 4
2y = -x
y  = (-1/2)x

this is your req. equation of line.
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Pls help asap due tomorrow
Svet_ta [14]

Based on the information given, it should be noted that the least number of marbles that can be in the bag will be 10 marbles.

<h2>Solution to the probability.</h2>

If the bag contains 60 marbles, the number of red marbles will be:

= 1/5 × 60

= 12 marbles.

The number of white marbles will be:

= 3/10 × 60

= 18 marbles.

The number of blue marbles will be:

= 60 - (12 + 18)

= 60 - 30

= 30 marbles.

Furthermore, when the bag has 4 red marbles and 6 white marbles, there'll be 10 blue marbles.

Lastly, to find the probability of choosing a blue marble, we've to multiply 1/2 by the number given.

Learn more about probability on:

brainly.com/question/25870256

4 0
2 years ago
How do you solve -4x - 10 &lt; 2
viva [34]
There is a way and you will find it but not from me
5 0
2 years ago
The population of a type of local dragonfly can be found using an infinite geometric series where a1 = 65 and the common ratio i
hram777 [196]
In mathematics, number sequencing of the same pattern are called progression. There are three types of progression: arithmetic, harmonic and geometric. The pattern in arithmetic is called common difference, while the pattern in geometric is called common ratio. Harmonic progression is just the reciprocal of the arithmetic sequence.

The common ratio is denoted as r. For values of r<1, the sum of the infinite series is equal to

S∞ = A₁/(1-r), where A1 is the first term of the sequence. Substituting A₁=65 and r=1/6:

S∞ = A₁/(1-r) = 65/(1-1/6)
S∞ = 78


6 0
3 years ago
What is the remainder in the synthetic division problem below?
NNADVOKAT [17]

Answer:

YES ITS C

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Find e^cos(2+3i) as a complex number expressed in Cartesian form.
ozzi

Answer:

The complex number e^{\cos(2+31)} = \exp(\cos(2+3i)) has Cartesian form

\exp\left(\cosh 3\cos 2\right)\cos(\sinh 3\sin 2)-i\exp\left(\cosh 3\cos 2\right)\sin(\sinh 3\sin 2).

Step-by-step explanation:

First, we need to recall the definition of \cos z when z is a complex number:

\cos z = \cos(x+iy) = \frac{e^{iz}+e^{-iz}}{2}.

Then,

\cos(2+3i) = \frac{e^{i(2+31)} + e^{-i(2+31)}}{2} = \frac{e^{2i-3}+e^{-2i+3}}{2}. (I)

Now, recall the definition of the complex exponential:

e^{z}=e^{x+iy} = e^x(\cos y +i\sin y).

So,

e^{2i-3} = e^{-3}(\cos 2+i\sin 2)

e^{-2i+3} = e^{3}(\cos 2-i\sin 2) (we use that \sin(-y)=-\sin y).

Thus,

e^{2i-3}+e^{-2i+3} = e^{-3}\cos 2+ie^{-3}\sin 2 + e^{3}\cos 2-ie^{3}\sin 2)

Now we group conveniently in the above expression:

e^{2i-3}+e^{-2i+3} = (e^{-3}+e^{3})\cos 2 + i(e^{-3}-e^{3})\sin 2.

Now, substituting this equality in (I) we get

\cos(2+3i) = \frac{e^{-3}+e^{3}}{2}\cos 2 -i\frac{e^{3}-e^{-3}}{2}\sin 2 = \cosh 3\cos 2-i\sinh 3\sin 2.

Thus,

\exp\left(\cos(2+3i)\right) = \exp\left(\cosh 3\cos 2-i\sinh 3\sin 2\right)

\exp\left(\cos(2+3i)\right) = \exp\left(\cosh 3\cos 2\right)\left[ \cos(\sinh 3\sin 2)-i\sin(\sinh 3\sin 2)\right].

5 0
2 years ago
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