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4 years ago

How to substitution math work

1 answer:

5 0

Well lets just we have

2x-3y=-2
4x+y=24
Well first we see we have "Y'" by itself which is great that means we can use substitution. So now we just set "Y"by itself. So subtract 4x on both sides and we get

y = -4x+24 now we just plug "Y" into the first equation
2x-3(-4x+24)=-2
Distribute the 3 and simplify and we get x=5

Plug (5) back into either top or bottom equations
4(5)+y=24
20+y=24
y=4

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Find an exact value.

Westkost [7]

Answer:

$\displaystyle \cos\left(-\frac{7\,\pi}{12}\right) = \frac{\sqrt{2} - \sqrt{6}}{4}$ .

Step-by-step explanation:

Convert the angle $\displaystyle \left(-\frac{7\, \pi}{12}\right)$ to degrees:

$\displaystyle \left(-\frac{7\, \pi}{12}\right) = \left(-\frac{7\, \pi}{12}\right) \times \frac{180^\circ}{\pi} = -105^\circ$ .

Note, that $\left(-105^\circ\right)$ is the sum of two common angles: $\left(-45^\circ\right)$ and $\left(-60^\circ\right)$ .

$\displaystyle \cos\left(-45^\circ\right) = \cos\left(45^\circ\right) = \frac{\sqrt{2}}{2}$ .
$\displaystyle \cos\left(-60^\circ\right) = \cos\left(60^\circ\right) = \frac{1}{2}$ .

$\displaystyle \sin\left(-45^\circ\right) = -\sin\left(45^\circ\right) = -\frac{\sqrt{2}}{2}$ .
$\displaystyle \sin\left(-60^\circ\right) = -\sin\left(60^\circ\right) = -\frac{\sqrt{3}}{2}$ .

By the sum-angle identity of cosine:

$\cos(A + B) = \cos(A)\cdot \cos(B) - \sin(A) \cdot \sin(B)$ .

Apply the sum formula for cosine to find the exact value of $\cos\left(-105^\circ \right)$ .

$\begin{aligned}\cos\left(-105^\circ \right) &= \cos\left(\left(-45^\circ\right) + \left(-60^\circ\right)\right) \\ &= \cos\left(-45^\circ\right) \cdot \cos\left(-60^\circ\right)\right) - \sin\left(-45^\circ\right) \cdot \sin\left(-60^\circ\right)\right) \\ &= \frac{\sqrt{2}}{2} \times \frac{1}{2} - \left(-\frac{\sqrt{2}}{2}\right)\times \left(-\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{2} - \sqrt{6}}{4}\end{aligned}$ .

$\displaystyle \left(-\frac{7\, \pi}{12}\right) = \left(-\frac{7\, \pi}{12}\right) \times \frac{180^\circ}{\pi} = -105^\circ$ . In other words, $\displaystyle \left(-\frac{7\, \pi}{12}\right)$ and $\left(-105^\circ\right)$ correspond to the same angle. Therefore, the cosine of $\displaystyle \left(-\frac{7\, \pi}{12}\right)\!$ would be equal to the cosine of $\left(-105^\circ\right)\!$ .

$\displaystyle \cos\left(-\frac{7\,\pi}{12}\right) = \cos\left(-105^\circ\right) = \frac{\sqrt{2} - \sqrt{6}}{4}$ .

3 0

3 years ago

What’s the median number

marta [7]

Vas happenin!!

The median is 94.7

Hope this helps

-Zayn Malik

5 0

3 years ago

Please help Need help

Anettt [7]

This problem can be readily solved if we are familiar with the point-slope form of straight lines:
y-y0=m(x-x0) ...................................(1)
where
m=slope of line
(x0,y0) is a point through which the line passes.

We know that the line passes through A(3,-6), B(1,2)

All options have a slope of -4, so that should not be a problem. In fact, if we check the slope=(yb-ya)/(xb-xa), we do find that the slope m=-4.

So we can check which line passes through which point:

a. y+6=-4(x-3)
Rearrange to the form of equation (1) above,
y-(-6)=-4(x-3) means that line passes through A(3,-6) => ok

b. y-1=-4(x-2) means line passes through (2,1), which is neither A nor B
****** this equation is not the line passing through A & B *****

c. y=-4x+6 subtract 2 from both sides (to make the y-coordinate 2)
y-2 = -4x+4, rearrange
y-2 = -4(x-1)
which means that it passes through B(1,2), so ok

d. y-2=-4(x-1)
this is the same as the previous equation, so it passes through B(1,2),
this equation is ok.

Answer: the equation y-1=-4(x-2) does NOT pass through both A and B.

8 0

3 years ago

A third-degree polynomial function f has real zeros -2, ½, and 3, and its leading coefficient negative. Write an equation for f.

Rom4ik [11]

F(x) = k(x+2)(2x-1)(x-3), where k is some constant
= k(2x^3-3x^2-11x+6)
= k(-2x^3+3x^2+11x-6)

k defines some vertical stretch, so there are an infinitely many solutions for f(x).

8 0

4 years ago

OX and OY are two straight lines which intersect at an acute angle of 60°. OX = 4.5cm and OY = 5cm . The point M is on OX such t

Neko [114]

Therefore the point P is at 3.46 cm from O and it lies on the angle bisector of ∠XOY

<h3>What is an Angle Bisector ?</h3>

The ray that bisects the angle into half is called Angle Bisector.

It is given that ∠XOY = 60 degree

the length of OX = 4.5 cm

OY =5 cm

The point M is on OX such that

OM = 2 MX

so The M is at 3 cm from O

The point P lies in the acute angle such that the distance between point P and OX and OY is always same and at 3 cm from M

According to the angle bisector theorem converse states that if a point is in the interior of an angle and is at equal distance from the sides then it lies on the bisector of that angle.

As it can be seen from the image that a point equidistant from the rays , at 3 cm from M will be at

By Pythagoras Theorem

3² +3² = OP²

OP = 2 $\sqrt{3}$ = 3.46 cm from O

To know more about Angle Bisector

brainly.com/question/12896755

#SPJ1

4 0

2 years ago