I'm not sure if there are options to your question, but this equation is complicated. It can only be graphed, not simplified, you can't find x or y, and cannot be written in slope-intercept form. Remember when I said it could only be graphed, well heres the info.
The slope is -2/3
The y-intercept is 0
and heres a little chart
X Y
0 0
1 -2/3
The points on the graph are
(0,0) (1, -2/3) (2, -4/3)
Answer:
You would put an open dot on both ends from -8 to 4 or just a distance of 12
Step-by-step explanation:
Given:
Right triangles:
To find:
The value of c and value of d
Solution:
<u>In the first right triangle:</u>
θ = 45°
Opposite side to θ = 5
Hypotenuse = c
![$\sin \theta=\frac{\text { Opposite side of } \theta}{\text { Hypotenuse }}](https://tex.z-dn.net/?f=%24%5Csin%20%5Ctheta%3D%5Cfrac%7B%5Ctext%20%7B%20Opposite%20side%20of%20%7D%20%5Ctheta%7D%7B%5Ctext%20%7B%20Hypotenuse%20%7D%7D)
![$\sin 45^\circ=\frac{5}{c}](https://tex.z-dn.net/?f=%24%5Csin%2045%5E%5Ccirc%3D%5Cfrac%7B5%7D%7Bc%7D)
The value of sin 45° = ![\frac{1}{\sqrt 2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5Csqrt%202%7D)
![$\frac{1}{\sqrt 2}=\frac{5}{c}](https://tex.z-dn.net/?f=%24%5Cfrac%7B1%7D%7B%5Csqrt%202%7D%3D%5Cfrac%7B5%7D%7Bc%7D)
Do cross multiplication, we get
![c=5\sqrt{2}](https://tex.z-dn.net/?f=c%3D5%5Csqrt%7B2%7D)
<u>In the second right triangle:</u>
θ = 60°
Opposite side to θ = 4
Adjacent side to θ = d
![$\tan \theta=\frac{\text { Opposite side of } \theta}{\text { Adjacent side of } \theta}](https://tex.z-dn.net/?f=%24%5Ctan%20%5Ctheta%3D%5Cfrac%7B%5Ctext%20%7B%20Opposite%20side%20of%20%7D%20%5Ctheta%7D%7B%5Ctext%20%20%7B%20Adjacent%20side%20of%20%7D%20%5Ctheta%7D)
![$\tan 60^\circ=\frac{4}{d}](https://tex.z-dn.net/?f=%24%5Ctan%2060%5E%5Ccirc%3D%5Cfrac%7B4%7D%7Bd%7D)
![$\sqrt{3} =\frac{4}{d}](https://tex.z-dn.net/?f=%24%5Csqrt%7B3%7D%20%3D%5Cfrac%7B4%7D%7Bd%7D)
Do cross multiplication, we get
![$d=\frac{4}{\sqrt{3} }](https://tex.z-dn.net/?f=%24d%3D%5Cfrac%7B4%7D%7B%5Csqrt%7B3%7D%20%7D)
The value of
and
.
<em><u>Question:</u></em>
A jeweler is dividing 3/8 of a pound of rubies among 4 lots. What part of a pound will each lot weigh? A.1/4 B.1/8 C.1/16 D.3/32
<em><u>Answer:</u></em>
Option D
Each lot weight
part of pound
<em><u>Solution:</u></em>
Given that,
A jeweler is dividing 3/8 of a pound of rubies among 4 lots
To find: part of a pound each lot weigh
From given,
![\frac{3}{8} \text{ of a pound } = 4 lots](https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B8%7D%20%5Ctext%7B%20of%20a%20pound%20%7D%20%3D%204%20lots)
Let "x" be the part of a pound each lot weigh
Therefore,
![\frac{3}{8} \text{ of a pound } = 4 lots\\\\"x" part = 1\ lots](https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B8%7D%20%5Ctext%7B%20of%20a%20pound%20%7D%20%3D%204%20lots%5C%5C%5C%5C%22x%22%20part%20%3D%201%5C%20lots)
This forms a proportion and solve by cross multiplying
![\frac{3}{8} \times 1 = 4 \times x\\\\x = \frac{3}{32}](https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B8%7D%20%5Ctimes%201%20%3D%204%20%5Ctimes%20x%5C%5C%5C%5Cx%20%3D%20%5Cfrac%7B3%7D%7B32%7D)
Thus, each lot weight
part of pound
Hey there! :)
Answer:
Third option. x = 2, and x = 4.
Step-by-step explanation:
Find the zeros of this quadratic equation by factoring:
f(x) = x² - 6x + 8
Becomes:
f(x) = (x - 4)(x - 2)
Set each factor equal to 0 to solve for the roots;
x - 4 = 0
x = 4
x - 2 = 0
x = 2
Therefore, the zeros of this equation are at x = 2, and x = 4.