Answers
y= -2/3x + 1
1= c
Step 1: Write slope-intercept form equation
To start answering this question, it helps to write out the example equation for slope-intercept form. Let’s do this now.
y=mx+b
Keep in mine that the y represents the answer, the m represents the slope, the x represents the variable, and the b represents the y-intercept.
Step 2: Determine the y-intercept
I find that it is easiest to determine the y-intercept first. This is the b in the equation. The y-intercept can be find by looking at the y-axis, and seeing where the line intersects with the axis. In this case, it seems to be at 1. Let’s substitute this into our equation now.
y=mx+b
y=mx+1
Step 3: Determine slope
For the final thing that we need to fill in, we need to find the slope. The slope is the m in the equation, and can be found by the formula of rise/run. Choose any two points, and see how far they are from each other up and down (rise), as well as left and right (run).
For the sake of this problem, we will work with (0,1) and (-3,3). There is a difference in the rise of 2 and in the run of 3. This makes the rise/ run be 2/3. Let’s substitute this now.
y=mx+1
y= 2/3x+1
The slope will also need to be negative since the line on the graph descends. Let’s change this to a negative now.
y= -2/3x+1
This is your answer. Hope this helps! Comment below for more questions.
Related Questions
Find the direction cosines and direction angles of the vector. (Give the direction angles correct to the nearest degree.) c, c, c , where c > 0
Answers
Answer:
cos(∝) = 1/√3
cos(β) = 1/√3
cos(γ) = 1/√3
∝ = 55°
β = 55°
γ = 55°
Step-by-step explanation:
Given the data in the question;
vector is z = < c,c,c >
the direction cosines and direction angles of the vector = ?
Cosines are the angle made with the respect to the axes.
cos(∝) = z < 1,0,0 > / |z|
so
cos(∝) = < c,c,c > < 1,0,0 > / √[c² + c² + c²] = ( c + 0 + 0 ) / √[ 3c² ]
cos(∝) = c / √[ 3c² ] = c / c√3 = 1/√3
∝ = cos⁻¹( 1/√3 ) = 54.7356° ≈ 55°
cos(β) = < c,c,c > < 0,1,0 > / √[c² + c² + c²] = ( 0 + c + 0 ) / √[ 3c² ]
cos(β) = c / √[ 3c² ] = c / c√3 = 1/√3
β = cos⁻¹( 1/√3 ) = 54.7356° ≈ 55°
cos(γ) = < c,c,c > < 0,0,1 > / √[c² + c² + c²] = ( 0 + 0 + c ) / √[ 3c² ]
cos(γ) = c / √[ 3c² ] = c / c√3 = 1/√3
γ = cos⁻¹( 1/√3 ) = 54.7356° ≈ 55°
Therefore;
cos(∝) = 1/√3
cos(β) = 1/√3
cos(γ) = 1/√3
∝ = 55°
β = 55°
γ = 55°
F(x) =-2x-4 find x if f(x)=14
Answers
Answer:
14=-2x-4
18=-2x
x=-9
Hope This Helps!!!
i need help solving this .
Answers
Answer:b
Step-by-step explanation:
Answer:
just be smart trust me u dont need us to give u the answer ur super smart
Step-by-step explanation:
make me brainliest to help people be encourage
find the exact value cos5pi/6
Answers
Answer:
[tex] - \frac{ \sqrt{3} }{2} [/tex]
Step-by-step explanation:
Unit circle
9. Mariah has 28 centimeters of reed
and 10 meters of reed for weaving
baskets. How many meters of reed
does she have? Write your answer as a
decimal and explain your answer.
Answers
PLSHELPASAPDFFFFFFFFFFFFFFFFFFFFFFFFFF
Answers
im struggling with the same one
if cosA=3√2/5,then show that cos2A=11/25
Answers
Answer:
Step-by-step explanation:
Cos 2A = 2Cos² A - 1
[tex]= 2*(\frac{3\sqrt{2}}{5})^{2}-1\\\\=2*(\frac{3^{2}*(\sqrt{2})^{2}}{5^{2}})-1\\\\=2*\frac{9*2}{25} - 1\\\\=\frac{36}{25}-1\\\\=\frac{36}{25}-\frac{25}{25}\\\\=\frac{11}{25}[/tex]
when price of indomie noodles was lowered from #50 to #40 per unit, quantity demanded increases from 400 to 600 units per week. calculate the coefficient of price elasticity of demand and determine whether by lowering price this firm has made a wise decision
Answers
Answer:
The price elasticity of demand is -10
Step-by-step explanation:
Given
[tex]p_1,p_2 = 50,40[/tex]
[tex]q_1,q_2 = 400,500[/tex]
Solving (a): The coefficient of price elasticity of demand (k)
This is calculated as:
[tex]k = \frac{\triangle q}{\triangle p}[/tex]
So, we have:
[tex]k = \frac{500 - 400}{40 - 50}[/tex]
[tex]k = \frac{100}{-10}[/tex]
[tex]k = -10[/tex]
Because |k| > 0, then we can conclude that the company made a wise decision.
A large container holds 4 gallons of chocolate milk that has to be poured into bottles. Each bottle holds 2 pints.
If the ratio of gallons to pints is 1: 8,
bottles are required to hold the 4 gallons of milk.
Answers
Answer:
64 Bottles
Step-by-step explanation:
that is the procedure above
if side of square is 4.05 find its area
Answers
Answer:
A
≈
16.4
please give brain listFind the solution of the differential equation that satisfies the given initial condition. (dP)/(dt)
Answers
Answer:
[tex]P = (\frac{1}{3}t^\frac{3}{2} + \sqrt 2 - \frac{1}{3})^2[/tex]
Step-by-step explanation:
Given
[tex]\frac{dP}{dt} = \sqrt{Pt[/tex]
[tex]P(1) = 2[/tex]
Required
The solution
We have:
[tex]\frac{dP}{dt} = \sqrt{Pt[/tex]
[tex]\frac{dP}{dt} = (Pt)^\frac{1}{2}[/tex]
Split
[tex]\frac{dP}{dt} = P^\frac{1}{2} * t^\frac{1}{2}[/tex]
Divide both sides by [tex]P^\frac{1}{2}[/tex]
[tex]\frac{dP}{ P^\frac{1}{2}*dt} = t^\frac{1}{2}[/tex]
Multiply both sides by dt
[tex]\frac{dP}{ P^\frac{1}{2}} = t^\frac{1}{2} \cdot dt[/tex]
Integrate
[tex]\int \frac{dP}{ P^\frac{1}{2}} = \int t^\frac{1}{2} \cdot dt[/tex]
Rewrite as:
[tex]\int dP \cdot P^\frac{-1}{2} = \int t^\frac{1}{2} \cdot dt[/tex]
Integrate the left hand side
[tex]\frac{P^{\frac{-1}{2}+1}}{\frac{-1}{2}+1} = \int t^\frac{1}{2} \cdot dt[/tex]
[tex]\frac{P^{\frac{-1}{2}+1}}{\frac{1}{2}} = \int t^\frac{1}{2} \cdot dt[/tex]
[tex]2P^{\frac{1}{2}} = \int t^\frac{1}{2} \cdot dt[/tex]
Integrate the right hand side
[tex]2P^{\frac{1}{2}} = \frac{t^{\frac{1}{2} +1 }}{\frac{1}{2} +1 } + c[/tex]
[tex]2P^{\frac{1}{2}} = \frac{t^{\frac{3}{2}}}{\frac{3}{2} } + c[/tex]
[tex]2P^{\frac{1}{2}} = \frac{2}{3}t^\frac{3}{2} + c[/tex] ---- (1)
To solve for c, we first make c the subject
[tex]c = 2P^{\frac{1}{2}} - \frac{2}{3}t^\frac{3}{2}[/tex]
[tex]P(1) = 2[/tex] means
[tex]t = 1; P =2[/tex]
So:
[tex]c = 2*2^{\frac{1}{2}} - \frac{2}{3}*1^\frac{3}{2}[/tex]
[tex]c = 2*2^{\frac{1}{2}} - \frac{2}{3}*1[/tex]
[tex]c = 2\sqrt 2 - \frac{2}{3}[/tex]
So, we have:
[tex]2P^{\frac{1}{2}} = \frac{2}{3}t^\frac{3}{2} + c[/tex]
[tex]2P^{\frac{1}{2}} = \frac{2}{3}t^\frac{3}{2} + 2\sqrt 2 - \frac{2}{3}[/tex]
Divide through by 2
[tex]P^{\frac{1}{2}} = \frac{1}{3}t^\frac{3}{2} + \sqrt 2 - \frac{1}{3}[/tex]
Square both sides
[tex]P = (\frac{1}{3}t^\frac{3}{2} + \sqrt 2 - \frac{1}{3})^2[/tex]
