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Greeley [361]
2 years ago
13

Write the equation of a line with a slope of 3 and a y-intercept of 1​

Mathematics
1 answer:
slega [8]2 years ago
8 0

Answer:

y=3x+1

Step-by-step explanation:

Your equation should start off as y=mx+b. You must solve for m and b, although they already give you m and b. m is the slope, which here is 3 and your y is 1. Making your equation, y=3x+1.

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HURRYYYYY<br> What is the value of the expression 4’4?
WINSTONCH [101]

Answer:

256

Step-by-step explanation:

Raise 4 to the power of 4

Basically, multiply 4 by itself four times. :)

4 0
3 years ago
Rp+s=t in terms of r, s,and t
expeople1 [14]

Answer:

p=(t-s)/r

Step-by-step explanation:

This is an equation so all we have to do is solve backwards step by step:

rp+s=t

rp+s-s=t-s

rp/r=(t-s)/r

p=(t-s)/r

8 0
3 years ago
Find the value of x if O is between D &amp; B and DO = 15, OB = 3x + 7, and DB = 9x - 2 x= ____
pishuonlain [190]

Answer:

x = 4

Step-by-step explanation:

DO + OB = DB

15 + 3x + 7 = 9x - 2

3x + 22 = 9x - 2

3x + 24 = 9x

24 = 6x

4 = x

7 0
3 years ago
This extreme value problem has a solution with both a maximum value and a minimum value. use lagrange multipliers to find the ex
noname [10]

L(x,y,z,\lambda)=10x+10y+2z+\lambda(5x^2+5y^2+2z^2-42)

L_x=10+10\lambda x=0\implies1+\lambda x=0

L_y=10+10\lambda y=0\implies1+\lambda y=0

L_z=2+4\lambda z=0\implies1+2\lambda z=0

L_\lambda=5x^2+5y^2+2z^2-42=0

\begin{cases}L_x=0\\L_y=0\\L_z=0\end{cases}\implies1+\lambda x=1+\lambda y=1+2\lambda z=0\implies x=y=2z

5x^2+5y^2+2z^2=5(2z)^2+5(2z)^2+2z^2=42z^2=42\implies z^2=1

z^2=1\implies z=\pm1\implies x=y=\pm2

There are two critical points, at which we have

f\left(2,2,1\right)=42\text{ (a maximum value)}

f\left(-2,-2,-1\right)=-42\text{ (a minimum value)}

3 0
3 years ago
Choose the correct equation for the line shown on the graph
Viefleur [7K]
It would be C y=-1/4x-1
5 0
2 years ago
Read 2 more answers
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