All categories

2 years ago

Y=mx+b b graph. how does changing the value of b affect the graph

Mathematics

2 answers:

Usimov [2.4K]2 years ago

6 0

Answer:

A.) It translates the graph up or down the y-axis by b units.

Step-by-step explanation:

astra-53 [7]2 years ago

3 0

Changing the value of b effects where y=mx+b is on the graph. b is the value of where the line crosses the y axis

You might be interested in

What is the radius of the base of the cone in the figure below?

liubo4ka [24]

Answer: approximately 12.17 units

==========================================================

Work Shown:

We'll use the tangent ratio to find r

tan(angle) = opposite/adjacent

tan(49) = 14/r

r*tan(49) = 14

r = 14/tan(49)

r = 12.1700143294271

r = 12.17

The radius is approximately 12.17 units long.

7 0

2 years ago

If a common logarithm is written without a base, what base should be used in evaluating it?

Mice21 [21]

10 is the default base for a logarithm
log of 50 is equal to log base 10 of 50

3 0

3 years ago

What is 54 as a percentage of 216?<br> %

OlgaM077 [116]

Answer:

54 as a percentage of 216% is 116.64.

Step-by-step explanation:

100% of 54 is 54 so 54 x 2 would be 108. We have solved 200% of it now for the 16%. The percentage of 54 of 16% would be 8.64. Now add up all we have learned.

108 + 8.64 = 116.64

6 0

1 year ago

This was the only major battle fought in New England during the War. Twelve hundred Patriots surrounded Boston, hoping to force

Mariana [72]

Step-by-step explanation:

Bunker Hill was the name of the battle!

8 0

2 years ago

Not sure what to do here can someone please help

kumpel [21]

Answer:

x = 2

Step-by-step explanation:

These equations are solved easily using a graphing calculator. The attachment shows the one solution is x=2.

<h3>Squaring</h3>

The usual way to solve these algebraically is to isolate radicals and square the equation until the radicals go away. Then solve the resulting polynomial. Here, that results in a quadratic with two solutions. One of those is extraneous, as is often the case when this solution method is used.

$\sqrt{x+2}+1=\sqrt{3x+3}\qquad\text{given}\\\\(x+2)+2\sqrt{x+2}+1=3x+3\qquad\text{square both sides}\\\\2\sqrt{x+2}=(3x+3)-(x+3)=2x\qquad\text{isolate the root term}\\\\x+2=x^2\qquad\text{divide by 2, square both sides}\\\\x^2-x-2=0\qquad\text{write in standard form}\\\\(x-2)(x+1)=0\qquad\text{factor}$

The solutions to this equation are the values of x that make the factors zero: x=2 and x=-1. When we check these in the original equation, we find that x=-1 does not work. It is an extraneous solution.

x = -1: √(-1+2) +1 = √(3(-1)+3) ⇒ 1+1 = 0 . . . . not true

x = 2: √(2+2) +1 = √(3(2) +3) ⇒ 2 +1 = 3 . . . . true . . . x = 2 is the solution

<h3>Substitution</h3>

Another way to solve this is using substitution for one of the radicals. We choose ...

$u=\sqrt{x+2}\qquad\text{requires $u\ge0$}\\\\u^2-2=x\qquad\text{solve for x}\\\\u+1=\sqrt{3(u^2-2)+3}\qquad\text{substitute for x in the original equation}\\\\(u+1)^2=3u^2-3\qquad\text{square both sides, simplify a little}\\\\2u^2-2u-4=0\qquad\text{subtract $(u+1)^2$}\\\\2(u-2)(u+1)=0\qquad\text{factor}$

Solutions to this equation are ...

u = 2, u = -1 . . . . . . the above restriction on u mean u=-1 is not a solution

The value of x is ...

x = u² -2 = 2² -2

x = 2 . . . . the solution to the equation

_____

<em>Additional comment</em>

Using substitution may be a little more work, as you have to solve for x in terms of the substituted variable. It still requires two squarings: one to find the value of x in terms of u, and another to eliminate the remaining radical. The advantage seems to be that the extraneous solution is made more obvious by the restriction on the value of u.

6 0

1 year ago