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3 years ago

find the constant m and b in the linear function f(x)=mx+b so that f(6)=9 and the straight line represents by f has slope -4

Mathematics

1 answer:

Charra [1.4K]3 years ago

4 0

Answer:

The constant m=-4, and b=33

Step-by-step explanation:

F(x)= MX+b

If f(6)=9 then x=6, m= -4

9= -4(6) + b

9= -24 + b

b= 9 + 24

b= 33

Therefore the constant m, which is the slope is -4 and b which is the intercept is 33

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Graph the points (-3,4) and (1,1). If you drew a line through the points, name three other points that would be on the line. How

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The easiest way to solve this problem is to find the equation of the line joining these two points, then get the values of the points on this line.

We have first point (x1,y1) = (-3,4) and second point (x2,y2) = (1,1).
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The slope (m) = (y2-y1) / (x2-x1) = (1-4) / (1--3) = -0.75

Then we will use one of these points to get the value of c as follows:
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The equation of this straight line is:
y = -0.75 x + 1.75

Now to get points on this line, we will assume values for either x or y and calculate the other as follows:
1- For x = 0:
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2- For y = 0:
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4 years ago

Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used.

Oduvanchick [21]

Answer:

see attached

Step-by-step explanation:

The Pythagorean theorem can be used to find the hypotenuse associated with each pair of legs. That tells you ...

c² = a² +b² . . . . . legs a, b; hypotenuse c

<h3>alternate form of Pythagorean theorem</h3>

For the purpose of this problem, it is convenient to consider a slightly different form of the equation.

For legs √a and √b, the hypotenuse √c is given by ...

(√c)² = (√a)² +(√b)²

c = a +b

That is ...

legs √a, √b ⇒ hypotenuse √(a+b)

<h3>application to this problem</h3>

Since the legs are (mostly) given in terms of square roots, the value under the radical for the hypotenuse is simply the sum of those:

legs: √1, √2 ⇒ hypotenuse √(1+2) = √3

legs: √2, √3 ⇒ hypotenuse √(2+3) = √5

legs: √5, √3 ⇒ hypotenuse √(5+3) = √8

legs: √5, √1 ⇒ hypotenuse √(5+1) = √6

_____

<em>Additional comment</em>

You may not see the leg lengths given as square roots very often. This is a rather unusual set of problems for hypotenuse length.

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