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2 years ago

13

3.) Given the graph of the line represented by the equation f(x) = -3x + b,

1 answer:

dsp732 years ago

4 0

Answer: (4)

Step-by-step explanation:

You might be interested in

Find 5 points that satisfy the equation y= 12+2x

Goshia [24]

We start by finding the intercept of the line: what does y equal when x=0? and what does x equal when y=0?
• intercept in x
y = 12 + 2x
0 = 12 + 2x
-12 = 2x
-6 = x
• intercept in y
y = 12 + 2x
y = 12 + 2(0)
y = 12 + 0
y = 12
Now we find three more points giving y a value and finding x
y = 12 + 2x
2 = 12 + 2x
2-12 = 2x
-10 = 2x
-5 = x

y = 12 + 2x
6 = 12 + 2x
6 - 12 = 2x
-6 = 2x
-3 = x

y = 12 + 2x
14 = 12 + 2x
14 - 12 = 2x
2 = 2x
1 = x

Notice how I gave y even numbers as values since we would have to divide with 2 at the end.

Sol. {(-6,0)(0,12)(-5,2)(-3,6)(1,14)}

8 0

3 years ago

Rewrite 7 − 8 using the additive inverse

DedPeter [7]

Answer:

-8 + 7

Step-by-step explanation:

7 - 8 = -1

-8 + 7 = -1

7 0

3 years ago

Idk how to do this 4x+y=1

ivolga24 [154]

4x+y=1
4x+y-1=1-1
4x+y-1=0

6 0

3 years ago

When you calculate the number of permutations of n distinct objects taken r at a time, what are you counting?.

expeople1 [14]

Answer:

The number of ordered arrangements of n objects taken r at a time.

Step-by-step explanation:

8 0

1 year ago

The region bounded by y=x^2+1, y=x, x=-1, x=2 with square cross sections perpendicular to the x-axis.

VLD [36.1K]

Answer:

The bounded area is 5 + 5/6 square units. (or 35/6 square units)

Step-by-step explanation:

Suppose we want to find the area bounded by two functions f(x) and g(x) in a given interval (x1, x2)

Such that f(x) > g(x) in the given interval.

This area then can be calculated as the integral between x1 and x2 for f(x) - g(x).

We want to find the area bounded by:

f(x) = y = x^2 + 1

g(x) = y = x

x = -1

x = 2

To find this area, we need to f(x) - g(x) between x = -1 and x = 2

This is:

$\int\limits^2_{-1} {(f(x) - g(x))} \, dx$

$\int\limits^2_{-1} {(x^2 + 1 - x)} \, dx$

We know that:

$\int\limits^{}_{} {x} \, dx = \frac{x^2}{2}$

$\int\limits^{}_{} {1} \, dx = x$

$\int\limits^{}_{} {x^2} \, dx = \frac{x^3}{3}$

Then our integral is:

$\int\limits^2_{-1} {(x^2 + 1 - x)} \, dx = (\frac{2^3}{2} + 2 - \frac{2^2}{2}) - (\frac{(-1)^3}{3} + (-1) - \frac{(-1)^2}{2} )$

The right side is equal to:

$(4 + 2 - 2) - ( -1/3 - 1 - 1/2) = 4 + 1/3 + 1 + 1/2 = 5 + 2/6 + 3/6 = 5 + 5/6$

The bounded area is 5 + 5/6 square units.

3 0

2 years ago