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3 years ago

I really need help!!

Mathematics

2 answers:

jonny [76]3 years ago

5 0

Answer:

#8. 2

#9. 7

#10. 2

#11. 2

#12. 0

#13. -4

#14. 17

#15. 1

alina1380 [7]3 years ago

5 0

Answer:#8=2

#9=-11

#10=8

#11=2

#12=0

#14=17

#15=1

Step-by-step explanation: #8-13 (Subtract the numbers using a number line, than add the numbers using a integer chips, after that calculate the sum of the differences) #14-15( normal solving stepping for integers(recommend using a T1-84 Calculator)

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In this composite function, (f◦g)(x) where f(x)=(X-3) and g(x)=X^2 what is (f◦g)(x) when x=1?

Alekssandra [29.7K]

Answer:

When x = 1, (f◦g)(x) is -2.

Step-by-step explanation:

Composite function:

The composite function of f and g is given by:

$(f \circ g)(x) = f(g(x))$

In this question:

$f(x) = x - 3, g(x) = x^2$

Composite function:

The composite function is:

$f(g(x)) = f(x^2) = x^2 - 3$

At x = 1

$(f \circ g)(1) = 1^2 - 3 = 1 - 3 = -2$

When x = 1, (f◦g)(x) is -2.

3 0

3 years ago

∆ABC has vertices A(–2, 0), B(0, 8), and C(4, 2)

Natali [406]

Answer:

Part 1) The equation of the perpendicular bisector side AB is $y=-\frac{1}{4}x+\frac{15}{4}$

Part 2) The equation of the perpendicular bisector side BC is $y=\frac{2}{3}x+\frac{11}{3}$

Part 3) The equation of the perpendicular bisector side AC is $y=-3x+4$

Part 4) The coordinates of the point P(0.091,3.727)

Step-by-step explanation:

Part 1) Find the equation of the perpendicular bisector side AB

we have

A(–2, 0), B(0, 8)

step 1

Find the slope AB

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

substitute the values

$m=\frac{8-0}{0+2}$

$m=4$

step 2

Find the slope of the perpendicular line to side AB

Remember that

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of their slopes is equal to -1)

therefore

The slope is equal to

$m=-\frac{1}{4}$

step 3

Find the midpoint AB

The formula to calculate the midpoint between two points is equal to

$M(\frac{x1+x2}{2},\frac{y1+y2}{2})$

substitute the values

$M(\frac{-2+0}{2},\frac{0+8}{2})$

$M(-1,4)$

step 4

Find the equation of the perpendicular bisectors of AB

the slope is $m=-\frac{1}{4}$

passes through the point $(-1,4)$

The equation in slope intercept form is equal to

$y=mx+b$

substitute

$4=(-\frac{1}{4})(-1)+b$

solve for b

$b=4-\frac{1}{4}$

$b=\frac{15}{4}$

$y=-\frac{1}{4}x+\frac{15}{4}$

Part 2) Find the equation of the perpendicular bisector side BC

we have

B(0, 8) and C(4, 2)

step 1

Find the slope BC

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

substitute the values

$m=\frac{2-8}{4-0}$

$m=-\frac{3}{2}$

step 2

Find the slope of the perpendicular line to side BC

Remember that

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of their slopes is equal to -1)

therefore

The slope is equal to

$m=\frac{2}{3}$

step 3

Find the midpoint BC

The formula to calculate the midpoint between two points is equal to

$M(\frac{x1+x2}{2},\frac{y1+y2}{2})$

substitute the values

$M(\frac{0+4}{2},\frac{8+2}{2})$

$M(2,5)$

step 4

Find the equation of the perpendicular bisectors of BC

the slope is $m=\frac{2}{3}$

passes through the point $(2,5)$

The equation in slope intercept form is equal to

$y=mx+b$

substitute

$5=(\frac{2}{3})(2)+b$

solve for b

$b=5-\frac{4}{3}$

$b=\frac{11}{3}$

$y=\frac{2}{3}x+\frac{11}{3}$

Part 3) Find the equation of the perpendicular bisector side AC

we have

A(–2, 0) and C(4, 2)

step 1

Find the slope AC

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

substitute the values

$m=\frac{2-0}{4+2}$

$m=\frac{1}{3}$

step 2

Find the slope of the perpendicular line to side AC

Remember that

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of their slopes is equal to -1)

therefore

The slope is equal to

$m=-3$

step 3

Find the midpoint AC

The formula to calculate the midpoint between two points is equal to

$M(\frac{x1+x2}{2},\frac{y1+y2}{2})$

substitute the values

$M(\frac{-2+4}{2},\frac{0+2}{2})$

$M(1,1)$

step 4

Find the equation of the perpendicular bisectors of AC

the slope is $m=-3$

passes through the point $(1,1)$

The equation in slope intercept form is equal to

$y=mx+b$

substitute

$1=(-3)(1)+b$

solve for b

$b=1+3$

$b=4$

$y=-3x+4$

Part 4) Find the coordinates of the point of concurrency of the perpendicular bisectors (P)

we know that

The point of concurrency of the perpendicular bisectors is called the circumcenter.

Solve by graphing

using a graphing tool

the point of concurrency of the perpendicular bisectors is P(0.091,3.727)

see the attached figure

5 0

3 years ago

Which is longer 3200 hours or 19 weeks

Katarina [22]

The 1 weeks is longer because the hours is like 53 days and the weeks are like 133 days

3 0

3 years ago

How is this number read out loud 0.034

Svetllana [295]

Answer:

0.034

Step-by-step explanation:

thirty-four thousandths

zero point zero three four

5 0

3 years ago

Read 2 more answers

To solve a system of inequalities so you can graph it how do you change these two equations into something like the two that are

Zigmanuir [339]

Explanation

Problem #2

We must find the solution to the following system of inequalities:

$\begin{gathered} 3x-2y\leq4, \\ x+3y\leq6. \end{gathered}$

(1) We solve for y the first inequality:

$-2y\leq4-3x.$

Now, we multiply both sides of the inequality by (-1), this changes the signs on both sides and inverts the inequality symbol:

$\begin{gathered} 2y\ge-4+3x, \\ y\ge\frac{3}{2}x-2. \end{gathered}$

The solution to this inequality is the set of all the points (x, y) over the line:

$y=\frac{3}{2}x-2.$

This line has:

• slope m = 3/2,

• y-intercept b = -2.

(2) We solve for y the second inequality:

$\begin{gathered} x+3y\leq6, \\ 3y\leq6-x, \\ y\leq-\frac{1}{3}x+2. \end{gathered}$

The solution to this inequality is the set of all the points (x, y) below the line:

$y=-\frac{1}{3}x+2.$

This line has:

• slope m = -1/3,

• y-intercept b = 2.

(3) Plotting the lines of points (1) and (2), and painting the region:

• over the line from point (1),

• and below the line from point (2),

we get the following graph:

Answer

The points that satisfy both inequalities are given by the intersection of the blue and red regions:

8 0

1 year ago

I really need help!!​

I really need help!!