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3 years ago

F(x) = 3x2 – 2x + 4 and g(x) = 5x2 + 6x – 8, find (f – g)(x)

1 answer:

7 0

(f - g)(x) = (3x² - 2x + 4) - (5x² + 6x - 8<span>)
</span>(f - g)(x) = 3x² - 2x + 4 - 5x² - 6x + 8
(f - g)(x) = -2x² - 8x + 12

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If someone could help me answer this and possibly give me a step-by-step for future reference that'd be great thanks.

andreyandreev [35.5K]

Answer:

Let's look at the Venn diagram and the data we have:

There are 12 pupils

9 have a brother

7 have a sister

2 have neither.

Now let's look at the diagram.

We know that the circle B represents the pupils who have a brother

Circle S represents the students that have a sister.

Then the intersection of the circles, represents the number of students that have both.

And the space outside the circles represents the pupils that do not have a brother nor a sister.

Then the first thing we can complete is a 2 in the bottom left corner, because we already know that there are 2 pupils that do not have brothers nor sisters.

Now let's find the number of students that have both, brothers and sisters:

There are 12 pupils.

9 have a brother

7 have a sister

2 have neither

if we add that, we get:

9 + 7 + 2 = 18

This is larger than 12, this means that we are counting some of the students more than once.

If X is the number of students that we are counting twice, we should have:

18 - X = 12

18 -12 = X = 6

There are 6 students who we are counting twice, and this happens because these 6 students have a brother and a sister.

Then in the intersection of both circles we should put a 6.

At the left of that (in the part that we have only circle B) we need to write the number of students that only have a brother, this is the number of students that have a brother minus the number of pupils that have a brother and a sister, this is:

9 - 6 = 3

We need to write a 3 in that square.

And in the last square (the one that is only one circle S) we need to write the number of pupils that only have a sister, this is calculated in the same way than before, this is:

7 - 6 = 1

We need to write a 1 in the rightmost square.

6 0

2 years ago

3 over 7 - 5 over 6. Simplify and turned into a mixed fraction simplified

valina [46]

3/7-5/6==>
get a common denominator...18/42-35/42==>
subtract nominators...-17/42
answer=-17 over 42 or -17/24

6 0

3 years ago

2x + 9

MAVERICK [17]

Answer:

I think is 12x2 +48x-27 but I'm confused

5 0

3 years ago

Read 2 more answers

Help pls help pls help

fiasKO [112]

Answer:

1/4 of a mile in 1/10 of an hour

you can set up a proportion

(1/4)/(1/10)=2/H

you can now simplify or cross multiply

simplifying the left side

(1/4) divided by (1/10) equals (1/4) times (10/1)

(1/4)*(10/1)=10/4=5/2

5/2=2/H

cross multiply

5H=2*2

5H=4

H=4/5 of an hour or (4/5)*60=48 minutes

another solution...

2/(1/4)=8 and 8*(1/10)=8/10=4/5 of an hour or 48 minutes again

4 0

1 year ago

Read 2 more answers

Pls show work! questions on pic

Lena [83]

Answer:

a.(10,23)

b.(7,30)

e.(20,20)

Step-by-step explanation:

We are given that

x=Width of garden

y=Length of garden

According to question

$x\geq 5$

$y\geq 20$

Perimeter of rectangle= $2(x+y)$

Where x=Width of rectangle

y=Length of rectangle

Fencing used = $2(x+y)$

$2(x+y)\leq 80$

First we change inequality into equality

$x=5$ ...(1)

$y=20$ ...(2)

$2(x+y)=80$

$x+y=40$ ...(3)

Substitute x=0 in equation (3)

$y=40$

Substitute y=0 in equation (3)

$x=40$

Substitute x=0 and y=0 in equation (3)

$2(x+y)\leq 80$

$0\leq 80$

It is true. Therefore, the shaded region below the line.

$0\geq 5$

$0\geq 20$

These are false equation .Therefore, the shaded region above the line.

Substitute x=5 in equation (3)

$5+y=40$

$y=40-5=35$

Substitute y=20 in equation (3)

$x+20=40$

$x=40-20=20$

Intersection point of equation (1) and (3) is (5,35) and intersect point of equation (2) and (3) is (20,20).

Now,

a.(10,23)

Substitute in

$2(x+y)\leq 80$

$2(10+23)\leq 80$

$66\geq 80$

10>5 and 23>20

it satisfied all inequality.

Hence, it is a solution.

b.(7,30)

7>5

30>20

$2(7+30)=74$

it satisfied all inequality.

Hence, it is a solution.

c.(18,25)

18>5

25>20

$2(18+25)=86>80$

It does not satisfied third inequality

Hence, it is not a solution.

d.(8,35)

8>5

35>20

$2(8+35)=86>80$

It does not satisfied third inequality

Hence, it is not a solution.

e.(20,20)

$2(20+20)=80$

20>5

20=20

it satisfied all inequality.

Hence, it is a solution.

4 0

3 years ago