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3 years ago

PLEASE HELP ASAP Due in 1 day!!!

Mathematics

1 answer:

Helen [10]3 years ago

7 0

Step-by-step explanation:

Q1 . (f+g)(x) = f(x) + g(x)

=4x-4 +2x^2 -3x

= 2x^2 + x -4

Q2. (f-g)(x) = f(x) - g(x)

= 2x^2−2 - (4x+1)

= 2x^2 -2 -4x -1

= 2x^2 - 4x -3

Q3. h(x)=3x−3 and g(x)=x^2+3

(h.g)(x) = h(x) × g(x)

= (3x-3) × (x^2 + 3)

=3x^3 -3x^2 + 9x -9

Q4.f(x)=x+4 and g(x)=x+6

(f/g)(x) = f(x) ÷ g(x)

= x+4 / x+6

the domain restriction is x>-6

x<-6

x doesn't equal (-6)

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Brads class collected 95 used books for the book drive. They packed the books in boxes of 23 books each. How many boxes did they

VashaNatasha [74]

Answer:

Step-by-step explanation:

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3 years ago

Hallie bikes 1 5/8 km before lunch 6 1/4 km How much farther does hallies bike after lunch than before lunch? Express your answe

kondaur [170]

Distance biked before lunch = 1 $\frac{5}{8}$ km (the number is written in mixed fraction)

⇒ Distance biked before lunch = $\frac{13}{8}$

Distance biked after lunch = 6 $\frac{1}{4}$ km (the number is written in mixed fraction)

⇒ Distance biked after lunch = $\frac{25}{4}$

We need to determine how much farther does Hallie bike after lunch than before lunch

Extra distance biked after lunch vs. before lunch = $\frac{25}{4}$ - $\frac{13}{8}$

⇒ Extra distance biked after lunch vs. before lunch = $\frac{50-13}{8}$

⇒ Extra distance biked after lunch vs. before lunch = $\frac{37}{8}$

⇒ Extra distance biked after lunch vs. before lunch = 4 $\frac{5}{8}$ (in mixed fraction)

Hence, Hallie biked 4 $\frac{5}{8}$ km extra after lunch vs. before lunch.

3 0

3 years ago

Jim is enclosing a rectangular garden with 170 feet of fencing. The length of the garden is 10 feet more than twice it’s width,

Darina [25.2K]

Answer:

$1,500\ ft^2$

Step-by-step explanation:

Let

L ----> the length of the rectangular garden in feet

w ---> the width of the rectangular garden in feet

step 1

Find the width

we know that

The perimeter of the rectangular garden is

$P=2(L+W)$

$P=170\ ft$

$170=2(L+W)$

Simplify

$85=(L+W)$ ----> equation A

$L=2W+10$ ----> equation B

substitute equation B in equation A and solve for W

$85=(2W+10+W)$

$85-10=3W$

$3W=75$

$W=25\ ft$

Find the value of L

$L=2(25)+10=60\ ft$

step 2

Find the area

we know that

The area of the rectangular garden is

$A=(LW)$

substitute the values

$A=(60)(25)=1,500\ ft^2$

6 0

3 years ago

Can you help me plz?

Ad libitum [116K]

With what because I cannot see the math or any subject of the answer you need

5 0

3 years ago

Find the distance between the points (–3, 2) and (0, 3).

Oksi-84 [34.3K]

Answer: $\sqrt{10}\ units$

Step-by-step explanation:

You need to use the following formula:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Given the points (-3, 2) and (0, 3), you can identify that:

$x_2=-3\\x_1=0\\\\y_2=2\\y_1=3$

Then, the final step is to substitutes these coordinates into the formula.

So, the distance between the given points is:

$d=\sqrt{(-3-0)^2+(2-3)^2}\\\\d=\sqrt{10}\ units$

6 0

3 years ago