The ordered pairs represent a function.

The area of a square garden is 50 m3. How long is the diagonal?

katen-ka-za [31]

Answer:

10 m

Step-by-step explanation:

If the area of a square is 50 m2, then the length of each side will be $\sqrt{50}$ m, to get the diagonal length, after dividing the square by joining the points of the diagonal with a line, it forms two right angles with length $\sqrt{50}$ m each.

the diagonal length represent the hypotenuse side of each of the right angle, knowing the pythagoras theorem:

$(hyp)^{2} = (opp)^{2} + (Adj)^{2}$ , so we have:

$(diagonal)^{2} = 50 + 50$

$(diagonal)^{2} = 100$

diagonal = 10 m

6 0

3 years ago

Match each function with the corresponding function formula when h(x)=5-3x and g(x)=-3+5

Grace [21]

Answer:

k(x) = (3g + 5h)(x) ⇒ (1)

k(x) = (5h - 3g)(x) ⇒ (3)

k(x) = (h - g)(x) ⇒ (2)

k(x) = (g + h)(x) ⇒ (4)

k(x) = (5g + 3h)(x) ⇒ (5)

k(x) = (3h - 5g)(x) ⇒ (6)

Step-by-step explanation:

* To solve this problem we will substitute h(x) and g(x) in k(x) in the

right column to find the corresponding function formula in the

left column

∵ h(x) = 5 - 3x

∵ g(x) = -3^x + 5

- Lets start with the right column

# k(x) = (3g + 5h)(x)

∵ g(x) = -3^x + 5

∵ 3g(x) = 3[-3^x + 5] = [3 × -3^x + 3 × 5]

- Lets simplify 3 × -3^x

take the negative out -(3 × 3^x), and use the rule a^n × a^m = a^(n+m)

∴ -3(3 × 3^x) = -(3^x+1)

∴ 3g(x) = -3^x+1 + 15

∵ h(x) = 5 - 3x

∵ 5h(x) = 5[5 - 3x] = [5 × 5 - 5 × 3x] = 25 - 15x

- Now substitute 3g(x) and 5h(x) in k(x)

∵ k(x) = (3g + 5h)(x)

∴ k(x) = -3^x+1 + 15 + 25 - 15x ⇒ simplify

∴ k(x) = 40 - 3^x+1 - 15x

∴ k(x) = 40 - 3^x+1 - 15x ⇒ k(x) = (3g + 5h)(x)

* k(x) = (3g + 5h)(x) ⇒ (1)

# k(x) = (5h - 3g)(x)

∵ 5h(x) = 25 - 15x

∵ 3g(x) = -3^x+1 + 15

∵ k(x) = (5h - 3g)(x)

∴ k(x) = 25 - 15x - (-3^x+1 + 15) = 25 -15x + 3^x+1 - 15 ⇒ simplify

∴ k(x) = 10 + 3^x+1 - 15x

∴ k(x) = 10 + 3^x+1 - 15x ⇒ k(x) = (5h - 3g)(x)

* k(x) = (5h - 3g)(x) ⇒ (3)

# k(x) = (h - g)(x)

∵ h(x) = 5 - 3x

∵ g(x) = -3^x + 5

∵ k(x) = (h - g)(x)

∴ k(x) = 5 - 3x - (-3^x + 5) = 5 - 3x + 3^x - 5 ⇒ simplify

∴ k(x) = 3^x - 3x

∴ k(x)= 3^x - 3x ⇒ k(x) = (h - g)(x)

* k(x) = (h - g)(x) ⇒ (2)

# k(x) = (g + h)(x)

∵ h(x) = 5 - 3x

∵ g(x) = -3^x + 5

∵ k(x) = (g + h)(x)

∴ k(x) = -3^x + 5 + 5 - 3x ⇒ simplify

∴ k(x) = 10 - 3^x - 3x

∴ k(x)= 10 - 3^x - 3x ⇒ k(x) = (g + h)(x)

* k(x) = (g + h)(x) ⇒ (4)

# k(x) = (5g + 3h)(x)

∵ g(x) = -3^x + 5

∵ 5g(x) = 5[-3^x + 5] = [5 × -3^x + 5 × 5] = 5(-3^x) + 25

∴ 5g(x) = -5(3^x) + 25

∵ h(x) = 5 - 3x

∵ 3h(x) = 3[5 - 3x] = [3 × 5 - 3 × 3x] = 15 - 9x

- Now substitute 5g(x) and 3h(x) in k(x)

∵ k(x) = (5g + 3h)(x)

∴ k(x) = -5(3^x) + 25 + 15 - 9x ⇒ simplify

∴ k(x) = 40 - 5(3^x) - 9x

∴ k(x) = 40 - 5(3^x) - 9x ⇒ k(x) = (5g + 3h)(x)

* k(x) = (5g + 3h)(x) ⇒ (5)

# k(x) = (3h - 5g)(x)

∵ 3h(x) = 15 - 9x

∵ 5g(x) = -5(3^x) + 25

∵ k(x) = (3h - 5g)(x)

∴ k(x) = 15 - 9x - [-5(3^x) + 25] = 15 - 9x + 5(3^x) - 25 ⇒ simplify

∴ k(x) = 5(3^x) - 9x - 10

∴ k(x) = 5(3^x) - 9x - 10 ⇒ k(x) = (3h - 5g)(x)

* k(x) = (3h - 5g)(x) ⇒ (6)

4 0

3 years ago

This question relates to concepts covered in Lectures 1 & 2. You can use any of the excel files posted to work through the q

liubo4ka [24]

Answer:

mean of this demand distribution = 100

Step-by-step explanation:

To find the mean of this demand distribution;

Mean = Expected vale = E[x]

for discrete provability function,

we say E[x] = ∑(x.p(x))

x p(x) x.p(x)

10 0.1 1

30 0.4 12

60 0.4 24

90 0.7 63

∴ ∑(x.p(x)) = ( 1 + 12 + 24 + 63 )

∑(x.p(x)) = 100

7 0

3 years ago

Which graph represents the function f(x) = 2 · 4^x ?

NARA [144]

Answer:

The fisrt one were it almost hits 3

Step-by-step explanation:

7 0

3 years ago

Water that flows over land is the definition of which term

Paraphin [41]

Answer:Surface

Step-by-step explanation:

6 0

3 years ago