F(x) = x2 to g(x) = (x + 3)2 + 4?

A rectangle width is one fourth it’s length, outs are is 9 square units. The equation I(1/4I)=9 can be used to find I, the lengt

OlgaM077 [116]

Answer:

The length of rectangle is 6 units and the width is 1.5 units

Step-by-step explanation:

Let

L -----> the length of rectangle

W ----> the width of rectangle

we know that

The area of rectangle is equal to

$A=LW$

we have

$A=9\ units^{2}$

so

$9=LW$ -----> equation A

A rectangle width is one fourth it’s length

so

$W=\frac{1}{4}L$ ----> equation B

substitute equation B in equation A and solve for L

$9=L(\frac{1}{4}L)$

$L^{2}=9*4$

$L^{2}=36$

take square root both sides

$L=6\ units$

Find the value of W

$W=\frac{1}{4}(6)$

$W=1.5\ units$

5 0

4 years ago

Please help will mark as brainliest

MakcuM [25]

The answer is A. it is in standard form. a=3 b=9 c=10

7 0

3 years ago

Please HELP for a graph equation!! 100 points!

KATRIN_1 [288]

Answer:

like -1/4

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Deja ordered 8 pizzas each pizza was 5.99 if 3/4 of them were cheese how much will she pay for the pizzas that aren’t cheese

lidiya [134]

Answer:

5.99 cause she only had 1 without cheese and they are all 5.99

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

The port of South Louisiana, located along 54 miles of the Mississippi River between New Orleans and Baton Rouge, is the largest

jeka94

Answer:

(a) The probability that the port handles less than 5 million tons of cargo per week is 0.7291.

(b) The probability that the port handles 3 or more million tons of cargo per week is 0.9664.

(c) The probability that the port handles between 3 million and 4 million tons of cargo per week is 0.2373.

(d) The number of tons of cargo per week that will require the port to extend its operating hours is 5.35 million tons.

Step-by-step explanation:

Let X = amount of cargo the port handles per week.

The random variable X is Normally distributed with parameters,

μ = 4.5 million

σ = 0.82 million

(a)

Compute the probability that the port handles less than 5 million tons of cargo per week as follows:

$P(X$

*Use a z-table for the probability.

Thus, the probability that the port handles less than 5 million tons of cargo per week is 0.7291.

(b)

Compute the probability that the port handles 3 or more million tons of cargo per week as follows:

$P(X\geq 3)=P(\frac{X-\mu}{\sigma}\geq \frac{3-4.5}{0.82})\\=P(Z>-1.83)\\=P(Z$

*Use a z-table for the probability.

Thus, the probability that the port handles 3 or more million tons of cargo per week is 0.9664.

(c)

Compute the probability that the port handles between 3 million and 4 million tons of cargo per week as follows:

$P(3$

Thus, the probability that the port handles between 3 million and 4 million tons of cargo per week is 0.2373.

(d)

It is provided that P (X < x) = 0.85.

Then, P (Z < z) = 0.85.

The value of z for this probability is:

z = 1.04.

Compute the value of x as follows:

$z=\frac{x-\mu}{\sigma}\\1.04=\frac{x-4.5}{0.82}\\x=4.5-+1.04\times 0.82)\\x=5.3528\approx5.35$

Thus, the number of tons of cargo per week that will require the port to extend its operating hours is 5.35 million tons.

5 0

4 years ago