All categories

2 years ago

HELP ME PLEASE ECPLAIN WHY

Mathematics

1 answer:

stiv31 [10]2 years ago

8 0

Answer:

A≈110.11

using formula 1/4 $\sqrt5(5+2\sqrt5)a2$

Step-by-step explanation:

You might be interested in

Where do i graph it

Anuta_ua [19.1K]

(-5,-10)E
(-5,-3)F
(-3,-10)G

7 0

2 years ago

Read 2 more answers

Game Stop is selling used games in bundles. They have one bundle of 8 games for $52, and one bundle of 10 games for $62.50. Whic

natka813 [3]

Answer:

Step-by-step explanation:

3 0

1 year ago

If ƒ(x) = x + 2 and g(x) = 3x2 + 7x, find ....

Tamiku [17]

Answer:

The correct answer is option C

(f o g)(x) = 3x² + 7x + 2

Step-by-step explanation:

Points to remember

Composite functions

Let f(x) and g(x) be the two functions then (f o g)(x) can be written as

(f o g)(x) = f(g(x))

To find the value of (f o g)(x)

Here f(x) =x + 2 and g(x) = 3x² + 7x

(f o g)(x) = f(g(x))

= f(3x² + 7x)

= 3x² + 7x + 2

Therefore the correct answer is option C

(f o g)(x) = 3x² + 7x + 2

7 0

2 years ago

D = 8a The variable a represents the number of attendees, and the variable d represents the cost in dollars. At most how many at

suter [353]

Answer: At most 9 attendees can be there.

Step-by-step explanation:

Given equation: d = 8a , where a represents the number of attendees, and the variable d represents the cost in dollars.

To find : the number of attendees, if Will budgets a total of $72 for his graduation picnic.

72=8a

⇒ 9 = a [divide both sides by 8]

∴ a= 9

Hence, at most 9 attendees can be there.

4 0

2 years ago

The 1992 world speed record for a bicycle (human-powered vehicle) was set by Chris Huber. His time through the measured 200 m st

Reil [10]

Answer:

a) 30.726m/s and b) 5.5549s

Step-by-step explanation:

a.) What was Chris Huber’s speed in meters per second(m/s)?

Given the distance and time, the formula to obtain the speed is

$v=\frac{d}{t}$ .

Applying this to our problem we have that

$v=\frac{200m}{6.509s}= 30.726m/s$ .

So, Chris Huber’s speed in meters per second(m/s) was 30.726m/s.

b) What was Whittingham’s time through the 200 m?

In a) we stated that $v=\frac{d}{t}$ . This formula implies that

$t=\frac{d}{v}$ .

First, observer that $19\frac{km}{h} =19,000\frac{m}{h}=\frac{19,000}{3,600}m/s= 5.2777m/s$ .

Then, Sam Whittingham speed was equal to Chris Huber’s speed plus 5.2777 m/s. So, $v=30.726\frac{m}{s} +5.2777\frac{m}{s}= 36.003 m/s.$

Then, applying 1) we have that

$t=\frac{200m}{36.003m/s}=5.5549s.$

So, Sam Whittingham’s time through the 200 m was 5.5549s.

5 0

2 years ago