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denis-greek [22]

3 years ago

9

The area of the rectangular flowerbed shown is 20.4 square meters. How many meters of edging are needed to go around the flowerb

ed? Explain

1 answer:

Temka [501]3 years ago

5 0

There are needed 24.7 meters to go around the flowerbed.

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Three times a number increased by eight is no more than a number decreased by four

Westkost [7]

Answer:

x is no more than -6

Step-by-step explanation:

3 0

3 years ago

Cookies are being sold in the cafeteria to raise money for the cheerleaders. One giant cookie costs $1 while each one bought aft

sladkih [1.3K]

Answer:

y = 0.75x + 1

Step-by-step explanation:

Here, we want to write an equation in slope-intercept form to represent this information

Since $1 is the cost of 1 only, this represents our y intercept

It means if we are not interested in purchasing more than 1, our cost will be $1

Now, for each one bought after, we have a rate of $0.75 per one

Mathematically, that will be $0.75 times the number bought after

Hence. we have that;

y = 0.75x + 1

if 1, x = 0; so we have cost of 1 as $1

7 0

2 years ago

If tan 2A = cot (A -18°), where 2A is an acute angle, find the value of A.

Reptile [31]

answer will be -24

A=-24

7 0

3 years ago

Read 2 more answers

Divide. Express your answer in simplest form.<br><br> 11/12 ÷ 3/4=

Tanzania [10]

Answer:

11/9

hope this helped :)

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

I built a storage shed in the shape of a rectangular box on a square base. The material that I used for the base cost $4 per squ

kumpel [21]

Answer:

side of 6.124 ft and height of 3.674 ft

Step-by-step explanation:

Let's s be the side of the square base and h be the height of the rectangular box.

The base and the roof would have an area of $s^2$ and cost of

$4s^2 + 2s^2 = 6s^2$

The sides would have an area of 4sh and cost of 4sh*2.5 = 10sh

So the total cost for the material is

$6s^2 + 10sh = 450$

$10sh = 450 - 6s^2$

$h = 45/s - 0.6s$

The volume of the shed has the following formula

$V = s^2h = s^2(45/s - 0.6s) = 45s - 0.6s^3$

To find the maximum value for V, we can take its first derivative, and set it to 0

$\frac{dV}{ds} = 45 - 1.2s^2 = 0$

$1.2s^2 = 45$

$s^2 = 45 / 1.2 = 37.5$

$s = \sqrt{37.5} = 6.124 ft$

h = 45/s - 0.6s = 3.674 ft

4 0

3 years ago