Evaluate<br> O 15<br> O1<br> What’s the answer?????

I hope that we pass this semester god bless :)) calming it

Umnica [9.8K]

Answer:

me too

Step-by-step explanation:

7 0

2 years ago

a rectangular patio is 9 yards 2 ft long and 5 yards 2 feet wide what is the perimeter of the patio in yards

qaws [65]

Answer:36 yards and 2 feet

Step-by-step explanation:

7 0

3 years ago

Write 56 as a product of primes.<br> Use index notation when giving your answer.

mixas84 [53]

Answer:

56 as a product of primes using index notation is: $\mathbf{2^3 \times 7}$

Step-by-step explanation:

We need to write 56 as a product of primes.

Prime Numbers:

<em>Numbers that are divisible by 1 or itself.</em>

Now, writing 56 as product of primes

56 = 2 x 2 x 2 x 7

Now, Use index notation when giving your answer.

Index Notation:

<em>Express elements in power form i.e. if we have 2 x 2 we can write it as 2^2</em>

So, answer will be:

$56 = 2 \times 2 \times 2 \times 7$

$=2^3 \times 7$

So, 56 as a product of primes using index notation is: $\mathbf{2^3 \times 7}$

3 0

2 years ago

4. An ice cream shop wants to design a super straw to serve with its extra thick milkshakes that is double both the width and th

Andre45 [30]

Answer:

A. $\\ 0.50cm^{2}$ ; B. $\\ 5.03cm^{3}$ ; C. 19min

Step-by-step explanation:

<h3>First Step: Determine the size of the new straw</h3>

The size of the new straw is "double both the width and thickness of a standard straw". "A standard straw is 4mm in diameter and 0.5mm thick". So, the new straw is, then:

$\\ 2 * 4mm = 8mm$ or, equivalently, $\\ \frac{1cm}{10mm} * 8mm = 0.8cm$ (<em>Diameter</em>).

$\\ 2 * 0.5mm = 1mm$ or, equivalently, $\\ \frac{1cm}{10mm} * 1mm = 0.1cm$ (<em>Thickness</em>).

<h3>Second Step: Determine the cross-sectional area of the new straw</h3>

The cross-section area is "a section made by a plane cutting anything transversely, especially at right angles to the longest axis" <em>Cross-section</em> (2020), In Dictionary.com. Therefore, the area here is that of a circle:

$\\ A_{cross_{section}}= \pi * r^2$ ,

Where $\\ \pi = 3.1415926535897932384626....$ , and represents the ratio between a circle's circumference and its diameter, and <em>r</em> is the circle's radius or its diameter divided by 2.

Then, the area is:

$\\ A_{cross_{section}} = \pi * (\frac{0.8cm}{2})^2 = \pi * (0.4cm)^2= 0.502655cm^2$ , rounded to <em>the nearest hundredth</em>:

$\\ A_{cross_{section}} = 0.50cm^2$

<h3>Third Step: Determine the maximum volume of milkshake that can be in the straw at one time</h3>

The new straw is 10cm long and its cross-section is, approximately,

$\\ A_{cross_{section}} = 0.502655cm^2$

Since the straw is a cylinder, and the volume of a cylinder is:

$\\ V_{straw} = \pi * r^2 * h = A_{cross_{section}} * h$ , where <em>h</em> is the height of the straw. In this case is 10cm long.

So, <em>the maximum volume of milkshake that can be in the straw at one time</em> is the volume of the straw:

$\\ V_{straw} = A_{cross_{section}}*h = 0.502655cm^2 * 10cm = 5.02655cm^3$ , rounded to the <em>nearest hundredth</em>:

$\\ V_{straw} = 5.03cm^3$

<h3>Fouth Step: Determine the minimum amount of time that it will take Corbin to drink the milkshake</h3>

A large milkshake is 950mL, and we know that:

$\\ 1000mL = 1000cm^3 = 1L$

So,

$\\ 1mL = 1cm^3$ , therefore, $\\ 950mL = 950cm^3$

"Corbin withdraws the full capacity of a straw 10 times a minute" or Corbin withdraws:

$\\ 10*V_{straw} = 10*5.02655cm^3 = 50.2655cm^3$ every minute.

The large milkshake is $\\ 950cm^3$ . If Corbin withdraws $\\ \frac{50.2655cm^3}{min}$ , <em>the minimum amount of time that it will take him to drink the milkshake</em> is: