H.C.F of 672 and 120 is

Jocelyn walks her dog 5 days each week.Each Walk is 2 1/4 miles long. What is the total number of miles that Jocelyn walks her d

igor_vitrenko [27]

Answer:the answer is c

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

How can you use a coordinate plane to solve perimeter and area problems?<br><br><br> Please help me

lana66690 [7]

Well, you can do
area=legnth times widht
perimiter=2(legnth+width)
then you just subsitute x for legnth and y for width then solve
area=xy
perimiter=2(x+y)
then graph and find the intersection to find the solution

6 0

2 years ago

Please help will mark brainly ! 11 points

pashok25 [27]

Answer:

3/10 > 1/5

Step-by-step explanation:

3/10 = 1.5/5

or

1/5 = 2/10

3/10 > 2/10

1.5/5 > 1/5

7 0

2 years ago

From a top of a building 20 m high if the angle of elevation top of a vertical pole is 30 degree.and the angle of depression of

Delvig [45]

Answer:

The height of the pole is 26.64 metres.

Step-by-step explanation:

Please kindly see the attached for explanation

7 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:

$\\ \frac{950cm^3}{\frac{50.2655cm^3}{min}} = \frac{950}{50.2655}*\frac{cm^3}{cm^3}*min = 18.8996min$ , since $\\ \frac{cm^3}{cm^3} = 1$ .

Rounded to the nearest minute, the minimum amount of time is 19min.

8 0

3 years ago

H.C.F of 672 and 120 is ​

H.C.F of 672 and 120 is