Select all number combinations that COULD make a triangle
4 cm, 5 cm, 7 cm
5 cm, 6 cm, 11 cm
2 cm, 9 cm, 10 cm
3cm, 4 cm, 10 cm

They are equal. 2/5 is 4/10.

Answer:

Option 2,    3(5y-11) - 2

Hope this helps!

Answer:

SSS - The two triangles share all 3 sides, proving they are congruent.

Step-by-step explanation:

Two sides of each triangle are marked in the diagram as equal. The third side, [tex]\overline{BD}[/tex] is shared between both sides, meaning that both triangles share all three sides. This is known as SSS (side-side-side) and is a proof of congruence.

Step-by-step explanation:

In triangle ABD In triangle CBD

AB. = BC. [Given]

AD. = CD. [Given]

BD. = BD. [Common side]

Therefore,

According to SSS Congruence condition

Triangle ABD is congruent to Triangle CBD

Answer:

c

Step-by-step explanation:

Answer:

Clint would have to run 127 yards next game to average the same number of yards as Jack.

Step-by-step explanation:

First, we find the boys' averages by adding all their yards and then dividing by the number of games.

For Jack, we get 490 yards in total, which we divide by 5, the number of games. We get a 98 yard average.

For Clint, we get 461 yards in total, which we divide by 5, the number of games. We get a 92.2 yard average.

We know Clint wants to average 98 yards at the end of the next game. That would make 6 games total. We can multiply 98 x 6 to get the total number of yards for a 98 yard average in 6 games. We get 588 yards in total. Clint already ran 461 yards in total, so we can subtract that from 588. We have 127 yards left. We now know that Clint needs to run 127 yards in the next game for a 98 yard average.

The number of yards that client needs to run in his next game to average the same number of yards as jack will be 127.

The mean is the average of a data set. Mean gives us an idea of that how small the amount of overall data have.

In other words, the mean is the foundation of the deviation of data from all data sets.

Mean = sum of data / Number of data

The average of Jack = (123 + 87 + 108 + 80 + 92)/5 = 98

The average of Clint = (88 + 101 + 96 + 92 + 84)/5 = 92.2

To make this average 98 let's say Clint's score in the next game is x.

Mean = (88 + 101 + 96 + 92 + 84 + x)/6 = 98

(92.2 x 5 + x) = 98 x 6

x = 127

Hence"Clint needed to add 127 to be mean as same as Jack".

For more about the mean,

https://brainly.com/question/17060266

#SPJ2

Answer:

i think c

Step-by-step explanation:

Answer:

Gross income for an individual—also known as gross pay when it's on a paycheck—is the individual's total pay from their employer before taxes or other deductions. This includes income from all sources and is not limited to income received in cash; it also includes property or services received.

Net income- In business and accounting, net income is an entity's income minus cost of goods sold, expenses, depreciation and amortization, interest, and taxes for an accounting period.

I think its B

Step-by-step explanation:

Answer: 52 and 793.25

Step-by-step explanation:

Answer:

1: 52

2: 105.80

Step-by-step explanation:

Answer: p= -3,2

q= -3,8

R= -8,8

s=-8,2

Step-by-step explanation:

Answer:

16.9

Step-by-step explanation:

We can use a proportion also in this case

51.6 : 23 = 38 : NO

NO = (38 x 23)/51.6 = 16.9

Answer:

55.5 is the I guess........

Answer: She needs to get 99 on her next test.

Step-by-step explanation:

Answer: The answer is 143.59

Step-by-step explanation:

I just subtracted 279.51 from 423.1! :)

With this prism, first separate it into a rectangular prism and a right triangle prism.

First, with the rectangular prism, it would have the units 5 * 5 * 6.3. Then with the right triangle prism, it would take the rest of the remaining space of it, being 3 * 5 * 6.3

The equation for the volume of a rectangular prism is L*W*H and the equation for the volume of a right triangle prism is (L*W*H) / 2.

With this, plug the numbers in for both prisms to get the volume of each of them:

Rectangular prism Volume

5*5*6.3

157.5

Right triangle prism Volume

(3*5*6.3) / 2

94.5 / 2

47.25

Lastly, combine these two values to get the total volume:

157.5 + 47.25 = 204.75 (Round Up) -->

204.8 cm^3 = Total Volume

Next, with the surface area of the prism. For this, lets combine all of the faces of the prism then add them all up.

First, lets do the bottom of the prism, or the base. It uses the lengths 8 cm and 6.3 cm. Lets do L * W to get the area of this face:

8 * 6.3 = 50.4

Next, lets do the slant side of it, which has the lengths 5.8 and 6.3.

5.8 * 6.3 = 36.54

Then, the top side with the lengths 5cm and 6.3 cm.

5.3 * 6.3 = 33.39

After that, the left side face that opposite to the slantly one:

5 * 6.3 = 31.5

Saving the most tedious part of it for last, the right trapezoids. Luckily, there is an equation for this:

1/2 x (Sum of parallel sides) x (perpendicular distance between the parallel sides).

So, within each of these right trapezoids, there's the parallel sides of 5 and 8. There's also a perpendicular side of 5cm. With this, we can plug this into the equation to solve for this part:

1/2 x (5+8) * (5)

1/2 x (13) * (5)

1/2 x (65)

32.5

Since there's two of them, times this by 2:

32.5 * 2 ---> 65.

Now, with the area of all of the faces, these can be added up for the total surface area:

50.4 + 36.54 + 33.39 + 31.5 + 65 ---> 216.83 (Round Down)-->

216.8 cm^2 = Total Surface Area

Answer:

D. 0.62

Step-by-step explanation:

I calculated it logically

Answer:

60,000

Step-by-step explanation:

Answer:

60,000

Step-by-step explanation:

Answer:

23

Step-by-step explanation:

6x3=18

18+5=23

Answer

this is wrong

Step-by-step explanation:

this is an equation not an expression. Equations have = and expressions dont. Please rewrite this and I will help.

Answer:

It is an arithmetic sequence and the common difference is 5

Step-by-step explanation:

4+5 = 9; 9+5=14 (adding is arithmetic, multiplying is geometric)

Answer:

65m = 6500 Cm.

Step-by-step explanation:

Answer:

x= -14

Step-by-step explanation:

Answer:

10/4

Step-by-step explanation:

=======================================================

Work Shown:

We'll apply these log rules

[tex]\text{Log Rule 1: } \ \ \ \ln\left(\frac{x}{y}\right) = \ln(x) - \ln(y)\\\\[/tex]

[tex]\text{Log Rule 2: } \ \ \ \ln\left(x^y\right) = y\ln(x)\\\\[/tex]

These log rules can be used for logs of any base, and not just natural logs.

[tex]z = \ln\left(\frac{\sqrt{x}}{y}\right)\\\\z = \ln\left(\sqrt{x}\right)-\ln\left(y\right) \text{ ... Use log rule 1}\\\\z = \ln\left(x^{1/2}\right)-\ln\left(y\right)\\\\z = \frac{1}{2}\ln\left(x\right)-\ln\left(y\right) \text{ ... Use log rule 2}\\\\z = \frac{1}{2}u-v\\\\[/tex]

Therefore,

[tex]\ln\left(\frac{\sqrt{x}}{y}\right) = \frac{1}{2}u-v\\\\[/tex]

when [tex]u = \ln(x) \ \text{ and } \ v = \ln(y)\\\\[/tex]

Answer:

the diameter is 7

Answer:

13. A

14. B

15. C

16. B

17. C

Step-by-step explanation:

Answer:

1.8 in.

Step-by-step explanation:

Hi there!

1) Determine the radius of the base

[tex]V=\frac{1}{3} \pi r^2h[/tex] where r is the radius of the base and h is the height of the cone

Plug in the known values (h=4.2, V=3.6)

[tex]3.6=\frac{1}{3} \pi r^2(4.2)[/tex]

Isolate r; divide both sides by 4.2

[tex]\frac{3.6}{4.2} =\frac{1}{3} \pi r^2\\\frac{6}{7} =\frac{1}{3} \pi r^2[/tex]

Multiply both sides by 3

[tex]\frac{6}{7}*3 =\pi r^2\\\frac{18}{7}=\pi r^2[/tex]

Divide both sides by π

[tex]\frac{18}{7\pi }=r^2[/tex]

Take the square root of both sides

[tex]\sqrt{\frac{18}{7\pi }} =r[/tex]

2) Determine the diameter

Now that we have the radius, to find the diameter, multiply r by 2:

[tex]d=\sqrt{\frac{18}{7\pi }}*2\\d=1.8[/tex]

Therefore, the diameter is 1.8 in. when rounded to the nearest tenth.

I hope this helps!

Answer:

1.8 in

Step-by-step explanation:

we know the equation for the volume of a cone is pi*r^2 h/3=3.6

pi r^2 *4.2=10.8

solve for r

r is about 0.9 so diameter is 1.8 in

Answer:

[tex]\displaystyle 2[/tex]

Step-by-step explanation:

[tex]\displaystyle \boxed{y = -2cos\:(\frac{\pi}{3}x - \frac{2}{3}\pi) + \frac{1}{2}} \\ \\ y = Acos(Bx - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow \frac{1}{2} \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \hookrightarrow \boxed{2} \hookrightarrow \frac{\frac{2}{3}\pi}{\frac{\pi}{3}} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \hookrightarrow \boxed{6} \hookrightarrow \frac{2}{\frac{\pi}{3}}\pi \\ Amplitude \hookrightarrow 2[/tex]

OR

[tex]\displaystyle \boxed{y = -2sin\:(\frac{\pi}{3}x - \frac{\pi}{6}) + \frac{1}{2}} \\ \\ y = Asin(Bx - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow \frac{1}{2} \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \hookrightarrow \boxed{\frac{1}{2}} \hookrightarrow \frac{\frac{\pi}{6}}{\frac{\pi}{3}} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \hookrightarrow \boxed{6} \hookrightarrow \frac{2}{\frac{\pi}{3}}\pi \\ Amplitude \hookrightarrow 2[/tex]

As you can see, the right-centred photograph displays the trigonometric graph of [tex]\displaystyle y = -2cos\:\frac{\pi}{3}x + \frac{1}{2},[/tex]in which you need to replase "sine" with "cosine", then figure out the appropriate C-term that will horisontally shift and map the graph onto the sine graph [photograph farthest to the left], accourding to the horisontal shift formula above. Also keep in mind that the −C gives you the OPPOCITE TERMS OF WHAT THEY REALLY ARE, so be very careful with your calculations. So, between the two photographs, we can tell that the cosine graph [centre right photograph] is shifted [tex]\displaystyle 2\:units[/tex]to the left, which means that in order to match the sine graph [photograph farthest to the left], we need to shift the graph FORWARD [tex]\displaystyle 2\:units,[/tex]which means the C-term will be positive, and by perfourming your calculations, you get [tex]\displaystyle \boxed{2} = \frac{\frac{2}{3}\pi}{\frac{\pi}{3}}.[/tex]So, the cosine graph of the sine graph, accourding to the horisontal shift, is [tex]\displaystyle y = -2cos\:(\frac{\pi}{3}x - \frac{2}{3}\pi) + \frac{1}{2}.[/tex]Intermediate, we have two sine graphs to deal with [centre left photograph and the photograph farthest to the left]. The left-centred photograph displays the trigonometric graph of [tex]\displaystyle y = -2sin\:\frac{\pi}{3}x + \frac{1}{2}.[/tex]So, just like before, we need to figure out the appropriate C-term that will horisontally shift the graph back into position ["position", meaning shifting this sine graph back in plase (matching the sine graph in the photograph on the left)]. So, between the two photographs, we can tell that the left-centred sine graph is shifted [tex]\displaystyle \frac{1}{2}\:unit[/tex]to the left, which means that in order to match the leftward sine graph, we need to shift the graph FORWARD [tex]\displaystyle \frac{1}{2}\:unit,[/tex]which means the C-term will be positive, and by perfourming your calculations, you get [tex]\displaystyle \boxed{\frac{1}{2}} = \frac{\frac{\pi}{6}}{\frac{\pi}{3}}.[/tex]So, when shifted back into original position, the sine graph is [tex]\displaystyle y = -2sin\:(\frac{\pi}{3}x - \frac{\pi}{6}) + \frac{1}{2}.[/tex]Now, with all that being said, in this case, sinse you ONLY have a graph to wourk with, you MUST figure the period out by using wavelengths. So, looking at where the graph WILL hit [tex]\displaystyle [-2\frac{1}{2}, \frac{1}{2}],[/tex]from there to [tex]\displaystyle [3\frac{1}{2}, \frac{1}{2}],[/tex]they are obviously [tex]\displaystyle 6\:units[/tex]apart, telling you that the period of the graph is [tex]\displaystyle 6.[/tex]Now, the amplitude is obvious to figure out because it is the A-term, and to be certain, look at the graph to see how low and high each crest extends beyond the midline. The midline is the centre of your graph, also known as the vertical shift, which in this case the centre is at [tex]\displaystyle y = \frac{1}{2},[/tex]in which each crest is extended two units beyond the midline, hence, your amplitude. Now... the sine graph in the photograph farthest to the right is the OPPOCITE of the sine graph in the photograph farthest to the left, and the reason for this is because of the negative inserted in front of the amplitude value. Whenever you insert a negative in front of the amplitude value of any trigonometric equation, the whole graph reflects over the midline. Keep this in mind moving forward. Now, with all that being said, no matter how far the graph shifts vertically, the midline will ALWAYS follow.

I am delighted to assist you at any time.

Answer:

whats this? can you provide more context?

Step-by-step explanation: