1. Find the measure of angle R.

Hi, I'm happy to help!

First off, we need to arrange an equation for what is being asked. 15 is 20% of her earnings and we don't know her total earnings.

15=20%×__

We can write this percentage as

15=0.2×__

We then change the equation to do the inverse operation, which is division, this makes it easier to solve.

15÷0.2=__

Now, we solve

15÷0.2=75

She earned $75

I hope this was helpful, keep learning! :D

Answer:

speed

1)the distance travelled by a body per unit time is known as speed.

2) speed is speed, or something how fast is going,

3) while velocity is speed how fast something is going in a specific direction.

average speed

1) average speed is the speed added up and divided by number of speed .

2)the average of all the instantaneous speed in a give time interval.

3) the average speed as the duration of the time interval approaches zero.

Answer:

y=32x-32    

Step-by-step explanation:

This time no algorithm or equation was needed i saw that if the equation was y=32x there would be ordered pairs (1,32) and (2,64) so to delay it by 1  on the x side just subtract the slope from the y - intercept sorry if that doesn't make sense

Answer:

25 gallons of sulfuric acid

Step-by-step explanation:

Find how much sulfuric acid is in the tank by finding 50% of 50 (the total gallons of solution):

50(0.5)

= 25

So, there are 25 gallons of sulfuric acid.

Answer:

80

Step-by-step explanation:

The Two-Tangent Theorem states that if two tangent segments are drawn to one circle from the same external point, then they are congruent

13+13+9+9+12+12+6+6= 80

Answer:

Hello,

Step-by-step explanation:

Le théorème est faux

since a=9 and b=16 then

[tex]\sqrt{9} +\sqrt{16} =3*4=7 \\\sqrt{9+16} =\sqrt{25} =5\\[/tex]

Answer:

x = 155º

Step-by-step explanation:

y = 25°    {Vertically opposite angles}

x + y = 180    {Co-interior angles are supplementary}

x + 25 = 180

Subtract 25 from both sides

x = 180 - 25

x = 155°

Hey there! I'm happy to help!

We see that Jack has already run 25 miles.

25

He also wants to run 2.5 miles every day. This means we will multiply 2.5 miles by the number of days he runs. We will call the number of days x.

25+2.5x

And he needs this to equal 275.

25+2.5=275

We subtract 25 from both sides.

2.5x=250

We divide both sides by 2.5

x=100

So, Jack will need to run 100 days to meet his goal.

Have a wonderful day and keep on learning! :D

Answer:

B. y= 6x2, y= -4.5x2, y= -x2

Step-by-step explanation:

y = 6x2, y = −4.5x2, y = −x2

This should be correct )

To solve this problem you must keep on mind the following information: By definition, a quadratic function has the following form:  

 y=ax^{2}+bx+c


Where a
 is the leading coefficient.

If the leading coeficient is closer to zero, the parabola is widest,if it has a larger positive or negative value, the parabola is narrowest.

Therefore, by knowing the information above, you have that the answer is:

y=- x^{2}  \\ y=-4.5 x^{2}  \\ y=6 x^{2}

[tex]\\ \sf\longmapsto 41sin\Theta=40[/tex]

[tex]\\ \sf\longmapsto sin\Theta=\dfrac{40}{41}[/tex]

Now

[tex]\boxed{\sf cos\Theta=\sqrt{1-sin^2\Theta}}[/tex]

[tex]\\ \sf\longmapsto cos\Theta=\sqrt{1-\left(\dfrac{40}{41}\right)^2}[/tex]

[tex]\\ \sf\longmapsto cos\Theta=\sqrt{1-\dfrac{1600}{1682}}[/tex]

[tex]\\ \sf\longmapsto cos\Theta=\sqrt{\dfrac{1681-1600}{1681}}[/tex]

[tex]\\ \sf\longmapsto cos\Theta=\sqrt{\dfrac{81}{1681}}[/tex]

[tex]\\ \sf\longmapsto cos\Theta=\dfrac{9}{41}[/tex]

We know

[tex]\boxed{\sf tan\Theta=\dfrac{Sin\theta}{Cos\Theta}}[/tex]

[tex]\\ \sf\longmapsto \dfrac{tan\Theta}{1-tan^2\Theta}[/tex]

[tex]\\ \sf\longmapsto \dfrac{\dfrac{sin\Theta}{cos\Theta}}{1-\dfrac{sin^2\Theta}{cos^2\Theta}}[/tex]

[tex]\\ \sf\longmapsto \dfrac{\dfrac{\left(\dfrac{40}{41}\right)}{\left(\dfrac{9}{41}\right)}}{1-\dfrac{\left(\dfrac{40}{41}\right)^2}{\left(\dfrac{9}{41}\right)^2}}[/tex]

[tex]\\ \sf\longmapsto \dfrac{\dfrac{{40}}{{9}}}{1-\dfrac{{40}^2}{{9}^2}}[/tex]

[tex]\\ \sf\longmapsto \dfrac{\dfrac{40}{9}}{\dfrac{9^2-40^2}{9^2}}[/tex]

[tex]\\ \sf\longmapsto \dfrac{\dfrac{40}{9}}{\dfrac{81-1600}{81}}[/tex]

[tex]\\ \sf\longmapsto \dfrac{\dfrac{40}{9}}{\dfrac{-1519}{81}}[/tex]

[tex]\\ \sf\longmapsto {\dfrac{40}{\cancel{9}}}\times \dfrac{\cancel{81}}{(-1519)}[/tex]

[tex]\\ \sf\longmapsto \dfrac{40\times 9}{(-1519)}[/tex]

[tex]\\ \sf\longmapsto - \dfrac{360}{1519}[/tex]

Answer:

Mapping C represents a function.

To be a function, there must be one and only one y value (output) for every input value (x value).

Mapping C matches this criteria, thus it is a function.

Let me know if this helps!

Answer:

x + y = 320

8x + 5y = 1900

Step-by-step explanation:

The equation x + y = 320 represents how 320 tickets were sold in total, from combining the adult and child tickets sold.

Since adult tickets are $8 each, the money from adult ticket sales can be represented by 8x.

The money from child ticket sales can be represented by 5y.

The equation 8x + 5y = 1900 represents how $1900 worth of tickets were sold.

So, the system of equations for the situation is:

x + y = 320

8x + 5y = 1900

Answer:

y = 220

x = 100

Step-by-step explanation:

x + y = 320

8x + 5y = 1900

x = 320 - y

8(320 - y) + 5y = 1900

2560 - 8y + 5y = 1900

2560 -3y = 1900

-2560        -2560

-------------------------

   -3y = -660

   -----    -------

    -3        -3

      y = 220

x + 220 = 320

    -220   -220

----------------------

        x = 100

Answer:

57/5

Step-by-step explanation:

what is that supposed to be ?

1.5 + 9.9 ?

I base my answer in that assumption.

1.5 = 3/2

9.9 = 99/10

so, we need 3/2 + 99/10.

how do we do that ?

by bringing both fractions to the same denominator (bottom parts) but without changing the value of the fractions. that means we need to multiply numerator and denominator (top and bottom parts) by the same factor. and that factor is determined by what is needed to transform the denominator.

we have denominators 2 and 10. what is the smallest common multiple ?

I think it is plainly visible : 10. as 2 also cleanly divides 10.

in other examples, if you don't see that right away, the simplest approach is to just multiply both denominators (in our case 2×10 = 20) and work with that result.

so, we are trying to bring both fractions to denominators of 10 (or, as mentioned 20), so that we can easily add the fractions.

let's start with 3/2.

what multiplication do we need to do to turn 2 into 10 ?

right, by multiplying by 5.

remember, we need to do the same thing to numerator and denominator to keep the value of the fraction unchanged.

so, we do the following

3/2 × 5/5

as you can see, multiplying something by 5/5 does not change any value, as it means we are just multiply by 1. but we can use that little trick to change the appearance of the original fraction.

so,

3/2 × 5/5 = (3×5) / (2×5) = 15/10

by the way, if we had wanted to change the denominator to 20, our factor would have been 10/10, and we would have gotten 30/20.

now to 99/10.

well, the denominator is already 10, so we don't need any transformation.

but if we had not seen that and went for 20 as desired denominator, we would have had to multiply by 2/2 giving us

198/20.

so, for the final sum :

15/10 + 99/10 = 114/10 = 57/5

and in case of 1/20th :

30/20 + 198/20 = 228/20 = 114/10 = 57/5

now, should that original problem have been

1/5 + 9/9 ?

the same principles apply.

bring both fractions to the same denominator.

9/9 = 1

and that can be transformed easily into any other denominator expression - like 5/5.

and then we have

1/5 + 5/5 = 6/5

but if we had not seen that simple approach, we could have turned both into fractions with denominator 5×9 = 45.

9/45 + 45/45 = 54/45 = 6/5

Answer:

12 type of bread 1 type of meat, 1 type of cheese and 1 type of bread.☺️☺️

Answer:

Answer is option C

Step-by-step explanation:

If you like my answer than please mark me brainliest thanks

Answer:

[tex]24x^{3} \sqrt{5x} -4x^{3} \sqrt{10x}[/tex]

Step-by-step explanation:

[tex]2\sqrt{8x^{3} } (3\sqrt{10x^{4} } -x\sqrt{5x^{2} } )[/tex]

[tex]=2.2x\sqrt{2x} (3x^{2} \sqrt{10} -x^{2} \sqrt{5} )[/tex]

[tex]=4x\sqrt{2x} (3x^{2} \sqrt{10} -x^{2} \sqrt{5} )[/tex]

[tex]=12x^{3} \sqrt{2xx10} -4x^{3} \sqrt{2xx5}[/tex]

[tex]=24x^{3} \sqrt{5x} -4x^{3} \sqrt{10x}[/tex]

OAmalOHopeO

Step-by-step explanation:

you're going to solve for y in the first equation and then substitute it in for the second equation and solve.

6 It is divisible by 2 and by 3.

4 it is divisible by 4...

The divisibility rules of 4 and 6 are completely different. In the divisibility rule of 4, if the last two digits are zeros or the number formed by the last two digits is exactly divisible by 4, then we can say that a number is divisible by 4.

HOPE ITS HELPFUL FOR U MATE

ω in terms of m and k is expressed as  [tex]\omega = \sqrt{\frac{k}{m} }[/tex]

The given expression is as follows;

     [tex]T_s = 2\pi \sqrt{\frac{m}{k} } , \ \ \\\\ T_s = \frac{2\pi}{\omega}[/tex]

To find:

From the given expression above make ω the subject of the formula;

[tex]T_s = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{m}{k} } \\\\ \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{m}{k} }\\\\ \frac{2\pi}{\omega} = \sqrt{4\pi^2\frac{m}{k} }\\\\square \ both \ sides \ of \ the \ equation;\\\\(\frac{2\pi}{\omega})^2 = 4\pi^2\frac{m}{k} \\\\\frac{4\pi^2}{\omega^2}= \frac{4\pi^2m}{k} \\\\\omega^2 4\pi^2m = k4\pi^2 \\\\divide \ both \ side \ by \ 4\pi ^2 \\\\\omega^2 m = k\\\\divide \ both \ sides \ by \ m\\\\\omega^2 = \frac{k}{m} \\\\[/tex]

[tex]take \ the \ square \ root \ of \ both \ sides \ of \ the \ equation\\\\\omega = \sqrt{\frac{k}{m} }[/tex]

Therefore, ω in terms of m and k is expressed as [tex]\omega = \sqrt{\frac{k}{m} }[/tex]

To learn more about subject of formula visit: https://brainly.com/question/15469690

Answer:

-3y-6

Step-by-step explanation:

[tex]-3(y+2)=-3y-6[/tex]

Answer:

ABC=BCD

so, X=60'

Z= 70'

Y=50'

Answer:

The terms are - 1, 7 and 15.

Step-by-step explanation:

Let the terms be a-d, a, a+d.

ATQ, a-d+a+a+d=21, a=7 and a+d-(a-d+a)=9. 2d-a=9, d=8. The terms are - 1, 7 and 15.

The first three terms will be :   -1, 7, 15

We have sum of first three terms of an arithmetic series is 21. It is given that if the sum of the first two terms is subtracted from the third term then it would be 9.

We have to find the three terms of the series.

An arithmetic progression or sequence with common difference [tex]d[/tex] is given as -

a, a + d, a + 2d, a + 3d .....

According to the question -

a + (a + d) + (a + 2d) = 21

3a + 3d = 21

3(a + d)=21

a + d = 7                             (Eqn. 1)

(a + 2d) - (a + a + d) = 9

a + 2d - a - a - d = 9

d - a = 9

a = d - 9

Substituting the value of 'a' in Eqn. 1, we get -

d - 9 + d = 7

2d - 9 = 7

2d = 16

d = 8

Therefore -

a = 8 - 9

a = -1

The three terms are -

a = -1

a + d = -1 + 8 = 7

a + 2d = -1 + 2 x 8 = 15

Hence, the first three terms will be - -1, 7, 15

To solve more questions on Arithmetic Sequences and finding the terms, visit the link below -

https://brainly.com/question/20384906

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Answer:

19)tu=12

20) true

Step-by-step explanation:

19) (22+5x)+(19-2x)=35

   3x+41=35

   3x=-6

  x=-2

TU=22+5*-2

     =12

(i) The number of students who registered for political science is 150

(ii) The number of students who registered for political science and geography but not economics is 50

(iii) The number of students who registered for economics and political science but not geography is 20

The Venn diagram for the question is shown in the attachment below.

E represents Economics

G represents Geography

and P represents Political science

From the question,

10 for economics only, that is, n(E∩G'∩P') = 10

150 from economics and geography, that is n(E∩G) = 150

90 from economics but not political science, that is, n(E∩P') = 90

210 from geography, that is, n(G) = 210

120 from political science and geography, n(P∩G) = 120

180 from economics, that is, n(E) = 180

Total of student registered for courses is 250, that is, n(ξ) = 250

Now from the given information,

n(E∩G∩P') = n(E∩P') - n(E∩G'∩P') = 90 - 10 = 80

n(E∩G∩P) =  n(E∩G) - n(E∩G∩P') = 150 - 80 = 70

n(E∩P∩G') = n(E) - [n(E∩G'∩P') + n(E∩G∩P') + n(E∩G∩P)]

= 180 - [10+80+70] = 180 - 160 = 20

n(P∩G∩E') = n(P∩G) - n(E∩G∩P) = 120 - 70 = 50

n(G∩E'∩P') = n(G) - [n(E∩G∩P') + n(E∩G∩P) + n(P∩G∩E')]

= 210 -[80+70+50] = 210 - 200 = 10

All of these are shown in the Venn diagram

(i) To determine the number of students who registered for political science, we will first determine n(P∩G'∩E'), that is, those who registered for political science only.

Let n(P∩G'∩E') = x

Then, using the Venn diagram, we can write that

10 + 80 + 20 + 70 + 10 + 50 + x = n(ξ) = 250

240 + x = 250

x = 250 - 240

x = 10

n(P∩G'∩E') = 10

∴ 10 registered for political science only

Now, number of students who registered for political science n(P) is

n(P) = n(E∩P∩G') + n(E∩G∩P) +  n(P∩G∩E')  + n(P∩G'∩E')

n(P) = 20 + 70 + 50 + 10

n(P) = 150

∴ 150 students registered for political science

(ii) For the number of students who registered for political science and geography but not economics, that is, n(P∩G∩E')

As determined above and as shown in the Venn diagram,

n(P∩G∩E') = 50

∴ 50 students registered for political science and geography but not economics

(iii) For the number of students who registered for economics and political science but not geography​, that is, n(E∩P∩G')

As determine above and as shown in the Venn diagram as well

n(E∩P∩G') = 20

∴ 20 students registered for economics and political science but not geography​

Hence,

(i) The number of students who registered for political science is 150

(ii) The number of students who registered for political science and geography but not economics is 50

(iii) The number of students who registered for economics and political science but not geography is 20

Learn more here: https://brainly.com/question/15311191

Answer:

Intergers

Step-by-step explanation:

A subset is a part of a whole math concept, idea, or sample space.

We need to find what math concept -5 is apart of.

-5 is part of the All Real Numbers section.

-5 is also part of integers as well.

Answer:

always

Step-by-step explanation:

they would all have the same gradient & be parallel to each other!

ie. A || B ,   B || C

in this case, A || C ^^

Answer:

always

Step-by-step explanation:

have the same gradient & be parallel to each other

Answer:

the last one

Step-by-step explanation:

The two on the right bounce up and down around -1. Two up and two down from -1. But only the one on the bottom right takes 3 pi to come back to the same place it started.

Explanation:

[tex]\displaystyle y = 2cos\:(\frac{2}{3}x - \frac{\pi}{2}) - 1 \\ y = 2sin\:\frac{2}{3}x - 1 \\ \\ 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 -1 \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \hookrightarrow \boxed{\frac{3}{4}\pi} \hookrightarrow \frac{\frac{\pi}{2}}{\frac{2}{3}} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \hookrightarrow \boxed{3\pi} \hookrightarrow \frac{2}{\frac{2}{3}}\pi \\ Amplitude \hookrightarrow 2[/tex]

OR

[tex]\displaystyle 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 -1 \\ Horisontal\:[Phase]\:Shift \hookrightarrow 0 \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \hookrightarrow \boxed{3\pi} \hookrightarrow \frac{2}{\frac{2}{3}}\pi \\ Amplitude \hookrightarrow 2[/tex]

You will need the above information to help you interpret the graph. So, first off, you MUST figure the period out by using wavelengths. So, looking at where the graph hits [tex]\displaystyle [-4\frac{1}{2}\pi, -1],[/tex] from there to [tex]\displaystyle [-1\frac{1}{2}\pi, -1],[/tex] they are obviously [tex]\displaystyle 3\pi\:units[/tex] apart, telling you that the period of the graph is [tex]\displaystyle 3\pi.[/tex] Now, the amplitude is obvious to figure out because it is the A-term, but of cource, if you want to be certain it is the amplitude, 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 = -1,[/tex] in which each crest is extended two units beyond the midline, hence, your amplitude. So, no matter how far the graph shifts vertically, the midline will ALWAYS follow.

*If you wanted to know the equation(s) of the graph, look above. If you have any questions on how the equation(s) came about, do not hesitate to ask.

I am delighted to assist you at any time.

Answer:

201, 206, 216, 228, 240

Step-by-step explanation: