MATHEMATICS HIGH SCHOOL

What is 30% of 60? A. 0.18 B. 1.8 C. 18 D. 180

Answers

Answer 1

Answer:

Answer:

C ) 18 hope this helps !!!

Answer 2

Answer:

Answer:

C. 18

Step-by-step explanation:

If you multiply 30%, or .3 by 60, you get 18.

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Simplify the expressions using the distributive property.8(7x + 3)5(5y - 2)

Answers

Simplify the expressions by using the distributive property.

$\begin{gathered} 8(7x+3)=8\cdot7x+3\cdot8 \n =56x+24 \end{gathered}$

For second expression,

$\begin{gathered} 5(5y-2)=5\cdot5y+5\cdot(-2) \n =25y-10 \end{gathered}$

The outlier, 78, is removed from the data set shown.23, 23, 25, 27, 34, 34, 35, 41, 45, 45, 78

How does this affect the range?

A.) The range is decreased by 4.

B.) The range is decreased by 16.

C.) The range is decreased by 22.

D.) The range is decreased by 33.

Answers

An outlier of a set of data is a data value that is too small or too large compared to other values in the set of data. Range of a set of data is the difference between the largest data value and the smallest data valu in the set of data. Range of 23, 23, 25, 27, 34, 34, 35, 41, 45, 45, 78 is 78 - 23 = 55 When the outlier 78 is removed, the range of the data set 23, 23, 25, 27, 34, 34, 35, 41, 45, 45 is 45 - 23 = 22 Thus, when the outlier 78 is removed, the range of the data set decreases by 55 - 22 = 33. Therefore, the correct answer to the question "The outlier, 78, is removed from the data set shown. 23, 23, 25, 27, 34, 34, 35, 41, 45, 45, 78 How does this affect the range?" is "The range is decreased by 33." (option D).

the awnser is d of course

Can someone help me with this? PLs i'm so confused!

Answers

1. E. $sine\ A = (a)/(c) = (opposite)/(hypotenuse) = (BC)/(AB) = (5)/(13)$

2. L. $cos\ A = (b)/(c) = (adjacent)/(hypotenuse) = (AC)/(AB) = (12)/(13)$

3. $tan\ A = (a)/(b) = (opposite)/(adjacent) = (BC)/(AC) = (5)/(12)$

4. Y. $sin\ B = (a)/(c) = (opposite)/(hypotenuse) = (BC)/(AB) = (5)/(13)$

5. W. $cos\ B = (b)/(c) = (adjacent)/(hypotenuse) = (AC)/(AB) = (12)/(13)$

6. $tan\ B = (b)/(a) = (adjacent)/(opposite) = (AC)/(BC) = (12)/(5) = 2(2)/(5)$

7. $sin\ A = (a)/(c) = (opposite)/(hypotenuse) = (BC)/(AB) = (1)/(2)$

8. W. $cos\ A = (b)/(c) = (adjacent)/(hypotenuse) = (AC)/(AB) = (√(3))/(2)$

9. I. $tan\ A = (a)/(b) = (opposite)/(adjacent) = (BC)/(AC) = (1)/(√(3)) = (1)/(√(3)) * (√(3))/(√(3)) = (√(3))/(3)$

10. $sin\ B = (a)/(c) = (opposite)/(hypotenuse) = (BC)/(AB) = (1)/(2)$

11. E. $cos\ B = (b)/(c) = (adjacent)/(hypotenuse) = (AC)/(AB) = (√(3))/(1) = √(3)$

12. I. $tan\ B = (a)/(b) = (opposite)/(adjacent) = (BC)/(AC) = (1)/(√(3)) = (1)/(√(3)) * (√(3))/(√(3)) = (√(3))/(3)$

13. U. $sin\ A = (a)/(c) = (hypotenuse)/(opposite) = (BC)/(AB) = (12)/(15) = (4)/(5)$

14. I. $cos\A = (b)/(c) = (adjacent)/(hypotenuse) = (AC)/(AB) = (9)/(15) = (3)/(5)$

15. $tan\ A = (a)/(b) = (opposite)/(adjacent) = (BC)/(AC) = (12)/(9) = (4)/(3) = 1(1)/(3)$

16. R. $sin\ B = (a)/(c) = (opposite)/(hypotenuse) = (BC)/(AB) = (4)/(√(65)) = (4)/(√(65)) * (√(65))/(√(65)) = (4√(65))/(65)$

17. M. $cos\ B = (b)/(c) = (adjacent)/(hypotenuse) = (AC)/(AB) = (7)/(4) = 1(3)/(4)$

18. N. $tan\ B = (a)/(b) = (opposite)/(adjacent) = (BC)/(AC) = (4)/(7)$

19. L. $sin\ A = (a)/(c) = (opposite)/(hypotenuse) = (BC)/(AB) = (16)/(34) = (8)/(17)$

20. H. $cos\ B = (b)/(c) = (adjacent)/(hypotenuse) = \fac{AC}{AB} = (30)/(34) = (15)/(17)$

21. $tan\ B = (a)/(b) = (opposite)/(adjacent) = (BC)/(AC) = (16)/(30) = (8)/(15)$

22. O. $sin\ A = (a)/(c) = (opposite)/(hypotenuse) = (BC)/(AB) = (1)/(√(2)) = (1)/(√(2)) * (√(2))/(√(2)) = (√(2))/(2)$

23. O. $cos\ A = (b)/(c) = (adjacent)/(hypotenuse) = (AC)/(AB) = (1)/(√(2)) = (1)/(√(2)) * (√(2))/(√(2)) = (√(2))/(2)$

24. N. $tan\ A = (a)/(b) = (opposite)/(adjacent) = (BC)/(AC) = (1)/(1) = 1$

Someone please help!

Answers

Hello!

Answer:

$\Large \boxed{\sf CD= 18}$

Step-by-step explanation:

→ We want to calculate CD.

We know that CD = DE.

→ So we have this equation:

$\sf 3x-6 = 2(x+1)$

→ Let's solve this equation to find the value of x to find CD:

◼ Simplify the right side:

$\sf 3x-6 = 2x+2$

◼ Add 6 to both sides:

$\sf 3x-6+6 = 2x+2+6$

◼ Simplify both sides:

$\sf 3x = 2x+8$

◼ Subtract 2x from both sides:

$\sf 3x-2x = 2x+8-2x$

◼ Simplify both sides:

$\sf x = 8$

→ Now, let's calculate CD, let's replace x by 8:

$\sf CD = 3x-6\n\n= 3 * 8 - 6\n\n= 24 - 6\n\n\boxed{\sf= 18}$

Check:

$\sf CD = DE\n\n3x-6 = 2(x+1)\n\n3 * 8 - 6 = 2(8+1)\n\n24 - 6 = 2 * 9\n\n18=18$

So the check is good.

Conclusion:

CD is equal to 18.

a teacher and 10 students are to be seated along a bench in the bleachers at a basketball game. In how many ways can this be done if the teacher must be seated in the middle and a difficult student must sit to the teachers immediate left?

Answers

Wow !

OK. The line-up on the bench has two "zones" ...

-- One zone, consisting of exactly two people, the teacher and the difficult student.
   Their identities don't change, and their arrangement doesn't change.

-- The other zone, consisting of the other 9 students.
   They can line up in any possible way.

How many ways can you line up 9 students ?

The first one can be any one of 9.   For each of these . . .
The second one can be any one of the remaining 8. For each of these . . .
The third one can be any one of the remaining 7. For each of these . . .
The fourth one can be any one of the remaining 6. For each of these . . .
The fifth one can be any one of the remaining 5. For each of these . . .
The sixth one can be any one of the remaining 4. For each of these . . .
The seventh one can be any one of the remaining 3. For each of these . . .
The eighth one can be either of the remaining 2. For each of these . . .
The ninth one must be the only one remaining student.

   The total number of possible line-ups is

   (9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) = 9! = 362,880 .

But wait ! We're not done yet !

For each possible line-up, the teacher and the difficult student can sit

-- On the left end,
-- Between the 1st and 2nd students in the lineup,
-- Between the 2nd and 3rd students in the lineup,
-- Between the 3rd and 4th students in the lineup,
-- Between the 4th and 5th students in the lineup,
-- Between the 5th and 6th students in the lineup,
-- Between the 6th and 7th students in the lineup,
-- Between the 7th and 8th students in the lineup,
-- Between the 8th and 9th students in the lineup,
-- On the right end.

That's 10 different places to put the teacher and the difficult student,
in EACH possible line-up of the other 9 .

So the total total number of ways to do this is

   (362,880) x (10) = 3,628,800 ways.

If they sit a different way at every game, the class can see a bunch of games
without duplicating their seating arrangement !

Answer:

Step-by-step explanation:

362,880

The graph shows two lines, A and B:A graph is shown with x- and y-axes labeled from 0 to 6 at increments of 1. A straight line labeled A joins the ordered pair 3, 0 and the ordered pair 0, 6. Another straight line labeled B joins the ordered pair 0, 0 and the ordered pair 5, 5.

Based on the graph, which statement is correct about the solution to the system of equations for lines A and B?

A (0, 6) is the solution to both lines A and B.
B(0, 6) is the solution to line B but not to line
C(2, 2) is the solution to both lines A and B.
D (2, 2) is the solution to line A but not to line

Answers

The correct answer for the question that is being presented above is this one: "A (0, 6) is the solution to both lines A and B." The statement that is correct about the solution to the system of equations for lines A and B is that (0, 6) is the solution to both lines A and B.

Answer:

C(2, 2) is the solution to both lines A and B.

Step-by-step explanation:

Line A is given as:

A straight line labeled A joins the ordered pair 3, 0 and the ordered pair 0, 6.

We know that the equation of a line passing through (a,b) and (c,d) is calculated as:

$y-b=(d-b)/(c-a)* (x-a)$