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sveticcg [70]
3 years ago
15

Can someone help me? The first person to get it correct gets brainliest.

Mathematics
1 answer:
Scilla [17]3 years ago
6 0
50 because 2(5)^2 equals to 2 (5x5) which equals 2(25) which is 50.
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Which expression is equivalent to 2cos^2(x/2)- cos(x)?<br> Edge 2020
olchik [2.2K]

Given:

The expression is

2\cos^2(\dfrac{x}{2})-\cos (x)

To find:

The expression which is equivalent to the given expression.

Solution:

We have,

2\cos^2(\dfrac{x}{2})-\cos (x)

We know that,

\cos x=2\cos^2 (\dfrac{x}{2})-1

Using this formula, the given expression can be written as

2\cos^2(\dfrac{x}{2})-\left(2\cos^2(\dfrac{x}{2})-1\right)

=2\cos^2(\dfrac{x}{2})-2\cos^2(\dfrac{x}{2})+1

=1

Therefore, the given expression is equivalent to 1.

4 0
3 years ago
Triangle EFG is similar to triangle HIJ. Find the measure of side JH. Round your
MAVERICK [17]

Answer:

JH = 97.9

Step-by-step explanation:

25/12 = x/47

12x = 1175

x = 97.917

8 0
3 years ago
Suppose segment PQ has one endpoint at P (0,0). If T (6,3) is a point 3/10 of the way from P to Q, find the coordinates of Q.
mixer [17]

Answer:

Q(20,10)

Step-by-step explanation:

If point T (6,3) is a point 3/10 of the way from P(0,0) to Q(x,y), then

\overrightarrow {PT}=\dfrac{3}{10}\overrightarrow {PQ}.

Find the coordinates of the vectors \overrightarrow {PT},\ \overrightarrow {PQ}:

\overrightarrow {PT}=(6-0,3-0)=(6,3);\\ \\\overrightarrow {PQ}=(x-0,y-0)=(x,y).

Thus,

(6,3)=\dfrac{3}{10}(x,y),\\ \\(x,y)=\dfrac{10}{3}(6,3)=(20,10).

3 0
4 years ago
A food server examines the amount of money earned in tips after working an 8-hour shift. The server has a total of $133 in denom
gogolik [260]

Answer:

x = 23 $1 bills

y = 12 $5 bills

z = 3 $10 bills

w = 1 $20 bill

Step-by-step explanation:

First of all we have to setup linear equations with the help of given information.

x = number of $1 bills

y = number of $5 bills

z = number of $10 bills  

w = number of $20 bills

The server has a total of $133 in denominations of $1, $5, $10, and $20 bills.

x + 5y + 10z + 20w = 133 eq. 1

The total number of paper bills is 39.

x + y + z + w = 39 eq. 2

The number of $5 bills is 4 times the number of $10 bills

y = 4z eq. 3

The number of $1 bills is 1 less than twice the number of $5 bills

x = 2y - 1 eq. 4

Now we have got 4 equations and four unknowns x, y, z, w

x + 5y + 10z + 20w = 133 eq. 1

x + y + z + w = 39 eq. 2

0 + y -4z + 0 = 0 eq. 3

x - 2y + 0 + 0 = -1 eq. 4

We have four equations and four unknowns x, y, z, w

We can solve this system of linear equations by various linear algebra methods like Crammer's rule, Gaussian Elimination etc. But if you notice eq. 3 and eq. 4 we have some zero entries in these equations and we might reduce the number of equations which will be easier and faster method to solve this system.

From eq. 3

0 + y -4z + 0 = 0

y = 4z

z = y/4

From eq. 4

x - 2y + 0 + 0 = -1

x = 2y - 1

Now put the values of x and z into eq. 1

x + 5y + 10z + 20w = 133

2y - 1 + 5y + 10(y/4) + 20w = 133

7y + (5/2)y + 20w = 133 + 1

9.5y + 20w = 134  eq. 5

Now put the values of x and z into eq. 2

x + y + z + w = 39

2y - 1 + y + y/4 + w = 39

3y + y/4 + w = 39 + 1

3.25y + w = 40  eq. 6

Now we have 2 equations ( eq. 5 and eq. 6) and 2 unknowns (y and w)

Solving these two equations simultaneously

9.5y + 20w = 134  eq. 5

3.25y + w = 40  eq. 6

Multiply eq. 6 by 20 and subtract it from eq. 5

9.5y + 20w = 134

65y + 20w = 800

-55.5y + 0 = -666

y = -666/-55.5 = 12

Put the value of y into eq. 5 to get the value of w

9.5(12) + 20w = 134

114 + 20w = 134

20w = 134 - 114

w = 20/20 = 1

Now we have got the value of y and we can substitute this value into eq. 3 and eq. 4 to get the values of x and z

From eq. 3

z = y/4 = 12/4 = 3

From eq. 4

x = 2y - 1 = 2(12) - 1 = 23

So we have solved the whole system of equations and got the following results

x = 23 $1 bills

y = 12 $5 bills

z = 3 $10 bills

w = 1 $20 bill

Verification:

Lets verify our results to check if we have got the correct numbers!

eq. 1  => x + 5y + 10z + 20w = 133

23 + 5(12) + 10(3) + 20(1) = 133

23 + 60 + 30 + 20 = 133

133 = 133 (hence satisfied)

eq. 2 => x + y + z + w = 39

23 + 12 + 3 + 1 = 39

39 = 39 (hence satisfied)

eq. 3 => 0 + y -4z + 0 = 0

0 + 12 -4(3) = 0

12 - 12 = 0

0 = 0 (hence satisfied)

eq. 4 => x - 2y + 0 + 0 = -1

23 - 2(12) + 0 + 0 = -1

23 - 24 = -1

-1 = -1 (hence satisfied)

5 0
4 years ago
(14) Geometry pls help<br><br> Check my ans<br><br> I really dont know on this one
mestny [16]
Close. It's actually choice D) 7 < n < 17
The value n must be less than 5+12 = 17. If its equal to 17 or larger, then a triangle isn't possible. 
3 0
4 years ago
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