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never [62]
2 years ago
14

I could use some help answering this math problem!!

Mathematics
1 answer:
Gnom [1K]2 years ago
3 0

Answer:

b

Step-by-step explanation:

2x is the same as 2 times x and if x represents the coaches uniform and the uniform is 2 times the cost of the players then the answer is b 2x

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The Human Resources manager at a company records the length
luda_lava [24]

Answer:

the correct answer is 0.84

give brainliest plzzz

6 0
3 years ago
FInd the GCF of the numbers using prime factorizations.
lesya [120]

Answer:

GCF(330, 75, 450, 225) = 15

Steps:

Prime factorization of the numbers:

330 = 2 × 3 × 5 × 11

75 = 3 × 5 × 5

450 = 2 × 3 × 3 × 5 × 5

225 = 3 × 3 × 5 × 5

GCF(330, 75, 450, 225)

= 3 × 5

= 15

3 0
3 years ago
I WILL GIVE BRAINLYIST IM BEING TIMED PLS HELP ASAP PLS PLS PLS PLS I WILL FAIL
nasty-shy [4]

Answer:

.35 per minute. single variables.

4 0
3 years ago
Which statement is true about point F? a.F is the midpoint of AA' because bisects AA'. b.F is the midpoint of EG because AA' bis
Dimas [21]

Answer:

The correct option is;

a. F is the midpoint of \overline{AA'} because line \overline{EG} bisects \overline{AA'}

Step-by-step explanation:

Here, since we have that the triangle is reflected across EG therefore the location of the point F which is along EG bisects the line \overline{AA'} as the  dimensions of the line from A to F must be equal to the dimension of the line that extends from A' to F

Therefore the point F is the midpoint of \overline{AA'} because line \overline{EG} bisects \overline{AA'}.

7 0
3 years ago
Use the diagram to complete the statement. Triangle J K L is shown. Angle K J L is a right angle. Angle J K L is 52 degrees and
zzz [600]

Answer:

\bold{sin(38^\circ)=cos(52^\circ)}

Step-by-step explanation:

Given that \triangle KJL is a right angled triangle.

\angle JKL = 52^\circ\\\angle KLJ = 38^\circ

and

\angle KJL = 90^\circ

Kindly refer to the attached image of \triangle KJL in which all the given angles are shown.

To find:

sin(38°) = ?

a) cos(38°)

b) cos(52°)

c) tan(38°)

d) tan(52°)

Solution:

Let us use the trigonometric identities in the given \triangle KJL.

We have to find the value of sin(38°).

We know that sine trigonometric identity is given as:

sin\theta =\dfrac{Perpendicular}{Hypotenuse}

sin(\angle JLK) = \dfrac{JK}{KL}\\OR\\sin(38^\circ) = \dfrac{JK}{KL}....... (1)

Now, let us find out the values of trigonometric functions given in options one by one:

cos\theta =\dfrac{Base}{Hypotenuse}

cos(\angle JLK) = \dfrac{JL}{KL}\\OR\\cos(38^\circ) = \dfrac{JL}{KL}....... (2)

By (1) and (2):

sin(38°) \neq cos(38°).

cos(\angle JKL) = \dfrac{JK}{KL}\\OR\\cos(52^\circ) = \dfrac{JK}{KL} ...... (3)

Comparing equations (1) and (3):

we get the both are same.

\therefore \bold{sin(38^\circ)=cos(52^\circ)}

6 0
3 years ago
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