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GarryVolchara [31]
3 years ago
15

How do you Simplify this expression in algebra? 9x + 2x (5 + 4) - 2(5x + 4x)

Mathematics
1 answer:
Minchanka [31]3 years ago
3 0

One can solve this by first using order of operations, and then combining like terms.


First is parenthesis


9x + 2x(9) - 2(9x)


Now we do the multiplication


9x + 18x - 18x


Now we combine like terms


9x

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How many pennies could you have if:
Gelneren [198K]

Answer:

The number of pennies owned is 7 pennies

Step-by-step explanation:

The given parameters are;

The number of pennies left over when we break the pennies into groups of 2s = 1 penny

The number of pennies left over when we break the pennies into groups of 3s = 1 penny

Let the number of pennies owned = c

What we are given are as follows;

2 × a = c - 1

3 × b = c - 1

2 × a = 3 × b

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Therefore, if we multiply 2 by 3, and 3 by 2 we get 6

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3 years ago
If tan A = 4/3 and sin B = 45/53 and angles A and B are in Quadrant I, find the value of tan(A+B)
lbvjy [14]

Answer:

First, find tan A and tan B.

cosA=35 --> sin2A=1−925=1625 --> cosA=±45

cosA=45 because A is in Quadrant I

tanA=sinAcosA=(45)(53)=43.

sinB=513 --> cos2B=1−25169=144169 --> sinB=±1213.

sinB=1213 because B is in Quadrant I

tanB=sinBcosB=(513)(1312)=512

Apply the trig identity:

tan(A−B)=tanA−tanB1−tanA.tanB

tanA−tanB=43−512=1112

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3 0
2 years ago
Two children own two-way radios that have a maximum range of 3 miles. One leaves a certain point at 1:00 P.M., walking due north
Grace [21]

Answer:

The answer is 21 minutes

Step-by-step explanation:

We use the equation Xf = Xo + vt

1) At 1:00 PM, child one leaves the starting point heading north at a constant velocity of 6 mi/hr or .1 [mi/min] (divide by 60 to convert from [mi/hr] to [mi/min])

2) He walks for 15 minutes before kid 2 starts walking. In 15 minutes he is able to cover 1.5 [mi]

  • x_{1f1} =x_{o} +v_{1} t_{1} \\x_{1f1} =0+.1(15)\\x_{1f1} =1.5 [mi]

3) Now, child 2 starts walking and we know that when the range reaches 3 miles, they won´t be able to communicate. So the sum of the final position of child 1 and child 2 must be 3[mi]

  • Child 1 final position => x_{1f} = x_{1f1} +v_{1} t\\x_{1f} =1.5+v_{1} t
  • Child 2 final position => x_{2f} =0+v_{2} t

4) Sum the equations and equate to 3

  • x_{1f} +x_{2f} =3

5) Substitute the values we already know

  • 1.5+v_{1} t+v_{2}t=3\\ 1.5+.1t+.15t=3\\1.5+.25t=3\\t=\frac{3-1.5}{.25} \\t=6 [min]

6) in 15 + 6 minutes they will be 3miles apart

7) In 21 minutes they will still be able to communicate with one another.

7 0
3 years ago
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