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xenn [34]
3 years ago
7

In the first half of a basketball game, a player scored 9 points on free throws and then scored a number of 2-point shots. In th

e second half, the player scored the same number of 3-point shots as the number of 2-point shots scored in the first half. Which expression represents the total number of points the player scored in the game?
2x + 3x + 9
2x + 3 + 9
2x + 3x + 9x
2 + 3x + 9
Mathematics
2 answers:
Romashka-Z-Leto [24]3 years ago
8 0

Answer:

2x+3x+9

Step-by-step explanation:

In first half,

Let number of 2-point shots=x

Points scored on free throws=9

Point scored in x number of 2-point shots=2x

In second half,

Number of 3-point shots=number of 2-point shots in first half=x

Point scored in x number of 3-point shots=3x

Therefore, the player scored total number of points in the game is given by

2x+3x+9

Hence, the expression which represents the total number of points scored by player in the game is given by

2x+3x+9

irinina [24]3 years ago
6 0
I had this same question yesterday it was c 2x+3x+9x
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Write down the explicit solution for each of the following: a) x’=t–sin(t); x(0)=1
Kay [80]

Answer:

a) x=(t^2)/2+cos(t), b) x=2+3e^(-2t), c) x=(1/2)sin(2t)

Step-by-step explanation:

Let's solve by separating variables:

x'=\frac{dx}{dt}

a)  x’=t–sin(t),  x(0)=1

dx=(t-sint)dt

Apply integral both sides:

\int {} \, dx=\int {(t-sint)} \, dt\\\\x=\frac{t^2}{2}+cost +k

where k is a constant due to integration. With x(0)=1, substitute:

1=0+cos0+k\\\\1=1+k\\k=0

Finally:

x=\frac{t^2}{2} +cos(t)

b) x’+2x=4; x(0)=5

dx=(4-2x)dt\\\\\frac{dx}{4-2x}=dt \\\\\int {\frac{dx}{4-2x}}= \int {dt}\\

Completing the integral:

-\frac{1}{2} \int{\frac{(-2)dx}{4-2x}}= \int {dt}

Solving the operator:

-\frac{1}{2}ln(4-2x)=t+k

Using algebra, it becomes explicit:

x=2+ke^{-2t}

With x(0)=5, substitute:

5=2+ke^{-2(0)}=2+k(1)\\\\k=3

Finally:

x=2+3e^{-2t}

c) x’’+4x=0; x(0)=0; x’(0)=1

Let x=e^{mt} be the solution for the equation, then:

x'=me^{mt}\\x''=m^{2}e^{mt}

Substituting these equations in <em>c)</em>

m^{2}e^{mt}+4(e^{mt})=0\\\\m^{2}+4=0\\\\m^{2}=-4\\\\m=2i

This becomes the solution <em>m=α±βi</em> where <em>α=0</em> and <em>β=2</em>

x=e^{\alpha t}[Asin\beta t+Bcos\beta t]\\\\x=e^{0}[Asin((2)t)+Bcos((2)t)]\\\\x=Asin((2)t)+Bcos((2)t)

Where <em>A</em> and <em>B</em> are constants. With x(0)=0; x’(0)=1:

x=Asin(2t)+Bcos(2t)\\\\x'=2Acos(2t)-2Bsin(2t)\\\\0=Asin(2(0))+Bcos(2(0))\\\\0=0+B(1)\\\\B=0\\\\1=2Acos(2(0))\\\\1=2A\\\\A=\frac{1}{2}

Finally:

x=\frac{1}{2} sin(2t)

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3 years ago
Mum is 25 years older than her son Max, but she is 1 year younger than
In-s [12.5K]

Answer:

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Step-by-step explanation:

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3 years ago
A unicycle travels 50 feet. If the radius of the tire is 24 INCHES, how many times does the tire rotate? A) about 8 times B) abo
Pie
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6 0
3 years ago
Read 2 more answers
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zzz [600]

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(4^p)^3 would become 4^3p

(4^1)^3 would become 4^3

When the bases are the same, you just simply add the exponents.

Let's just ignore the bases so it'll look less confusing.

3p + 3 = 15

Subtract 3 from both sides.

3p = 12

Divide both sides by 3.

<em>p = 4</em>

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Answer:

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ans done

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2 years ago
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