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Fed [463]
3 years ago
10

Morton made 42 out of 56 free throws last season.

Mathematics
1 answer:
3241004551 [841]3 years ago
3 0

Answer:

90

Step-by-step explanation:

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Solve the system by substitution.<br><br> 6x + 3y = -6<br> 2x + y = -2
Mariana [72]
Solutions 

First Lets solve for this equation 6x + 3y = -6&#10; 

6x + 3y = -6&#10; 

6x+3(-2-2x)=-6&#10;&#10; &#10; 

-6=-6 

Therefore this given equation is a <span>dependent system. 
</span>
<span>Solve for </span>2x + y = -2 

2x+y=-2 

y=-2-2x&#10;&#10; &#10;&#10; &#10; 

There are too many solutions to these equations. 


8 0
3 years ago
Read 2 more answers
B is the midpoint of ac. bc = 3x - 1. ac = x + 3. find the value of ab
Ilia_Sergeevich [38]

Answer:

ab=bc because b devide ac in two equal sizes

ac=ab+bc

ac=2×ab

ac =2×bc

lets substitute ac and bc there we are given

x+3=2(3x-1)

x+3=6x-2

-5x/-5=-5/-5

x=1

ab=3(1)-1

ab=2

8 0
1 year ago
On sale , the original price of every jacket in a store is reduced by 20% . Later , this sale price is further reduced by anothe
dimulka [17.4K]

Answer:

40%

Step-by-step explanation:

20% + 20% = 40%

Hope this helps!

4 0
3 years ago
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)

7 0
3 years ago
15 miles in 20 minutes what is the speed in miles per hour
Readme [11.4K]

15 / 20 = 0.75. The answer is 0.75

8 0
3 years ago
Read 2 more answers
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