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romanna [79]
3 years ago
6

Help plzzzzz it's a 9th grade question ​

Mathematics
1 answer:
Semenov [28]3 years ago
4 0
X is equal to 7 and y is equal to 63
you get this by equaling the two y= equations to each other so 9x=2x+49 then you would solve algebraically to get x so 7x=49.
X=7
Then you would go back to the two original y= equations and place the value of x into the equation to find the value of y so
Y=9(7)
Y=63 and if you want to check it to make sure its right you can place it into the other y= equation and the values should be the same and in this case they are so the values are correct .
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Which of the following is the total number of cubic units needed to fill a space figure​
attashe74 [19]

Answer:

Volume is the required answer.

Step-by-step explanation:

The number of cubic units needed to fill the space inside a three-dimensional figure.

For example:- air in room, fluids in container, water in a bottle, etc.

I got this answer with complete explanation from Gauthmath app. They have the tutors who can answer our questions. In my opinion it's a very good app to get help. You don't even need to wait. I think you should check it out once.

<em>Hope </em><em>my </em><em>answer </em><em>helps.</em><em> </em><em>Best </em><em>of </em><em>luck!</em><em> </em><em />

8 0
3 years ago
Need help use screenshot(SAT Prep) In △ABC, AB = BC = 20 and DE ≈ 9.28. Approximate BD. bd equals?
Lilit [14]

Answer:

5.36

Step-by-step explanation:

Given that:

<BAD = <CAE, therefore, BD = EC

Let's take x to be the length of BD = EC

BD + DE + EC = BC

BC = 20,

BD = EC = x

DE ≈ 9.28

Thus,

x + 9.28 + x = 20

x + x + 9.28 = 20

2x + 9.28 = 20

Subtract 9.28 from both sides

2x + 9.28 - 9.28 = 20 - 9.28

2x = 10.72

Divided both sides by 2 to solve for x

\frac{2x}{2} = \frac{10.72}{2}

x = \frac{10.72}{2}

x = 5.36

BD ≈ 5.36

4 0
4 years ago
Jada bikes 2 miles in 12 minutes.
drek231 [11]
A 1/6 miles per minute
6 0
3 years ago
Read 2 more answers
Prove that if a and b are nonzero integers, a divides b, and a b is odd, then a is odd.
hammer [34]

It is proved that the if a and b are nonzero integers, a divides b, and a b is odd, then a is odd.

According to the statement

we have to prove that the if a and b are nonzero integers, a divides b, and a b is odd, then a is odd.

And for this proof we use the contradiction

So,

Proof by contradiction is a common proof technique that is based on a very simple principle: something that leads to a contradiction can not be true, and if so, the opposite must be true.

So for this purpose,

Assume a is even, so a = 2k for some integer k. Now let a and b be integers such that a divides b and a + b is odd.

Since a divides b, b = an for integer n, and in turn b = 2nk, which means b is even and hence a + b is also even. But this contradicts our initial assumption, so a must be odd.

So, It is proved that the if a and b are nonzero integers, a divides b, and a b is odd, then a is odd.

Learn more about integers here

brainly.com/question/17695139

#SPJ4

7 0
2 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
4 years ago
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