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sergey [27]
3 years ago
10

The measure of a side of a square is x units. A new square is formed with each side 6 units longer than the original squares sid

e. Write an expression to represent the area of the new square.
Mathematics
1 answer:
xenn [34]3 years ago
4 0
We know that
[area old square]=x*x--------> x²

[area new square]=(x+6)*(x+6)-------> (x+6)²-----> A=x²+12x+36

the answer is  
An <span>expression to represent the area of the new square is (</span>x²+12x+36)
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How do I write the equation 9x-6y=-72 in slope-intercept form?
inna [77]
Y=mx+b 72+9x=6y 12+9x/6=y 12 I is the y intercept and 9/6 is the slope. 9 being the rise or y value and 6 being the run or x value.
6 0
3 years ago
Determine which graph shows Y is a function of x PLEASE HELP ASAPPP ​
ANEK [815]

Answer: The Horizontal line at 5

Step-by-step explanation: For Y to be a function, there cannot be different points on the same x value. The horizontal line is the only one that has one y value per x value.

Hope this helps!

6 0
3 years ago
Read 2 more answers
3 ≤ 3 <br> 3 is less than or equal to 3.<br> Is this disjunction true or false?
solniwko [45]

Answer:

true....... but only because it can be Lee's than OR equal to...

Step-by-step explanation:

brainliest pls! :)

5 0
3 years ago
In baseball, a player pitches a ball from the mound to a catcher behind the plate. A pitch that passes over the plate above the
matrenka [14]

To solve this question, you just need to count all the probability of the options.

The probability that a pitch not over the plate is a strike is zero. So, P(A | D) = 0.

True. It is 0/0+20= 0

The probability that a pitch not over the plate is a ball is 1. So, P(B | D) = 1.

True, it is 20/20+0= 1

The probability that a pitch over the plate is a strike is 10:15. So, ...

Incomplete but it sounds to be true. It should be 10/10+5= 10/15 = 2/3

The probability that a pitch over the plate is a ball is 5:10. So, P(B | C) = 0.5.

7 0
4 years ago
Read 2 more answers
Que is on pic.i can't able to type in text.
ad-work [718]
It's not difficult to compute the values of A and B directly:

A=\displaystyle\int_1^{\sin\theta}\frac{\mathrm dt}{1+t^2}=\tan^{-1}t\bigg|_{t=1}^{t=\sin\theta}
A=\tan^{-1}(\sin\theta)-\dfrac\pi4

B=\displaystyle\int_1^{\csc\theta}\frac{\mathrm dt}{t(1+t^2)}=\int_1^{\csc\theta}\left(\frac1t-\frac t{1+t^2}\right)\,\mathrm dt
B=\left(\ln|t|-\dfrac12\ln|1+t^2|\right)\bigg|_{t=1}^{t=\csc\theta}
B=\ln\left|\dfrac{\csc\theta}{\sqrt{1+\csc^2\theta}}\right|+\dfrac12\ln2

Let's assume 0, so that |\csc\theta|=\csc\theta.

Now,

\Delta=\begin{vmatrix}A&A^2&B\\e^{A+B}&B^2&-1\\1&A^2+B^2&-1\end{vmatrix}
\Delta=A\begin{vmatrix}B^2&-1\\A^2+B^2&-1\end{vmatrix}-e^{A+B}\begin{vmatrix}A^2&B\\A^2+B^2&-1\end{vmatrix}+\begin{vmatrix}A^2&B\\B^2&-1\end{vmatrix}
\Delta=A(-B^2+A^2+B^2)-e^{A+B}(-A^2-A^2B-B^3)+(-A^2-B^3)
\Delta=A^3-A^2-B^3+e^{A+B}(A^2+A^2B+B^3)

There doesn't seem to be anything interesting about this result... But all that's left to do is plug in A and B.
3 0
3 years ago
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