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Makovka662 [10]
3 years ago
9

Substitution for y=3x-34 and y=2x-5

Mathematics
1 answer:
Romashka-Z-Leto [24]3 years ago
3 0
3x-34 = 2x -5
3x-2x = 34 -5
x = 29
plug the 29 back into the second equation ( or the first , your choice )
y = 2 (29) -5 = 58-5= 53
(29, 53)
I think u might need to show the work so here you go
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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
Suppose that Y and Z are points on the number line.
Cerrena [4.2K]

Answer:

16-17 is the answer

although I am not sure

6 0
3 years ago
Translate and solve: Three more than the difference of 6s and 5s is equal to the difference of 1/12 and 1/6
PilotLPTM [1.2K]

Answer:

a. (6s-5s)+3=\frac{1}{12}-\frac{1}{6}

b. s=-\frac{37}{12}

Step-by-step explanation:

You need to remember that:

1. The word "more" indicates Addition.

2. The "Difference" is the result of a Subtraction.

a. Then, knowing this information, you know that:

- "The difference of 6s and 5s" must be expressed as:

6s-5s

- Then, "Three more than the difference of 6s and 5s " is:

(6s-5s)+3

- "The difference of  \frac{1}{12} and \frac{1}{6} " can be expressed in this form:

\frac{1}{12}-\frac{1}{6}

Therefore, "Three more than the difference of 6s and 5s is equal to the difference of  \frac{1}{12} and \frac{1}{6} " is:

(6s-5s)+3=\frac{1}{12}-\frac{1}{6}

b. Solve for "s" in order to find its value:

s+3=-\frac{1}{12}\\\\12(s+3)=-1\\\\12s+36=-1\\\\12s=-1-36\\\\s=-\frac{37}{12}

8 0
3 years ago
Find the exact value of x
Galina-37 [17]

Answer:

8.94 (Rounded would be 9)

Step-by-step explanation:

By using the Pythagorean Theorem formula,

A^2+B^2 = C^2

You have the hypotenuse (c)  = 12

and you have a leg (a) = 8

and you're trying to find another leg (b) = x

In order to find this, you'll have to change up the formula in order to find (B) instead of (C)

B^2 = C^2 - A^2

(B) = (12)^2 - (8)^2

(B) = 144 - 64

(B) = 80

Square root of 80

B = 8.9

5 0
2 years ago
Read 2 more answers
Use the points (2,-5) and (-1,4) and determine the slope using the slope formula
Rama09 [41]
(2,-5),(-1,4)
slope = (4 - (-5) / (-1 - 2) = (4 + 5) / -3 = -9/3 = -3 <==
6 0
3 years ago
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