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sasho [114]
3 years ago
14

What’s the answers to math problems I have lots of troubles in math but the teach won’t help me ?

Mathematics
1 answer:
leva [86]3 years ago
3 0

If your teacher isn’t teaching you I suggest you talk to a higher ranked staff member instead.

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Felix said quadrilateral ABCD is a parallelogram because<br> BE = DE. What was his error?
Kamila [148]

Answer:

There is no angle E in the quadilateral presented in the question. Also, the angles he used are not on opposite side.

Step-by-step explanation:

To classify a quadrilateral is a parallelogram, a quadrilateral with both pairs of opposite sides must be parallel or congruent, that is in a quadrilateral ABCD, angle AB must equal CD and angle AD must equal BC.

Hence, when Felix said quadrilateral ABCD is a parallelogram because of BE = DE. His error was that "there is no angle E in the quadrilateral presented in the question and the angle he used are not on opposite side"

7 0
3 years ago
Divide 5/6 by negative 2/3
ohaa [14]

Answer:

-1.25 :)

Step-by-step explanation:

8 0
3 years ago
I'm stuck on this part of my pretest, I jut need to know where the correct angles go to.
Zolol [24]

Refer to the attachment :)

6 0
3 years ago
Here's a graph of a linear function. Write the equation that describes that function
kifflom [539]
Y = 2/3x + 3 is the answer
8 0
3 years ago
I know you’re supposed to change the bounds and break up the integral, but for some reason, I can’t get the 44/3. Can someone ex
tatyana61 [14]

First, look for the zeroes of the integrand in the interval [0, 6] :

x² - 6x + 8 = (x - 4) (x - 2) = 0   ⇒   x = 2   and   x = 4

Next, split up [0, 6] into sub-intervals starting at the zeroes we found. Then check the sign of x² - 6x + 8 for some test points in each sub-interval.

• For x in (0, 2), take x = 1. Then

x² - 6x + 8 = 1² - 6•1 + 8 = 3 > 0

so x² - 6x + 8 > 0 over this sub-interval.

• For x in (2, 4), take x = 3. Then

x² - 6x + 8 = 3² - 6•3 + 8 = -1 < 0

so x² - 6x + 8 < 0 over this sub-interval.

• For x in (4, 6), take x = 5. Then

x² - 6x + 8 = 5² - 6•5 + 8 = 3 > 0

so x² - 6x + 8 > 0 over this sub-interval.

Next, recall the definition of absolute value:

|x| = \begin{cases}x & \text{for }x \ge0 \\ -x & \text{for }x < 0\end{cases}

Then from our previous analysis, this definition tells us that

|x^2 - 6x + 8| = \begin{cases}x^2 - 6x + 8 & \text{for }0

So, in the integral, we have

\displaystyle \int_0^6 |x^2-6x+8| \, dx = \left\{\int_0^2 - \int_2^4 + \int_4^6\right\} (x^2 - 6x + 8) \, dx

Then

\displaystyle \int_0^2 (x^2 - 6x + 8) \, dx = \left(\frac13 x^3 - 3x^2 + 8x\right) \bigg|_0^2 = \frac{20}3 - 0 = \frac{20}3

\displaystyle \int_2^4 (x^2 - 6x + 8) \, dx = \left(\frac13 x^3 - 3x^2 + 8x\right) \bigg|_2^4 = \frac{16}3 - \frac{20}3 = -\frac43

\displaystyle \int_4^6 (x^2 - 6x + 8) \, dx = \left(\frac13 x^3 - 3x^2 + 8x\right) \bigg|_4^6 = 12 - \frac{16}3 = \frac{20}3

and the overall integral would be

20/3 - (-4/3) + 20/3 = 44/3

3 0
3 years ago
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