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DedPeter [7]
3 years ago
14

I’m on my homework and I can’t figure out the answer

Mathematics
1 answer:
Levart [38]3 years ago
3 0
The answer would be the 4th one, if we consider x and HD video, and if we consider y a classic video. then it would be 5x + 2y = 31, 3x + 5y = 30! Hope this helped! Brainliest is always appreciated.
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If a = 1, find the values of b, c and d that make the given expression equivalent to the expression below.
vlada-n [284]

Answer:

To do this, you need to multiply out the expressions. This is a bit tedious, but remember like FOIL for binomials, for these trinomials you must multiply each term. If you need a step-by-step, I'd be happy to provide it. Let me know.

Once you have simplified the expression, you get

-x-9/2x-4

But, the problem stipulates that a must equal 1. We can equivalently factor out the negative sign and put it on the denominator with no change to write

x+9/-(2x-4) = x+9/-2x+4

So, seeing where each coefficient corresponds between the two expressions, you get a = 1, b = 9, c = –2, and d = 4.

8 0
3 years ago
How many triangles can be formed
Nastasia [14]

Answer:

B. One

Step-by-step explanation:

Since the two angles are already given, the value of the third angle is already fixed. The third angle can be found by the sum of both angles subtracted from 180. Since 2 angles and lengths are already given, there can only be 1 triangle.

Hope this helps!

If not, I am sorry.

6 0
1 year ago
Determine the open t-intervals on which the curve is concave downward or concave upward. (Enter your answer using interval notat
Rashid [163]

Solution :

We have been given a parametric curve :

x = sin t , y = cos t , 0 < t < π

In order to determine concavity of the given parametric curve, we need to evaluate its second derivative first.

Therefore,

$\frac{dx}{dt} = \cos t, \ \frac{dy}{dt}= - \sin t$

$\therefore \frac{dy}{dx}= \frac{- \sin t}{\cos t}$

        $=- \tan t$

Taking double derivatives of the above equation:

$\frac{d^2y}{dx^2}= - \frac{d}{dx}(\tan t) $

      $= - \sec^2 t \frac{dt}{dx}$

     $= - \sec^2 t \left(\frac{1}{\cos t}\right)$

    $= - \sec^3 t$

For the concave up, we have

$\frac{d^2y}{dx^2} > 0$

$\Rightarrow - \sec^3 t > 0$

∴ $t \  \epsilon \left( \frac{\pi }{2}, \pi \right)$

For the concave down, we have

$\frac{d^2y}{dx^2} < 0$

$\Rightarrow - \sec^3 t < 0$

$t \  \epsilon \left( 0,\frac{\pi }{2} \right)$

8 0
3 years ago
I need help number 13 and 14, how can I find x?
Alex73 [517]
#13.
x = 50

#14
x = 142
7 0
3 years ago
Do this only on number line ​
Tatiana [17]

#4

  • -2/8=-1/4
  • 3/6=1/2

Refer to the attachment

#2

Mid point:-

  • -2/5+1/2/2
  • -4+5/10/2
  • 1/10/2
  • 1/20

4 0
2 years ago
Read 2 more answers
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