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Serga [27]
3 years ago
5

Solve for p. 4p + 19 + 13p = –17 − p

Mathematics
2 answers:
puteri [66]3 years ago
8 0

<em>So</em><em> </em><em>the</em><em> </em><em>va</em><em>lue</em><em> </em><em>of</em><em> </em><em>p</em><em> </em><em>is</em><em> </em><em>-</em><em>2</em><em>.</em>

<em>Look</em><em> </em><em>at</em><em> </em><em>the</em><em> </em><em>attached</em><em> </em><em>picture</em>

<em>Hope</em><em> </em><em>it</em><em> </em><em>will</em><em> </em><em>help</em><em> </em><em>you</em>

<em>Good</em><em> </em><em>luck</em><em> </em><em>on</em><em> </em><em>your</em><em> </em><em>assignment</em>

Marina CMI [18]3 years ago
7 0

Answer:

p=-2

Step-by-step explanation:

4p + 19 + 13p = –17 − p

4p+13p=17p

17p + 19 = -17 - p

18p=-36

p=-2

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For what value of k will the lines x+2y=0, 3x-4y-10=0 and 5x+ky-7=0 are concurrent?​
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After solve the equations we get value of k=3

Step-by-step explanation:

We need to find value of k for which the lines x+2y=0, 3x-4y-10=0 and 5x+ky-7=0 are concurrent.

If the lines are concurrent, they pass through same point.

Let:

x+2y=0--eq(1)\\ 3x-4y-10=0--eq(2)\\ 5x+ky-7=0--eq(3)

First solving equation 1 and 2 to find values of x and y

From eq(1) we find value of x and put it in eq(2)

From \ eq(1) x+2y=0\\x=-2y\\Put x=-2y \ in \ eq(2)\\3x-4y-10=0\\3(-2y)-4y-10=0 \\-6y-4y=10\\-10y=10\\y=\frac{10}{-10}\\y=-1

After solving we get value of y=-1

Now putting in eq(1) to get value of x

x+2y=0\\x+2(-1)=0\\x-2=0\\x=2

So, Value of x= 2

Now put value of x=2 and y=-1 into eq(3) to find value of k

5x+ky-7=0\\5(2)+k(-1)-7=0\\10-k-7=0\\-k+3=0\\-k=-3\\k=3

So, After solve the equations we get value of k=3

4 0
3 years ago
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