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dlinn [17]
3 years ago
8

Help me!!!! Please!!!!! Math!!!!

Mathematics
1 answer:
Nitella [24]3 years ago
3 0
If f(x)=3/x+2-_/x-3

To find f(19),we would insert (19) as x into the equation:

f(19)=3/(19)+2-_/19-3
=3/21-_/16
=1/7-4
=-27/7

Hope it helps!
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4d = -16 <br><br> what’s the value of d
prohojiy [21]

4d =  - 16 \\  \\ d = \cancel  \frac{ - 16}{4}  \\  \\ d =  - 4

<h3>Hope This Helps You </h3>
7 0
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Please help!! I'm really confused
Readme [11.4K]

Answer:

Point a is correct.

Step-by-step explanation:

hope I helped.

8 0
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The graph 4x^2-4x-1 is shown. Use the grpah to find the estimates for the solutions of 4x^2-4x-1=0 and 4x^2 - 4x-1=2
Darina [25.2K]

Answer:

a) The estimates for the solutions of 4\cdot x^{2}-4\cdot x -1 = 0 are x_{1}\approx -0.25 and x_{2} \approx 1.25.

b) The estimates for the solutions of 4\cdot x^{2}-4\cdot x -1 = 2 are x_{1}\approx -0.5 and x_{2} \approx 1.5

Step-by-step explanation:

From image we get a graphical representation of the second-order polynomial y = 4\cdot x^{2}-4\cdot x -1, where x is related to the horizontal axis of the Cartesian plane, whereas y is related to the vertical axis of this plane. Now we proceed to estimate the solutions for each case:

a) 4\cdot x^{2}-4\cdot x -1 = 0

There are two approximate solutions according to the graph, which are marked by red circles in the image attached below:

x_{1}\approx -0.25, x_{2} \approx 1.25

b) 4\cdot x^{2}-4\cdot x -1 = 2

There are two approximate solutions according to the graph, which are marked by red circles in the image attached below:

x_{1}\approx -0.5, x_{2} \approx 1.5

5 0
3 years ago
GEOMETRY. PLEASE HELP. I'M STUMPED!!!
tatiyna
Given the triangles ABC and PQR.
Angle A = Angle P
Angle B = Angle Q
Angle C = Angle R

Angle B = 3v+4
Angle Q = 8v-6
Let's find v

3v+4 = 8v-6
-5v = -10
v = 2

Angle B = 3v + 4
Angle B = 3(2) + 4
Angle B = 10.

The correct answer is letter C. 10


8 0
3 years ago
Math homework need help with this one. Thank you
Sliva [168]
Answer:

At 25 weeks Quert will have more pibs than Gespil.
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