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makvit [3.9K]
4 years ago
10

I WILL GIVE BRAINLIEST TO WHOEVER IS CORRECT

Mathematics
2 answers:
Vadim26 [7]4 years ago
7 0
It's (-2,3), 2nd choice is yours to make.
KatRina [158]4 years ago
3 0
-2, 3 is the correct answer
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PLZ HELP I WILL MARK BRAINLIEST <br> Use the given values to complete each table
Ne4ueva [31]
Ok this is a pretty easy one.
2=22
3=20
4=18
6 0
3 years ago
Read 2 more answers
Which coordinate pair is outside the circle and inside the rectangle as shown below? 10 9 8 o in N 6 5 2 0 1 2 3 4 5 6 7 8 9 10
docker41 [41]
It’s the 1st one 5,5
5 0
2 years ago
the coordinates of one point of a line segment are (-2,-7). The line segment is 12 units long. Give three possible coordinates o
valentinak56 [21]

(-2,-19) , (-2, 5), ( 2,10)

4 0
3 years ago
Rewrite the expression in standard form ( use the fewest number of symbols and characters possiable). a. 5g•7h b.3•4•5•m•n
CaHeK987 [17]

Answer:

a. 35 *gh or 35 gh

b. 60 *mn or 60 mn

Step-by-step explanation:

The fewest number of symbols and characters can be obtained by multiplying the constant numbers with the constant numbers and variables with the variables.

a. The co efficients 5 and 7 are multiplied to get 35 and g is multiplied with h to get gh. So the answer is 35*gh. It can be further simpliefied into 35 gh

b. The co efficients 3,4 and 5 are multiplied to get 60 and m is multiplied with n to get mn. The answer is 60* mn. On further simplificaation we get 60 mn.

6 0
3 years ago
The point P(1,1/2) lies on the curve y=x/(1+x). (a) If Q is the point (x,x/(1+x)), find the slope of the secant line PQ correct
lukranit [14]

Answer:

See explanation

Step-by-step explanation:

You are given the equation of the curve

y=\dfrac{x}{1+x}

Point P\left(1,\dfrac{1}{2}\right) lies on the curve.

Point Q\left(x,\dfrac{x}{1+x}\right) is an arbitrary point on the curve.

The slope of the secant line PQ is

\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{\frac{x}{1+x}-\frac{1}{2}}{x-1}=\dfrac{\frac{2x-(1+x)}{2(x+1)}}{x-1}=\dfrac{\frac{2x-1-x}{2(x+1)}}{x-1}=\\ \\=\dfrac{\frac{x-1}{2(x+1)}}{x-1}=\dfrac{x-1}{2(x+1)}\cdot \dfrac{1}{x-1}=\dfrac{1}{2(x+1)}\ [\text{When}\ x\neq 1]

1. If x=0.5, then the slope is

\dfrac{1}{2(0.5+1)}=\dfrac{1}{3}\approx 0.3333

2. If x=0.9, then the slope is

\dfrac{1}{2(0.9+1)}=\dfrac{1}{3.8}\approx 0.2632

3. If x=0.99, then the slope is

\dfrac{1}{2(0.99+1)}=\dfrac{1}{3.98}\approx 0.2513

4. If x=0.999, then the slope is

\dfrac{1}{2(0.999+1)}=\dfrac{1}{3.998}\approx 0.2501

5. If x=1.5, then the slope is

\dfrac{1}{2(1.5+1)}=\dfrac{1}{5}\approx 0.2

6. If x=1.1, then the slope is

\dfrac{1}{2(1.1+1)}=\dfrac{1}{4.2}\approx 0.2381

7. If x=1.01, then the slope is

\dfrac{1}{2(1.01+1)}=\dfrac{1}{4.02}\approx 0.2488

8. If x=1.001, then the slope is

\dfrac{1}{2(1.001+1)}=\dfrac{1}{4.002}\approx 0.2499

7 0
3 years ago
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