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soldier1979 [14.2K]
2 years ago
15

True or False?In Exercise,determine whether the statement is true or false given that f(x) = In x.if it is false,explain why or

given an example that shows it is false.
f(0) = 0
Mathematics
1 answer:
EastWind [94]2 years ago
7 0

Answer:

False

Step-by-step explanation:

This is false. We can disprove this by plugging in the number into the function f(x) = lnx

f(0) = ln0

Since e^{- \infty} = 0, ln0 = -\infty

This is not 0 so f(0) cannot be equal to 0, thus disproving the statement.

You might be interested in
Which ordered pairs make the open sentence true? 3x+y<14 Select all the correct answers. (7, −7) (−1, 6) (3, 6) (4, −1) (6, 0
Jet001 [13]

Answer:

(4, - 1)

(-1, 6)

Step-by-step explanation:

Given the expression :

3x+y<14

Test the opprions to find out the true statement.

(7, - 7)

3(7) + (-7) < 14

21 - 7 < 14

15 < 14 (not true)

(-1, 6)

3(-1) + 6 < 14

3 + 6 < 14

9 < 14 ( true)

(3, 6)

3(3) + 6 < 14

9 + 6 < 14 ( not true)

(4, - 1)

3(4) - 1 < 14

12 - 1 < 14 ( true)

(6,0)

3(6) + 0 < 14

18 < 14 ( not true)

4 0
2 years ago
2 Here are two equations:
MatroZZZ [7]

a. (3,4) is only the solution to Equation 1.

(4, 2.5) is the solution to both equations

(5,5) is the solution to Equation 2

(3,2) is not the solution to any equation.

b. No, it is not possible to have more than one (x,y) pair that is solution to both equations

Step-by-step explanation:

a. Decide whether each neither of the equations,

i (3,4)

ii. (4,2.5)

ill. (5,5)

iv. (3,2)

To decide whether each point is solution to equations or not we will put the point in the equations

Equations are:

Equation 1: 6x + 4y = 34

Equation 2: 5x – 2y = 15

<u>i (3,4) </u>

Putting in Equation 1:

6(3) + 4(4) = 34\\18+16=34\\34=34\\

Putting in Equation 2:

5(3) - 2(4) = 15\\15-8 = 15\\7\neq 15

<u>ii. (4,2.5)</u>

Putting in Equation 1:

6(4) + 4(2.5) = 34\\24+10=34\\34=34\\

Putting in Equation 2:

5(4) - 2(2.5) = 15\\20-5 = 15\\15=15

<u>ill. (5,5)</u>

6(5) + 4(5) = 34\\30+20=34\\50\neq 34

Putting in Equation 2:

5(5) - 2(5) = 15\\25-10 = 15\\15=15

<u>iv. (3,2)</u>

6(3) + 4(2) = 34\\18+8=34\\26\neq 34

Putting in Equation 2:

5(3) - 2(2) = 15\\15-4 = 15\\11\neq 15

Hence,

(3,4) is only the solution to Equation 1.

(4, 2.5) is the solution to both equations

(5,5) is the solution to Equation 2

(3,2) is not the solution to any equation.

b. Is it possible to have more than one (x, y) pair that is a solution to both

equations?

The simultaneous linear equations' solution is the point on which the lines intersect. Two lines can intersect only on one point. So a linear system cannot have more than one point as a solution

So,

a. (3,4) is only the solution to Equation 1.

(4, 2.5) is the solution to both equations

(5,5) is the solution to Equation 2

(3,2) is not the solution to any equation.

b. No, it is not possible to have more than one (x,y) pair that is solution to both equations

Keywords: Linear equations, Ordered pairs

Learn more about linear equations at:

  • brainly.com/question/10534381
  • brainly.com/question/10538663

#LearnwithBrainly

8 0
3 years ago
It takes Dimitri 9 minutes to make a simple bracelet and 20 minutes to make a deluxe bracelet. He has been making bracelets for
frosja888 [35]
Last one 7 and 3
9(7)+20(3)>120
63+ 60
123>120
Says 123 minutes is longer than 120 that's why it is that answer
7 0
2 years ago
Read 2 more answers
Need helps asap, im tryna get all of my work done today. So someone answer this fastt
xeze [42]

(4,-5) is Point C

(-5,-4) is point H

(1,-1) is point G

(3,4) is point P

(5,0) is point Z

3 0
3 years ago
Read 2 more answers
Consider the following system of equations:
Delicious77 [7]

Answer:

The system has infinitely solutions

Step-by-step explanation:

we have

-\frac{1}{3}x^{2}=-\frac{5}{6}+\frac{1}{3}y^{2}

\frac{1}{3}x^{2}+\frac{1}{3}y^{2}=\frac{5}{6}

Multiply by 3 both sides

x^{2}+y^{2}=\frac{5}{2} ----> equation A

The equation A is a circle centered at origin with radius r=\sqrt{5/2}\ units

and

5y^{2} =\frac{25}{2}-5x^{2}

5x^{2}+5y^{2} =\frac{25}{2}

Divide by 5 both sides

x^{2}+y^{2} =\frac{5}{2} ----> equation B

The equation B is a circle centered at origin with radius r=\sqrt{5/2}\ units

Equation A and Equation B are the same

Therefore

The system has infinitely solutions

7 0
3 years ago
Read 2 more answers
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