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2 years ago

What is the area of a square that has 2/4 units on each side

Mathematics

1 answer:

soldi70 [24.7K]2 years ago

3 0

Answer:

1/4 of 0.25

Step-by-step explanation:

2/4 x 2/4 = 1/4

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Find the second derivative y= x^2lnx

Dimas [21]

First we use product rule
y=x^2lnx
dy/dx = x^2 d/dx (lnx) + lnx d/dx (x^2)
dy/dx = x^2 (1/x) + lnx (2x)
dy/dx = x + 2xlnx

now taking second derivative:
d2y/dx2 = 1 + 2[x (1/x) + lnx (1)]
d2y/dx2 = 1 + 2[1+lnx]
1+2+2lnx
3+2lnx is the answer

3 0

3 years ago

Solve the following pairs of simultaneous equations.

Sergeeva-Olga [200]

Answer and step-by-step explanation:

a) -4x + 5y = -9 (2)

2x + 3y = 21

Multiply the second equation by 2, you will get:

4x + 6y = 42 (1)

Use (1) plus with (2), you will get:

4x + (-4x) + 6y + 5y = -9 + 42

11y = 33

y = 33/11

y = 3

4x + 18 = 42

4x = 24

x = 6

b) 2x + 5y = -6

3x + 2y = 2

Multiply first equation by 3, you will get:

6x + 15y = -18 (1)

Multiply the second equation by 2 and you will get:

6x + 4y = 4 (2)

Subtract (1) by (2), we got:

6x -6x + 15y - 4y = -18 - 4

11y = -22

y = -22/11

y = -2

6x + 4y = 4

6x -8 = 4

6x = 12

x = 2

c) 2x - 2y = -4

5x + 4y = 17 (2)

Multiply the first euation by 2, we got:

4x - 4y = -4 (3)

Use (3) minus for (2), we get:

4x - 5x -4y + 4y = -4 - 17

-x = -21

x = 21

4x - 4y = -4

84 - 4y = -4

-4y = -4 + 84

-4y = 80

y = 80 / -4

y = -20

d) 2x - 3y = 6

4x - 5y = 9 (1)

Multiply the first equation by 2, we got:

4x - 6y = 12 (2)

Use (1) minus for (2), we got:

4x -4x - 5y - 6y = 9 -12

-11y = -3

y = -3/-11

y = 3/11

2x - 3y = 6

2x - 9/11 = 6

2x = 66/11 + 9/ 11

2x = 75/11

x = 75/11 * 1/2

x = 75 / 22

e) x/2 + y/3 = 2

x/4 + y/3 = 4

Use the first equation subtract the second equation, we got:

x/2 - x/4 + y/3 - y/3 = 2 - 4

2x/4 -x/4 = -2

x/ 4 = -2

x = -2 * 4

x = -8

x/2 + y/3 = 2

-8 / 2 + y/3 = 2

y / 3 = 2 + 4

y/3 = 6

y = 6*3

y = 18

f) x/3 + y/2 = 3/2 (1)

x/2 + y/5 = -1/2 (2)

I don't know how to solve this, sr

Hope this helped :3

3 0

3 years ago

Find the area of the shape below.

m_a_m_a [10]

Answer:

$A_{shape}=A_{triangle}+A_{square}\\\\A_{shape}=(\frac{bh}{2}+s^2)\ units^2$

Step-by-step explanation:

I assume that it is a composed figure formed by a triangle and a square.Since you did not give the dimensions of the triangle and the dimensions of the square, I will giveyou a general explanation about how to solve this exercise.

The area of the shape will be the sum of the the area of the square and the area of the triangle.

Then, you need to remember that the area of a triangle can be calculated with the following formula:

$A_{triangle}=\frac{bh}{2}$

Where "b" is the base and "h" is the height.

And the area of a square can be found with this formula:

$A_{square}=s^2$

Where "s" is the length of any side of the square.

Therefore, the area of the shape will be

$A_{shape}=A_{triangle}+A_{square}\\\\A_{shape}=(\frac{bh}{2}+s^2)\ units^2$

3 0

3 years ago

5th grade Pre algebra. Pls help I’m rlly stuck will give brainliest c:

anzhelika [568]

Step-by-step explanation:

so ,

Question 6 :-

5 × (3+4) - 4

first solve the Bracket ......

5 × (7) - 4

after multiplication ......

35 - 4

= 31

so , Answer is 1st one ....

Question 7 :-

8 + 1/2 × (6-2) -1

so , here also solve the Bracket first .....

8 + 1/2 × (4) - 1

after multiplication .....

8 + 2 - 1

after + and - .......

10 - 1

= 9 .......

so , Answer is 2nd one .....

4 0

2 years ago

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2 years ago