All categories

3 years ago

What are the solution(s) to the quadratic equation 9x2 = 4?

1 answer:

3 0

Multiply 9 by 2

18 = 4

Then subtract 4 from both sides

18 - 4 = 0 will become

14 = 0

Since 14 ≠ 0, there are no solutions

You might be interested in

Simplify the expression by combining like terms. 16+8−3+6−9

lyudmila [28]

Answer=18
First step combine the positive like terms
16+8+6
|
|
\/
30
Second step is to combine negative like terms with 30
30-3-9

When you do this you add 9 and 3 together since they are similar to each other so it equals 12 there for the equation becomes.....
30-12

Final answer 18

4 0

3 years ago

william is thinking of two numbers both numbers are square numbers greater than 1 the sum of numbers is 100 write down two numbe

zheka24 [161]

Answer:

Step-by-step explanation:

At first,

Let start writing the squares of number 0 to 10,

0²=0

1²=1

2²=4

3²=9

4²=16

5²=25

6²=36

7²=49

8²=64

9²=81

10²=100

Now,

According to your question,

The two numbers should be square numbers
They should be greater than 1.
Their sum should be 100.

Hence, your given conditions matches with the numbers 64 and 36.

Therefore, the numbers are 64 and 36.

64 and 36 are greater than 1.
64 is a square of 8 and 36 is the square of 6.
64+36=100

3 0

3 years ago

I need help with reflections on a graph, I'll mark brainliest. Please help me.

kvv77 [185]

Answer: 1) (0,-3) 2) (4,-2) 3) (2,3) 4) (-2,2)

Step-by-step explanation: Hope this helps!

3 0

3 years ago

1. Write the form of the partial fraction decomposition of the rational expression. Do not solve for the constants.

german

Answer:

Step-by-step explanation:

To write the form of the partial fraction decomposition of the rational expression:

We have:

$\mathbf{\dfrac{8x-4}{x(x^2+1)^2}= \dfrac{A}{x}+\dfrac{Bx+C}{x^2+1}+\dfrac{Dx+E}{(x^2+1)^2}}$

Using partial fraction decomposition to find the definite integral of:

$\dfrac{2x^3-16x^2-39x+20}{x^2-8x-20}dx$

By using the long division method; we have:

$x^2-8x-20 | \dfrac{2x}{2x^3-16x^2-39x+20 }$

$- 2x^3 -16x^2-40x$

$x+ 20$

So;

$\dfrac{2x^3-16x^2-39x+20}{x^2-8x-20}= 2x+\dfrac{x+20}{x^2-8x-20}$

By using partial fraction decomposition:

$\dfrac{x+20}{(x-10)(x+2)}= \dfrac{A}{x-10}+\dfrac{B}{x+2}$

$= \dfrac{A(x+2)+B(x-10)}{(x-10)(x+2)}$

x + 20 = A(x + 2) + B(x - 10)

x + 20 = (A + B)x + (2A - 10B)

Now; we have to relate like terms on both sides; we have:

A + B = 1 ; 2A - 10 B = 20

By solvong the expressions above; we have:

$A = \dfrac{5}{2}$ $B = \dfrac{3}{2}$

Now;

$\dfrac{x+20}{(x-10)(x+2)} = \dfrac{5}{2(x-10)} + \dfrac{3}{2(x+2)}$

Thus;

$\dfrac{2x^3-16x^2-39x+20}{x^2-8x-20}= 2x + \dfrac{5}{2(x-10)}+ \dfrac{3}{2(x+2)}$

Now; the integral is:

$\int \dfrac{2x^3-16x^2-39x+20}{x^2-8x-20} \ dx = \int \begin {bmatrix} 2x + \dfrac{5}{2(x-10)}+ \dfrac{3}{2(x+2)} \end {bmatrix} \ dx$

$\mathbf{\int \dfrac{2x^3-16x^2-39x+20}{x^2-8x-20} \ dx = x^2 + \dfrac{5}{2}In | x-10|\dfrac{3}{2} In |x+2|+C}$

3. Due to the fact that the maximum words this text box can contain are 5000 words, we decided to write the solution for question 3 and upload it in an image format.

Please check to the attached image below for the solution to question number 3.

4 0

2 years ago

The solution of a system of two linear equations yields the equation 0=3. Describe what the graph of the system looks like.

Alenkasestr [34]

Well TBH, does 0 really equal 3. no that means that they are parallel, there is no point of intersection.

4 0

3 years ago