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Ede4ka [16]
3 years ago
6

Replace the values of mand n to show the solutions of this equation.

Mathematics
1 answer:
umka2103 [35]3 years ago
8 0
(m + n)^2 + 6( m+n) - 5 = 0
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Heather collects shells. The picture shows the 18 shells that she found on her family beach vacation. She wants to use all of
IRISSAK [1]
Answer:
D) 9 postcards with 2 shells each.

Explanation:
9 x 2 = 18
Therefore, 9 postcards with 2 shells each is the correct answer.

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6 0
3 years ago
Please help!
laila [671]
Because
\frac{f(x)}{x-3} = 2x^{2} + 10x - 1
therefore
f(x) = (x-3)(2x² + 10x - 1) + k, where k =  constant.

Because f(3) = 4, therefore k =4.
The polynomial is
f(x) = 2x³ + 10x² - x - 6x² - 30x + 3 + 4
      = 2x³ + 4x² - 31x + 7

Answer:  f(x) = 2x³ + 4x² - 31x + 7

6 0
3 years ago
Solve by taking the square root of both sides 3(x-1)^2-162=0
Natali5045456 [20]
We know that: <span>3(x-1)^2-162=0
or (x-1)</span>²= 162:3
and (x-1)²= 54
we <span>take the square root of both sides 
* x-1=</span>√54= 3√6 or x=1+3√6
* x-1= -<span>√54= -3√6 or x=1-3√6
This equation has 2 solutions</span>
7 0
3 years ago
Read 2 more answers
What is 14ab+10a written in factored form?
svp [43]

Answer:

2a(7b+5)

Step-by-step explanation:

yeah-ya....... right?

5 0
2 years ago
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Check whether the function yequalsStartFraction cosine 2 x Over x EndFraction is a solution of x y prime plus yequalsnegative 2
Jobisdone [24]

The question is:

Check whether the function:

y = [cos(2x)]/x

is a solution of

xy' + y = -2sin(2x)

with the initial condition y(π/4) = 0

Answer:

To check if the function y = [cos(2x)]/x is a solution of the differential equation xy' + y = -2sin(2x), we need to substitute the value of y and the value of the derivative of y on the left hand side of the differential equation and see if we obtain the right hand side of the equation.

Let us do that.

y = [cos(2x)]/x

y' = (-1/x²) [cos(2x)] - (2/x) [sin(2x)]

Now,

xy' + y = x{(-1/x²) [cos(2x)] - (2/x) [sin(2x)]} + ([cos(2x)]/x

= (-1/x)cos(2x) - 2sin(2x) + (1/x)cos(2x)

= -2sin(2x)

Which is the right hand side of the differential equation.

Hence, y is a solution to the differential equation.

6 0
4 years ago
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