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MArishka [77]
4 years ago
8

How do i solve this quadratic equation x²+6x+c

Mathematics
1 answer:
Anuta_ua [19.1K]4 years ago
6 0

recall in a perfect square trinomial, the middle term is the tale-tell guy, we know the middle term is the product of 2 and the two other guys without the exponent, so in this case 6x = 2*√x² * √c.


\bf 6x=2\cdot \sqrt{x^2}\cdot \sqrt{c}\implies 6x=2\sqrt{x^2c}\implies 6x=2x\sqrt{c}\\\\\\\cfrac{6x}{2x}=\sqrt{c}\implies 3=\sqrt{c}\implies 3^2=c\implies 9=c

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Write the next three multiples of 9 <br> 36,
GrogVix [38]

Answer:

45,54,63

Step-by-step explanation:

The multiples of 9 are

9,19,27,36,45,54,63,72,81

The three multiples are 36 are

45,54,63

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3 years ago
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The shaded area of the picture shows what part of the yard Rashon mowed in 15 minutes. He wonders how long it will take him to m
irina [24]

Answer:60

Step-by-step explanation:

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two terms of an arithmetic sequence are a6= 40 and a20= -16. write and explicit rule for the nth term
Lana71 [14]

Answer:

Tn = 64-4n

Step-by-step explanation:

The nth term of an AP is expressed as:

Tn = a+(n-1)d

a is the common difference

n is the number of terms

d is the common difference

Given the 6th term a6 = 40

T6 = a+(6-1)d

T6 = a+5d

40 = a+5d ... (1)

Given the 20th term a20 = -16

T20 = a+(20-1)d

T20 = a+19d

-16 = a+19d... (2)

Solving both equation simultaneously

40 = a+5d

-16 = a+19d

Subtracting both equation

40-(-16) = 5d-19d

56 = -14d

d = 56/-14

d = -4

Substituting d = -4 into equation

a+5d = 40

a+5(-4) = 40

a-20 = 40

a = 20+40

a = 60

Given a = 60, d = -4, to get the nth term of the sequence:

Tn = a+(n-1)d

Tn = 60+(n-1)(-4)

Tn = 60+(-4n+4)

Tn = 60-4n+4

Tn = 64-4n

7 0
4 years ago
I need answers to this
ololo11 [35]

Answer:

x = -5, and y = -6

Step-by-step explanation:

Suppose that we have two equations:

A = B

and

C = D

combining the equations means that we will do:

First we multiply both whole equations by constants:

k*(A = B) ---> k*A = k*B

j*(C = D) ----> j*C = j*D

And then we "add" them:

k*A + j*C = k*B + j*D

Now we have the equations:

-x  - y = 11

4*x - 5*y = 10

We want to add them in a given form that one of the variables cancels, so we can solve it for the other variable.

Then we can take the first equation:

-x  - y = 11

and multiply both sides by 4.

4*(-x - y = 11)

Then we get:

4*(-x - y) = 4*11

-4*x - 4*y = 44

Now we have the two equations:

-4*x - 4*y = 44

4*x - 5*y = 10

(here we can think that we multiplied the second equation by 1, then we have k = 4, and j = 1)

If we add them, we get:

(-4*x - 4*y) + (4*x - 5*y) = 10 + 44

-4*x - 4*y + 4*x - 5*y = 54

-9*y = 54

So we combined the equations and now ended with an equation that is really easy to solve for y.

y = 54/-9 = -6

Now that we know the value of y, we can simply replace it in one of the two equations to get the value of x.

-x - y = 11

-x - (-6) = 11

-x + 6 = 11

-x = 11 -6 = 5

-x = 5

x = -5

Then:

x = -5, and y = -6

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Korolek [52]
300 of his coins are not quarters. 
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