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3 years ago

Need help rn 30 points

Mathematics

1 answer:

Ilia_Sergeevich [38]3 years ago

7 0

Answer: X = ( $-\frac{5}{4}\\$ , 6)

Step-by-step explanation:

x(4x - 19) = 30

4x^2 - 19x = 30

4x^2 - 19x - 30 = 0

ax^2 + bx + c = 0

$X{1}=$ $\frac{-b+\sqrt{b^{2} -4ac}}{2a}\\$ = $\frac{19+\sqrt{(-19)^{2} -4(4)(-30)}}{2(4)}\\$ = 6

$X{2}=$ $\frac{-b-\sqrt{b^{2} -4ac}}{2a}\\$ = $\frac{19-\sqrt{(-19)^{2} -4(4)(-30)}}{2(4)}\\$ = $-\frac{5}{4}$ = -1.25

X = ( $-\frac{5}{4}\\$ , 6)

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Solve systems of equations by algebraically x-2y=7 2x+5y =5

Natali [406]

Answer:

(5,-1) or x=5 y=-1

Step-by-step explanation:

I used the substitution method to solve this!

1. Pick one of your equations and solve for one of the variables. I chose the first equation and solved for x.

x-2y=7

(Move the -2y to the other side of the equation in order to get the x by itself. You do the opposite, so it becomes +2y.)

x=2y+7

2. Now take your second equation and plug in what you got for x into the x variable.

2(2y+7)+5y=5

(Multiple 2 by everything inside of the parentheses.)

4y+14+5y=5

(We want to get the y by itself, so move the 14 to the other side.)

4y+5y=-14+5

(Combine all the like terms.)

9y=-9

(Divide the 9 from the y. What you do to one side you must do to the other.)

y=-1

3. Since you have one variable solved for. Now take the first equation and plug in your y.

x-2(-1)=7

(Multiple -2 by -1)

x+2=7

(Move the 2 to the other side in order to get the x by itself.)

x=5

4. If needed, plug in your x and y values into the equations in order to check your answer.

Hope this could help!

8 0

3 years ago

Read 2 more answers

X squared minus 25 equals 0

Yuri [45]

$x^2-25=0\\
x^2=25\\
x=\sqrt{25}\\ x=-5 \vee x=5$

or

$x^2-25=0\\
(x+5)(x-5)=0\\
x=-5 \vee x=5
$

7 0

4 years ago

For each equation, determine whether it has no solutions, exactly one solution, or is true for all values of x (and has infinite

nignag [31]

Answer:

Equation 1 has exactly one solution.
Equation 2 has infinitely many solutions.
Equation 3 has no solution.

Step-by-step explanation:

We are given three equations to solve. First, let's solve the equations for x.

Equation 1

$\displaystyle{6x+8=7x+13}\\\\7x + 13 = 6x + 8\\\\x + 13 = 8\\\\\bold{x = -5}$

Therefore, we determined that for the first equation, x = -5. We can check our solution by substituting it back into the original equation.

$\displaystyle{6(-5)+8=7(-5)+13}\\\\-30 + 8 = -35 + 13\\\\-22 = -22 \ \checkmark$

Since we got a true statement, there are no other values of x for which we get a true statement. Let's test this with the opposite value: positive 5.

$6(5)+8=7(5)+13\\\\30 + 8 = 35 + 13\\\\38 = 48 \ \text{X}$

Therefore, for Equation 1, there is exactly one solution.

Equation 2

$6 x + 8 = 2 ( 3 x + 4 )\\\\6x + 8 = 6x + 8\\\\0 + 8 = 8\\\\8 = 8 \ \checkmark$

We get a true statement by solving for x (which ends up canceling out of the equation entirely). Therefore, we can check any value in place of x to see if we get a true statement. For this instance, I will use -3.

$6(-3) + 8 = 2 ( 3(-3) + 4 )\\\\-18 + 8 = 2(-9+4)\\\\-18 + 8 = 2(-5)\\\\-18 + 8 = -10\\\\-18 = -18 \ \checkmark$

We still get a true statement, so Equation 2 has infinitely many solutions.

Equation 3

$6 x + 8 = 6 x + 13\\\\0 + 8 = 13\\\\8 \neq 13$

We get a false statement. Therefore, Equation 3 has no solution.

3 0

3 years ago

Read 2 more answers

URGENT !!!

Sedaia [141]

Answer:

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Step-by-step explanation:

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3 years ago

Please help I don't understand i'm in the lowest class?

Anna71 [15]

Answer:

Step-by-step explanation:

6 0

2 years ago