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2 years ago

HELP PLS IM FAILING DAYUMMMM

Mathematics

2 answers:

cupoosta [38]2 years ago

3 0

Answer:

x=3

y=-4

Step-by-step explanation:

-7x-2y=-13 ---(1)

x=2y+11

sub (2) into (1), -7(2y+11)-2y=-13

-14y-2y-77=-13

-16y=64

y=-4

put y=-4 into (2), x=2(-4)+11

x=3

btw, next time, do not use words like(failing,exam,free points,etc in ur caption or content),otherwise,the question would be deleted.

photoshop1234 [79]2 years ago

3 0

-7x -2y = 13, and then to solve it we put in (2y + 11) where it says x.

So
-7(2y +11) -2y = 13 , then distribute the -7

-14y -77 -2y = 13
-16y -77 = 13
-16y = 64
y = -4

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Which of the following constants can be added to x2 - 3x to form a perfect square trinomial?

jeka57 [31]

Answer:

2 whole 1/4

Step-by-step explanation:

When forming a perfect square trinomial you need to "complete the square".

All of the steps to completing the square when solving an equation:

1. The leading coefficient must be 1.

2. Divide b by 2.

3. Square (b/2)

4. Add (b/2)^2 to both sides to keep the polynomial balanced.

5. You can now write the perfect square trinomial and solve.

x^2 - 3x

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5 0

2 years ago

Read 2 more answers

The line $y = 3$ intersects the graph of $y = 4x^2 + x - 1$ at the points $A$ and $B$. The distance between $A$ and $B$ can be w

raketka [301]

Answer:

Step-by-step explanation:

Let's find the points $A$ and $B$ .

We know that the $y$ -coordinates of both are $3$ .

So let's first solve:

$3=4x^2+x-1$

Subtract 3 on both sides:

$0=4x^2+x-1-3$

Simplify:

$0=4x^2+x-4$

I'm going to use the quadratic formula, $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ , to solve.

We must first compare to the quadratic equation, $ax^2+bx+c=0$ .

$a=4$

$b=1$

$c=-4$

$\frac{-1 \pm \sqrt{1^2-4(4)(-4)}}{2(4)}$

$\frac{-1 \pm \sqrt{1+64}}{8}$

$\frac{-1 \pm \sqrt{65}}{8}$

Since the distance between the points $A$ and $B$ is horizontal. We know this because they share the same $y-coordinate$ .This means we just need to find the positive difference between the $x$ -values we found for the points of $A$ and $B$ .