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3 years ago

14

Y= 1/3 (x + 4)(x - 2)

1 answer:

Setler79 [48]3 years ago

7 0

Answer:x = -1 +√ 3(3 + y)

Step-by-step explanation: solve the equation for x by finding a, b and c of the quadratic then applying the quadractic formula

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Anyone help me solve for y.

riadik2000 [5.3K]

Answer:

y = - 4

Step-by-step explanation:

Given

y = 2x + ( - 2) , that is

y = 2x - 2 ← substitute x = - 1

y = 2(- 1) - 2 = - 2 - 2 = - 4

This is confirmed by the point (- 1, - 4 ) on the graph

3 0

3 years ago

Read 2 more answers

A company makes bicycles in two sizes. The first bicycle size weighs 18.5 pounds. The second bicycle sizes weighs 1/14 as much a

QveST [7]

The second bicycle would weigh 1.32142857143 pounds or rounded to 1.32.
You would find this by multiply 18.5 by 1/14

7 0

3 years ago

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PLEASE HELP!!<br> Convert each linear equation to standard form.<br> y-4= -2(x+3)

hjlf

So first, this is point slope form. The first step is to distribute.)
$y-4=-2x-6$
Then add 4 to both sides.
$y=-2x-2$
Then add 2x and it becomes
$2x+y=-2$

4 0

3 years ago

Read 2 more answers

the first three steps of completing the square to solve the quadratic equation X2+4X-6=0 are shown below. what re the next 3 ste

Vilka [71]

Answer:

see the explanation

Step-by-step explanation:

<u><em>The complete question is</em></u>

The first three steps of completing the square to solve the quadratic equation x2 +4x -6= 0, are shown below.

Step 1: x2 +4x= 6

Step 2: x2 +4x+4=6+4

Step 3: (x +2)^2=10

What are the next 3 steps?

we have

step 4

Take square root both sides

$\sqrt{(x+2)^2} =\sqrt{10}$

step 5

Simplify

$x+2=\pm\sqrt{10}$

step 6

Solve for x

subtract 2 both sides

$x=-2\pm\sqrt{10}$

$x_1=-2+\sqrt{10}$

$x_2=-2-\sqrt{10}$

3 0

3 years ago

Let Q(x, y) be the predicate "If x < y then x2 < y2," with domain for both x and y being R, the set of all real numbers.

Levart [38]

Answer:

a) Q(-2,1) is false

b) Q(-5,2) is false

c)Q(3,8) is true

d)Q(9,10) is true

Step-by-step explanation:

Given data is $Q(x,y)$ is predicate that $x$ then $x^{2}$ . where $x,y$ are rational numbers.

a)

when $x=-2, y=1$

Here $-2$ that is $x$ satisfied. Then

$(-2)^{2}$

$4$ this is wrong. since $4>1$

That is $x^{2}$ $>y^{2}$ Thus $Q(x,y)$ $=Q(-2,1)$ is false.

b)

Assume $Q(x,y)=Q(-5,2)$ .

That is $x=-5, y=2$

Here $-5$ that is $x$ this condition is satisfied.

Then

$(-5)^{2}$

$25$ this is not true. since $25>4$ .

This is similar to the truth value of part (a).

Since in both $x$ satisfied and $x^{2} >y^{2}$ for both the points.

c)

if $Q(x,y)=Q(3,8)$ that is $x=3$ and $y=8$

Here $3$ this satisfies the condition $x$ .

Then $3^{2}$

$9$ This also satisfies the condition $x^{2}$ .

Hence $Q(3,8)$ exists and it is true.

d)

Assume $Q(x,y)=Q(9,10)$

Here $9$ satisfies the condition $x$

Then $9^{2}$

$81$ satisfies the condition $x^{2}$ .

Thus, $Q(9,10)$ point exists and it is true. This satisfies the same values as in part (c)

6 0

3 years ago