Ask question

Ask question

All categories

3 years ago

13

Solve the system of equations y=x^2-3x+2 y=-9x-3

1 answer:

iragen [17]3 years ago

5 0

Answer:

Step-by-step explanation:

The slope-intercept form is

y

=

m

x

+

b

, where

m

is the slope and

b

is the y-intercept.

y

=

m

x

+

b

Find the values of

m

and

b

using the form

y

=

m

x

+

b

.

m

=

1

b

=

0

The slope of the line is the value of

m

, and the y-intercept is the value of

b

.

Slope:

1

y-intercept:

0

You might be interested in

What fraction is less than 1/2. 3/4,1/6,6/12,4/10,5/8,5/2

vazorg [7]

1/6! Any fraction that represents less than a half of a whole is less than 1/2. You could also convert them to the same denominator and see the difference. Make sure whatever you do to the first fraction, you do to the other!

6 0

3 years ago

Read 2 more answers

What are the solutions of the equation (2x + 3)2 + 8(2x + 3) + 11 = 0

Tom [10]

Answer:

$x=\frac{-7-\sqrt{5}}{2}$ or $x=\frac{-7+ \sqrt{5}}{2}$

Step-by-step explanation:

The given equation is

$(2x+3)^2+8(2x+3)+11=0$

Let us treat this as a quadratic equation in $(2x+3)$ .

where $a=1,b=8,c=11$

The solution is given by the quadratic formula;

$(2x+3)=\frac{-b\pm \sqrt{b^2-4ac} }{2a}$

We substitute these values into the formula to obtain;

$(2x+3)=\frac{-8\pm \sqrt{8^2-4(1)(11)} }{2(1)}$

$(2x+3)=\frac{-8\pm \sqrt{64-44} }{2}$

$(2x+3)=\frac{-8\pm \sqrt{20} }{2}$

$(2x+3)=\frac{-8\pm2\sqrt{5} }{2}$

$(2x+3)=-4\pm \sqrt{5}$

$(2x+3)=-4-\sqrt{5}$ or $(2x+3)=-4+ \sqrt{5}$

$2x=-3-4-\sqrt{5}$ or $2x=-3-4+ \sqrt{5}$

$2x=-7-\sqrt{5}$ or $2x=-7+ \sqrt{5}$

$x=\frac{-7-\sqrt{5}}{2}$ or $x=\frac{-7+ \sqrt{5}}{2}$

5 0

3 years ago

Gino is buying wood screws at the corner hardware store. The table shows different numbers of bags of screws and their correspon

likoan [24]

Answer:

I do not knoww <333

Step-by-step explanation:

7 0

3 years ago

Find the solution of the square root of the quantity of x plus 2 plus 4 equals 8, and determine if it is an extraneous solution

Iteru [2.4K]

Consider the equation:

$\sqrt{x+2}+4 = 8$

Subtracting '4' from both the sides of the equation, we get as

$\sqrt{x+2}+4-4= 8-4$

$\sqrt{x+2}= 4$

Squaring on both the sides of the equation, we get

$(\sqrt{x+2})^2 = (4)^2$

$x+2 = 16$

Subtracting '2' from both the sides of the equation, we get

$x+2-2=16-2$

x=14

Since, An extraneous solution is a solution that arises from the solving process that is not really a solution at all. But, in this equation x=14 is the solution of the given equation.

Hence, it is not an extraneous solution.

4 0

3 years ago

Read 2 more answers

GEOMETRY PLEASE HELP!! NEED HELP WITH PROOFS

FrozenT [24]

4. ΔBDE ~ ΔBAC; Side-Angle-Side (SAS) Similarity Postulate
5. ∠B ≅ ∠B; Reflexive Property of Equality

4 0

3 years ago

Read 2 more answers