Write y= -2x-3 in standard form

Will give brainliest if you help!!

jonny [76]

Answer:

No solution

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

4(3-2x)=4(x-6) what is x

erastova [34]

Here are the steps
(4*3)+(4*-2x)=(4x)+(4*-6)
...12+-8x=16x-24x
...-8x+24x-16x=-12
0x=-12
x=-12/0
dividing by 0.
so there is no solution.

3 0

3 years ago

Marty is taking piano lessons. She is going to practice every afternoon. The 1st week she practiced for 11 minutes each day. The

defon

Answer:

$m=7w+4$

Step-by-step explanation:

<u>Linear Modeling</u>

It consist is setting up a linear relationship between two variables, given some experimental data. Only 2 points are needed to set up the equation of a line, but if more than 2 points are used, then the result should use statistical approaches like linear regression to find the best-fit line.

For the question at hand, Marty practices his piano lessons 11 minutes the week #1. It provides the first point (1,11). He practices 25 minutes per day on the third week. It gives us another point (3,25). This is enough to find the equation of a line. The general formula for a line, having two points (m1,w1) (m2,w2) is

$\displaystyle m-m_1=\frac{m_2-m_1}{w_2-w_1}(w-w_1)$

Let's plug in our values

$\displaystyle m-11=\frac{25-11}{3-1}(w-1)$

Simplifying:

$m-11=7w-7$

$\boxed{m=7w+4}$

3 0

4 years ago

- Polynomial Functions -For each function, state the vertex; whether the vertex is a maximum or minimum point; the equation of t

vladimir2022 [97]

EXPLANATION

Given the function f(x) = (x-6)^2 + 1

$\mathrm{The\: vertex\: of\: an\: up-down\: facing\: parabola\: of\: the\: form}\: y=ax^2+bx+c\: \mathrm{is}\: x_v=-\frac{b}{2a}$

Expanding (x-6)^2 + 1 by applying the Perfect Square Formula:

$=x^2-12x+37$ $\mathrm{The\: parabola\: params\: are\colon}$ $a=1,\: b=-12,\: c=37$ $x_v=-\frac{b}{2a}$ $x_v=-\frac{\left(-12\right)}{2\cdot\:1}$ $\mathrm{Simplify}$ $x_v=6$ $y_v=6^2-12\cdot\: 6+37$

Simplify:

$y_v=1$

$\mathrm{Therefore\: the\: parabola\: vertex\: is}$ $\mleft(6,\: 1\mright)$ $\mathrm{If}\: a$ $\mathrm{If}\: a>0,\: \mathrm{then\: the\: vertex\: is\: a\: minimum\: value}$ $a=1$ $\mathrm{Minimum}\mleft(6,\: 1\mright)$ $\mathrm{For\: a\: parabola\: in\: standard\: form}\: y=ax^2+bx+c\: \mathrm{the\: axis\: of\: symmetry\: is\: the\: vertical\: line\: that\: goes\: through\: the\: vertex}\: x=\frac{-b}{2a}$

Expanding (x-6)^2 + 1 by applying the Perfect Square Formula:

$y=x^2-12x+37$ $\mathrm{Axis\: of\: Symmetry\: for}\: y=ax^2+bx+c\: \mathrm{is}\: x=\frac{-b}{2a}$ $a=1,\: b=-12$ $x=\frac{-\left(-12\right)}{2\cdot\:1}$ $\mathrm{Refine}$

Axis of simmetry : x=6

The quadratic function has the same shape than the parent function y=x^2 because there is NOT a coefficient within x.

3 0

1 year ago

Find the intersection of the lines 2x+5y=8 and 6x+y=10 in two ways by elimination and by substitution, step by step please.

Law Incorporation [45]

Answer:

The lines intersect at x = 1.5 and y = 1

Step-by-step explanation:

We need to find the intersection of the lines 2x+5y=8 and 6x+y=10.

We need to find the values of x and y by elimination and by substitution.

a) By Elimination:

2x+5y = 8 (1)

6x + y = 10 (2)

Multiply eq(2) with 5 and subtract eq(1) from(2)

30x + 5y = 50

2x + 5y = 8

- - -

___________

28x = 42

x = 1.5

Now putting value of x in eq(2)

6x + y = 10

6(1.5) + y = 10

9 + y = 10

=> y = 10 - 9

y = 1

so, (x,y) = (1.5,1)

The lines intersect at x = 1.5 and y = 1

b) By substitution

2x+5y = 8 (1)

6x + y = 10 (2)

Finding value of y in equation 2 and substituting in eq(1)

y = 10 -6x

2x + 5(10 - 6x) = 8

2x + 50 - 30x = 8

-28x = 8-50

-28x = -42

x = -42/-28

x = 1.5

Now finding value of y by substituting value of x

6x + y = 10

6x = 10-y

x = 10 - y /6

2x + 5y = 8

2(10-y/6) + 5y = 8

10-y/3 + 5y = 8

10 -y +15y/3 = 8

10 +14y = 8*3

+14 y = 24 -10

+14 y = 14

y = 14/14

y = 1

So, (x,y) = (1.5,1)

The lines intersect at x = 1.5 and y = 1

8 0

4 years ago