All categories

4 years ago

Find the linear equation with slope = 7 and y-intercept = -4

Mathematics

2 answers:

muminat4 years ago

6 0

The answer would be
y = 7x - 4

LekaFEV [45]4 years ago

3 0

Y=7x+(-4) is the slope intercept equation

You might be interested in

Please help 7r+18=-2(6-6r)

ivolga24 [154]

Step-by-step explanation:

7r+18=-2(6-6r)

7r+18= -12+12r

7r = -12-18+12r

-30+12r = 7r

-30 = 7r-12r

-5r = -30

r = 30/5

r = 6

hope it helps you ❤️

7 0

3 years ago

Read 2 more answers

Find all the zeros of 2x4+x3-14x2-19x-6 two of its zeros are -2 and -1

GalinKa [24]

Answer:

$-2,\ -1,\ -\dfrac{1}{2},\ 3.$

Step-by-step explanation:

Consider polynomial $2x^4+x^3-14x^2-19x-6.$

If x=-2 is its zero, then you can divide the polynomial $2x^4+x^3-14x^2-19x-6$ by $x+2$ and get

$2x^4+x^3-14x^2-19x-6=(x+2)(2x^3-3x^2-8x-3).$

If x=-1, then the polynomial $2x^4+x^3-14x^2-19x-6$ can be rewritten as

$2x^4+x^3-14x^2-19x-6=(x+2)(x+1)(2x^2-5x-3).$

The quadratic polynomial has roots

$x_{1,2}=\dfrac{-(-5)\pm\sqrt{(-5)^2-4\cdot2\cdot(-3)}}{2\cdot 2}=\dfrac{5\pm \sqrt{25+24}}{4}=\dfrac{5\pm \sqrt{49}}{4}=3,-\dfrac{1}{2}.$

Then the polynomial $2x^4+x^3-14x^2-19x-6$ has zeros $-2,\ -1,\ -\dfrac{1}{2},\ 3.$

7 0

3 years ago

Let X have the uniform distribution U(0, 2) and let the conditional distribution of Y , given that X = x, be U(0, x). Find the j

Elina [12.6K]

Answer:

$f(x,y) = \frac{1}{x} \frac{1}{2}= \frac{1}{2x} , 0\leq x \leq 2 , 0\leq y \leq x$

$E(Y|x) = \int_{x=y}^2 y \frac{1}{x} dx= y ln x \Big|_{x=y}^2 =y ln 2 -y ln y = y(1-lny) \$

Step-by-step explanation:

We have two random variables X and Y. $X \sim Unif(0,2)$ and given that X=x, Y has uniform distribution (0,x)

From the definition of the uniform distribution we have the densities for each random variable given by:

$f_X (x) =\frac{1}{2} , 0\leq x\leq 2$

$f_{Y|X} (y|x) = \frac{1}{x}, 0\leq y \leq x$

And on this case we can find the joint density with the following formula:

$f(x,y) = f_{Y|X}(y|x) f_X (x)$

And multiplying the densities we got this:

$f(x,y) = \frac{1}{x} \frac{1}{2}= \frac{1}{2x} , 0\leq x \leq 2 , 0\leq y \leq x$

Now with the joint density we can find the expected value E(Y|x) with the following formula:

$E(Y|x) = \int y f_{Y|X}(y|x)dx$

And replacing we got:

$E(Y|x) = \int_{x=y}^2 y \frac{1}{x} dx= y ln x \Big|_{x=y}^2 =y ln 2 -y ln y = y(1-lny) \$

5 0

4 years ago

Find the degree of each algebraic expression

morpeh [17]

Step-by-step explanation:

The degree of an algebraic expression is the largest exponent of the variable present. In expressions with multiple variables, the exponents of each variables are added.

First Expression;

pq: Degree = 1 + 1 = 2

p²q: Degree = 2 + 1 = 3

p²q²: Degree = 2 + 2 = 4

The degree of this expression is 4

Second Expression;

2y²z: Degree = 2 + 1 = 3

10yz: Degree = 1 + 1 = 2

The degree of this expression is 3

4 0

3 years ago

PLZZ HELP I’m bad at math.

ipn [44]

Answer:

y< $\frac{-1}{3}$ x+1 is inequality shown on graph.

Step-by-step explanation:

The graph shows representation of inequality,

Here, Line passes through points say, A(3,0) and B(0,1)

The slope of line is given by m= $\frac{Y2-Y1}{X2-X1}$

m= $\frac{Y2-Y1}{X2-X1}$

m= $\frac{1-0}{0-3}$

m= $\frac{-1}{3}$

and Y-intercept is c=1

The equation of line is given by y=mx+c

y= $\frac{-1}{3}$ x+1 is equation of line in graph.

To decide which side you are going to shaded area on the graph

Take test of origin.

If (0,0) is statisfy the equation then, shaded the graph towards the origin.

Let suppose,

Take, y< $\frac{-1}{3}$ x+1

0< $\frac{-1}{3}$ (0)+1

0<1

TRUE,

Therefore, y< $\frac{-1}{3}$ x+1 is inequality shown on graph.

4 0

4 years ago