All categories

3 years ago

Write in y=mb+b form

Mathematics

1 answer:

kotykmax [81]3 years ago

6 0

Answer:

y=5x+4

Step-by-step explanation:

5 is the rate of change and 4 is where the line intersects y

You might be interested in

Kelly spends $75.98 on items for a party. She buys thank you cards that cost $3.18 (including tax). She also buys gift bags that

vladimir2022 [97]

B. 7 because if you do $75.98/$10.40=7

6 0

3 years ago

Rationalize the denominator and simplify. 6 /5-3

sattari [20]

Answer:

-9/5

Step-by-step explanation:

Step 1: Take the LCM in the denominator

6/5-3

multiply the numerator and denominator by 5 to make the denominator 5.

6-3 (x5)

5 1 (x5)

=6-15

= -9

6 0

3 years ago

Ignore circled one and work it out please

mash [69]

Answer:

a. $\frac{(x-7)}{x^2+x-30}$

Step-by-step explanation:

$\huge \frac{x^2-3x-28}{(x^2+10x+24)(x-5)} \\\\\huge {=\frac{x^2-7x+4x-28}{(x^2+6x+4x+24)(x-5)}}\\\\\huge {=\frac{x(x-7)+4(x-7)}{\{x(x+6)+4(x+6)\}(x-5)}}\\\\\huge {=\frac{(x-7)(x+4)}{(x+6)(x+4)(x-5)}}\\\\\huge {=\frac{(x-7)}{(x+6)(x-5)}}\\\\\huge {=\frac{(x-7)}{x^2+(6-5)x+6(-5)}}\\\\\huge {=\frac{(x-7)}{x^2+x-30}}$

3 0

3 years ago

Can someone please find the vertex?

crimeas [40]

You have the correct axis of symmetry value. Nice work.

Plug that x value into the equation to get...
y = -2x^2 - 12x - 10
y = -2*(-3)^2 - 12*(-3) - 10 <<-- replace every x with -3; then use PEMDAS
y = -2*(9) - 12*(-3) - 10
y = -18 + 36 - 10
y = 18 - 10
y = 8

The vertex is the ordered pair (-3, 8)

Note: the axis of symmetry is the vertical line through the vertex

8 0

3 years ago

I am having trouble with this relative minimum of this equation.

Norma-Jean [14]

Answer:

So the approximate relative minimum is (0.4,-58.5).

Step-by-step explanation:

Ok this is a calculus approach. You have to let me know if you want this done another way.

Here are some rules I'm going to use:

$(f+g)'=f'+g'$ (Sum rule)

$(cf)'=c(f)'$ (Constant multiple rule)

$(x^n)'=nx^{n-1}$ (Power rule)

$(c)'=0$ (Constant rule)

$(x)'=1$ (Slope of y=x is 1)

$y=4x^3+13x^2-12x-56$

$y'=(4x^3+13x^2-12x-56)'$

$y'=(4x^3)'+(13x^2)'-(12x)'-(56)'$

$y'=4(x^3)'+13(x^2)'-12(x)'-0$

$y'=4(3x^2)+13(2x^1)-12(1)$

$y'=12x^2+26x-12$

Now we set y' equal to 0 and solve for the critical numbers.

$12x^2+26x-12=0$

Divide both sides by 2:

$6x^2+13x-6=0$

Compaer $6x^2+13x-6=0$ to $ax^2+bx+c=0$ to determine the values for $a=6,b=13,c=-6$ .

$a=6$

$b=13$

$c=-6$

We are going to use the quadratic formula to solve for our critical numbers, x.

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

$x=\frac{-13 \pm \sqrt{13^2-4(6)(-6)}}{2(6)}$

$x=\frac{-13 \pm \sqrt{169+144}}{12}$

$x=\frac{-13 \pm \sqrt{313}}{12}$

Let's separate the choices:

$x=\frac{-13+\sqrt{313}}{12} \text{ or } \frac{-13-\sqrt{313}}{12}$

Let's approximate both of these:

$x=0.3909838 \text{ or } -2.5576505$ .

This is a cubic function with leading coefficient 4 and 4 is positive so we know the left and right behavior of the function. The left hand side goes to negative infinity while the right hand side goes to positive infinity. So the maximum is going to occur at the earlier x while the minimum will occur at the later x.

The relative maximum is at approximately -2.5576505.

So the relative minimum is at approximate 0.3909838.

We could also verify this with more calculus of course.

Let's find the second derivative.

$f(x)=4x^3+13x^2-12x-56$