All categories

2 years ago

I NEED HELP ASAP... PLEASE HELP

Mathematics

1 answer:

Evgen [1.6K]2 years ago

6 0

The first one is y=6(x-2)^2+4 and the second one is y=-3/2(x+6)^2+4

You might be interested in

Jack’s dog has a rectangular pen.The length is 2 feet longer than the width. The width is 6 feet. What is the distance around th

Marizza181 [45]

Answer:

L=W+X X=2 feet, W=width, L=length

if you were to solve it then you would get L=8

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

What are the zeros of the quadratic function f(x) = 2x2 – 10x – 3?

Natasha2012 [34]

Answer:

$x=\frac{50+\sqrt{31}}{2},\frac{50-\sqrt{31}}{2}$ are zeroes of given quadratic equation.

Step-by-step explanation:

We have been a quadratic equation:

$2x^2-10x-3$

We need to find the zeroes of quadratic equation

We have a formula to find zeroes of a quadratic equation:

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

General form of quadratic equation is $ax^2+bx+c$

On comparing general equation with b given equation we get

a=2,b=-10,c=-3

On substituting the values in formula we get

$D=\sqrt{(-10)^2-4(2)(-3)}$

$\Rightarrow D=\sqrt{100+24}=\sqrt{124}$

Now substituting D in $x=\frac{b^2\pm\sqrt{D}}{2a}$ we get

$x=\frac{(-10)^2\pm\sqrt{124}}{2\cdot 2}$

$x=\frac{100\pm\sqrt{124}}{4}$

$x=\frac{100\pm2\sqrt{31}}{4}$

$x=\frac{50\pm\sqrt{31}}{2}$

Therefore, $x=\frac{50+\sqrt{31}}{2},\frac{50-\sqrt{31}}{2}$

5 0

3 years ago

Read 2 more answers

A lighthouse is 90 feet above sea level. The angle of depression from the top of the lighthouse to the center of the rock is 35

Marysya12 [62]

Notice the picture below

so.... the angle of depression, is above in dashed lines.... however, the angle of depression, is identical to the angle of elevation from the water, since, both angles are "alternate interior" angles

now, using the 45-45-90 rule, then we know the swimmer is 90 feet from the lighthouse, then just use the tangent ratio to get "x"