Help please thank you.

ZanzabumX [31]

Log₄(y+2) = 3, transform it into exponent form:

(y+2) = 4³
y+2 = 64
y= 62

5 0

3 years ago

The coordinates of the center are 0,-3 and length of the diameter is 6 find equation so I know it’s x^2+(y+3)^2= I just don’t kn

velikii [3]

Answer:

hsuaiabshsisnsbkjwnsjns

Step-by-step explanation:

1828392919287382928whshsisnsjjs

sorry, i can't answer that question

6 0

2 years ago

Read 2 more answers

What is the value of x?<br> 38<br> 21<br> 21<br> Drawing not to scale

Snezhnost [94]

Where is it?
step by step:

3 0

2 years ago

What is the measure of o

Helga [31]

Answer:

D

Step-by-step explanation:

We need the formula for arc length (in radians) to solve this problem. The formula is:

$s=r \theta$

Where

s is the arc length

r is the radius of the circle

$\theta$ is the intercepted angle

In this diagram, we can see that $\theta$ is beign intercepted by the big arc with measure 20 cm. So arc length, s, is 20

Also, if we look, we see that the radius is 5 cm, so r = 5

Now, we substitute into formula and find $\theta$ :

$s=r\theta\\20=5\theta\\\theta=\frac{20}{5}\\\theta=4$

This is given in radians, so $\theta$ is 4 radians

D is the correct answer choice.

6 0

2 years ago

Find the vertex of the graph of the function.

lapo4ka [179]

Answer: The correct options are (1) (5,10), (2) (3,-3), (3) x = -1, (4) $y=(x+2)^2+3$ , (5) 21s and (6) 0, -1, and 5.

Explanation:

Te standard form of the parabola is,

$f(x)=a(x-h)^2+k$ .....(1)

Where, (h,k) is the vertex of the parabola.

(1)

The given equation is,

$f(x)=(x-5)^2+10$

Comparing this equation with equation (1),we get,

$h=5$ and $k=10$

Therefore, the vertex of the graph is (5,10) and the fourth option is correct.

(2)

The given equation is,

$f(x)=3x^2-18x+24$

$f(x)=3(x^2-6x)+24$

To make perfect square add $(\frac{b}{2a})^2$ , i.e., $9$ . Since there is factor 3 outside the parentheses, so subtract three times of 9.

$f(x)=3(x^2-6x+9)+24-3\times 9$

$f(x)=3(x-3)^2-3$

Comparing this equation with equation (1),we get,

$h=3$ and $k=-3$

Therefore, the vertex of the graph is (3,-3) and the fourth option is correct.

(3)

The given equation is

$f(x)=4x^2+8x+7$

$f(x)=4(x^2+2x)+7$

To make perfect square add $(\frac{b}{2a})^2$ , i.e., $1$ . Since there is factor 4 outside the parentheses, so subtract three times of 1.

$f(x)=4(x^2+2x+1)+7-4$

$f(x)=4(x+1)^2+3$

Comparing this equation with equation (1),we get,

$h=-1$ and $k=3$

The vertex of the equation is (-1,3) so the axis is x=-1 and the correct option is 2.

(4)

The given equation is,

$y=x^2+4x+7$

To make perfect square add $(\frac{b}{2a})^2$ , i.e., $2^2$ .

$f(x)=x^2+4x+4+7-4$

$f(x)=(x+2)^2+3$

Therefore, the correct option is 4.

(5)

The given equation is,

$h=-16t^2+672t$

It can be written as,

$h=-16(t^2-42t)$

It is a downward parabola. so the maximum height of the function is on its vertex.

The x coordinate of the vertex is,

$x=\frac{b}{2a}$

$x=\frac{42}{2}$

$x=21$

Therefore, after 21 seconds the projectile reach its maximum height and the correct option is first.

(6)

The given equation is,

$f(x)=3x^3-12x^2-15x$

$f(x)=3x(x^2-4x-5)$

Use factoring method to find the factors of the equation.

$f(x)=3x(x^2-5x+x-5)$

$f(x)=3x(x(x-5)+1(x-5))$

$f(x)=3x(x-5)(x+1)$

Equate each factor equal to 0.

$x=0,-1,5$

Therefore, the zeros of the given equation is 0, -1, 5 and the correct option is 2.

3 0

3 years ago