Y=-4x+9. (-5,?) (2,?) (4,?) Solve for y using those numbers as x

Evaluate 15/r when r=5

strojnjashka [21]

So do 15/5 and the answer is 3.

3 0

4 years ago

Find f(2) if f(x) = (x + 1)2

Svetradugi [14.3K]

The answer is X=6
Plug in 2 to the equation

7 0

4 years ago

Read 2 more answers

Measures of two supplementary angles are consecutive odd integers find the angles

torisob [31]

<h3>☀️ <u>Solution</u><u>:</u></h3>

Supplementary angles are those angles which add upto 180°. They may or may not be linear pair.

We have,

Two consecutive odd angles.
They are supplementary $.$

Let,

One of the angles be x
Then, other angle will be x + 2°

Then,

➝ x + x + 2° = 180°

➝ 2x + 2° = 180°

➝ 2x = 178°

➝ x = 89°

Other angle = x + 2° = 91°

Answer - 89° and 91°

⚘ Hence, solved !!

<u>━━━━━━━━━━━━━━━━━━━━</u>

7 0

3 years ago

Read 2 more answers

Based on the graph, what is the perimeter of an octagon?

Ghella [55]

Answer:

The perimeter of the regular octagon is 104 units

Step-by-step explanation:

we know that

The segment LM is the length of one side of the octagon

we have the coordinates

L(-6,-2) and M(6,3)

step 1

Find the distance LM

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

substitute the values

$d_L_M=\sqrt{(3+2)^{2}+(6+6)^{2}}$

$d_L_M=\sqrt{(5)^{2}+(12)^{2}}$

$d_L_M=\sqrt{169}$

$d_L_M=13\ units$

step 2

Find the perimeter

we know that

A regular octagon has eight equal sides

so

To find out the perimeter, multiply by 8 the length of the segment LM

$P=8(13)=104\ units$

5 0

3 years ago

Someone help me with this

puteri [66]

Answer:

A) t = 1 second to reach the highest point

B) h = 96 feet

C) t = 3.449 seconds to hit the ground

Step-by-step explanation:

We notice that the equation that describes the position of the object in terms of the time, is a parabola (quadratic) with negative leading coefficient. therefore this is a parabola with arms pointing down, and with vertex at the top. It is at that top vertex that the maximum altitude of the object is reached.

We need therefore to find the position of the vertex (time "t" is the horizontal coordinate and height "h" is the vertical coordinate).

We know that the vertex (maximum of a parabola of this type - normally described by $y=ax^2+bx+c$ ) is given by $x_{vertex} = -\frac{b}{2*a}$ . Therefore in our case, understanding that time "t" is equivalent to the variable "x", and eight "h" is equivalent to the variable "y", that "-16" is $a$ , and "32" is $b$ , we have that the maximum occurs for:

$t_{max}=-\frac{32}{2*(-16)} = \frac{-32}{-32} = 1$

which means at one second from the launching.

We then can find the answer to part B by replacing t with "1" second in the formula for the height "h":

$h(1)=-16*(1)^2+32(1)+80=-16+32+80=96$

That is : a height of 96 feet.

To find the answer for part C: the time to hit the ground, we solve for the variable "t" in the given expression for the height, by setting the height to zero (object touching the ground) and using the quadratic formula:

$0=-16t^2+32t+80\\t=\frac{-32+/-\sqrt{32^2-4*(-16)*80} }{2*(-16)} \\ t=\frac{-32+/-\sqrt{6144} }{-32}$

which gives us two solutions: t = -1.449 seconds, and t = 3.449 seconds

Since the negative time does not have physical meaning in our case (time before the object was launched), we adopt the second answer: t = 3.449 seconds

5 0

3 years ago