Find the total surface area of the<br> following square pyramid:<br> 6 cm<br> 8 cm<br> SA = [?] cm2

The vertex of a parabola is (-1.5, -12.5), and its y-intercept is (0, -8).

Alecsey [184]

The general equation of a parabola is y=ax^2+bx+c At the y-intercept, x=0 and y= -8: -8 = a(0)^2 + b(0) + c. Thus, c = -8. So, our equation becomes

y = ax^2 + bx - 8. Next, substitute -1.5 for x and -12.5 for y. Then,
-12.5 = a(-1.5)^2 + b(-1.5) - 8. This simplifies to -4.5 = a(2.25) - 1.5b.

Next, take advantage of the info that the vertex is at x= -1.5.
The formula for the vertex is x=-b/(2a). Letting this formula = -1.5,

-1.5 = -b/(2a). We can then solve for b: 1.5 = b/(2a), or 3a = b.

Now go back to the equation we derived previously: -4.5 = a(2.25) - 1.5b.
Substitute 3a for b:

-4.5 = a(2.25) - 1.5(3a). Then -4.5 = -2.25a, and a = 4.5/2.25 = 2.

Last, substitute a = 2 into 3a=b to determine the value of b.

b=3(2) = 6.

Therefore, your equation is y=2x^2 + 6x - 8.

Check this result. Substitute the coordinates of the vertex (-1.5,-12.5) into this equation. Is the equation still true? If so, your equation correctly represents this parabola.

4 0

3 years ago

Read 2 more answers

Can you please help me

Sedbober [7]

Answer:

1. (x-2)(x-15)

2 (3x-4)(x+2)

3. cannot factor

Step-by-step explanation:

math papa greatest website for math equations.

6 0

2 years ago

The price of a box of 15 cloud markers is $12.70. The price of a box of 42 cloud markers is $31.60. All prices are without tax,

stich3 [128]

Answer:

Here, The price of one box having N markers = Price of N markers + packaging charge

Let M be the price of one Marker and P' be the price of packaging of one box which is same for any size.

Since, The price of a box of 15 cloud markers is $12.70.

That is , 15 M + P' = 12.70 ---------(1)

Also, The price of a box of 42 cloud markers is $31.60,

That is, 42 M + P' = 31.60 ---------(2),

Equation (2) - Equation (1),

27 M = 18.90,

⇒ M = 0.7

By putting the value of P in equation (1),

We get,

10.5 + P' = 12.70 ⇒ P' = 2.2,

Thus, the price of one marker, M = 0.7

And, the price of packaging one box, P' = 2.2

Thus, the price of a box of 50 marker = 50 × M + P' = 50×0.7 + 2.2 = 35 + 2.2 = $ 37.2

And, the equation that shows the price(P) of a box of N markers,

P = 0.7 N + 2.2

3 0

3 years ago

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Pls help me with my math

givi [52]

Answer:

The definition for the given piecewise-defined function is: $\boxed{\displaystyle\sf\ Option\:D:\:\: f(x) = \begin{cases}\displaystyle\sf\ x + 2 & \sf\:{if\:\:x \leq -1} \\\displaystyle\sf\ 2x + 4 & \sf\:{if\:\:x > -1}\end{cases}}$ .

Step-by-step explanation:

<h3>General Concepts:</h3>

Piecewise-defined functions.
Interval notations.

<h3>What is a piecewise-defined function?</h3>

A piecewise-defined function represents specific rules over different intervals of the domain.

<h3>Symbols used in expressing interval notations:</h3>

Open interval: This means that the endpoint is <em>not</em> included in the interval.

We can use the following symbols to indicate the <u>exclusion</u> of endpoints in the interval:

Left or right parenthesis, "( )" (or both).

Greater than (>) or less than (<) symbols.
Open dot " $\circ$ " is another way of expressing the exclusion of an endpoint in the graph of a piecewise-defined function.

Closed interval: This implies the inclusion of endpoints in the interval.

We can use the following symbols to indicate the <u>inclusion</u> of endpoints in the interval:

Open- or closed brackets (or both), "[ ]."

Greater than or equal to (≥) or less than or equal to (≤) symbols.
Closed circle or dot, "•" is another way of expressing the <em>inclusion</em> of the endpoint in the graph of a piecewise-defined function.

<h2>Determine the appropriate function rule that defines different parts of the domain. </h2>

The best way to determine which piecewise-defined function represents the graph is by observing the <u>endpoints</u> and <u>orientation</u> of both partial lines.

Open circle on (-1, 2): The graph shows that one of the partial lines has an <em>excluded</em> endpoint of (-1, 2) extending towards the <u>right</u>. This implies that its domain values are defined when x > -1.

Closed circle on (-1, 1): The graph shows that one of the partial lines has an <em>included</em> endpoint of (-1, 1) extended towards the <u>left</u>. Hence, its domain values are defined when x ≤ -1.

Based on our observations from the previous step, we can infer that x > -1 or x ≤ -1 apply to piecewise-defined functions A or D. However, only one of those two options represent the graph.

<h2>Solution:</h2><h3>a) Test option A:</h3>

$\boxed{\displaystyle\sf Option\:A)\:\:\:f(x) = \begin{cases}\displaystyle\sf\ 2x + 2 & \sf\:{if\:\:x \leq -1} \\\displaystyle\sf\ x + 4 & \sf\:{if\:\:x > -1}\end{cases}}$

<h3>Piece 1: If x ≤ -1, then it is defined by f(x) = 2x + 2. </h3>

We must choose a domain value that falls within the interval of x ≤ -1 whose output is included is included in the graph of the partial line with a <u>closed dot</u>.

Substitute x = -2 into f(x) = 2x + 2:

f(x) = 2x + 2

f(-2) = 2(-2) + 2
f(-2) = -4 + 2
f(-2) = -2 ⇒ <em>False statement</em>.

⇒ The output value of f(-2) = -2 is <u>not</u> included in the graph of the partial line whose endpoint is at (-1, 1).

<h3>Piece 2: If x > -1, then it is defined by f(x) = x + 4. </h3>

We must choose a domain value that falls within the interval of x > -1 whose output is included in the graph of the partial line with an <u>open dot</u>.

Substitute x = 0 into f(x) = x + 4:

f(x) = x + 4
f(0) = (0) + 4
f(0) = 4 ⇒ <em>True statement</em>.

⇒ The output value of f(0) = 4 <u>is</u> included in the graph of the partial line whose endpoint is at (-1, 2).

Conclusion for Option A:

Option A is not the correct piecewise-defined function because one of the pieces, f(x) = 2x + 2, does not specify the interval (-∞, -1].

<h3>b) Test option D:</h3>

$\boxed{\displaystyle\sf Option\:D)\:\:\:f(x) = \begin{cases}\displaystyle\sf\ x + 2 & \sf\:{if\:\:x \leq -1} \\\displaystyle\sf\ 2x + 4 & \sf\:{if\:\:x > -1}\end{cases}}$

<h3>Piece 1: If x ≤ -1, then it is defined by f(x) = x + 2. </h3>

We must choose a domain value that falls within the interval of x ≤ -1 whose output is included is included in the graph of the partial line with a <u>closed dot</u>.

Substitute x = -2 into f(x) = x + 2:

f(x) = x + 2
f(-2) = (-2) + 2
f(-2) = 0 ⇒ <em>True statement</em>.

⇒ The output value of f(-2) = 0 <u>is</u> included the graph of the partial line whose endpoint is at (-1, 1).

<h3>Piece 2: If x > -1, then it is defined by f(x) = 2x + 4.</h3>

We must choose a domain value that falls within the interval of x > -1 whose output is included is included in the graph of the partial line with an <u>open dot</u>.

Substitute x = 0 into f(x) = 2x + 4:

f(x) = 2x + 4
f(0) = 2(0) + 4
f(0) = 0 + 4 = 0 ⇒ <em>True statement</em>.

⇒ The output value of f(0) = 4 <u>is</u> included in the graph of the partial line whose endpoint is at (-1, 2).

<h2>Final Answer: </h2>

We can infer that the piecewise-defined function that represents the graph is:

$\boxed{\displaystyle\sf\ Option\:D:\:\: f(x) = \begin{cases}\displaystyle\sf\ x + 2 & \sf\:{if\:\:x \leq -1} \\\displaystyle\sf\ 2x + 4 & \sf\:{if\:\:x > -1}\end{cases}}$ .

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Learn more about piecewise-defined functions here:

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