Please help.. what is the function rule for the perimeter P of a building with a rectangular base if the width w is two times the length L?
A.) P=2L B.) P=6L C.) P=6w

Answer: C - Y = -16(x- 2)^2 + 68

Step-by-step explanation:

Edge 2023

Final answer:

The correct quadratic equation that models the ball's vertical motion is y = –16(x – 2)² + 68, since it reflects the ball reaching its maximum height at 2 seconds and the vertex of the parabola being the highest point of the ball's path.

Explanation:

The question involves finding an equation that models the height of a tennis ball thrown straight up into the air after a certain number of seconds. Given that the ball reaches its maximum height at 2 seconds, we can determine which equation best describes the ball's vertical motion. The correct equation will show the ball peaking at 2 seconds and then descending symmetrically in a parabolic path.

The equation that correctly models the motion of the ball in this context is y = –16(x – 2)² + 68. This quadratic equation is in the vertex form of a parabola, where the maximum height is represented by the vertex point (2, 68), and the vertex is the highest point of the parabola since the coefficient of the squared term is negative.

When x is 2 seconds, the equation simplifies to y = –16(0)²+ 68, which results in y = 68, demonstrating that the maximum height of 68 is indeed reached at 2 seconds after the ball is thrown, thus supporting that this is the correct quadratic equation for the scenario.

The answer is y= -2x

40% = 80,000

20% = 40,000

x5 = 100% = 200,000

Answer:

(x÷9)−7 is your answer.

Answer

Find out the altitude of the equilateral triangle .

To proof

By using the trignometric identity.

[tex]tan\theta = \frac{Perpendicular}{base}[/tex]

As shown in the diagram

and putting the values of the angles , base and perpendicular

[tex]tan 60^{\circ} = \frac{a}{4\sqrt{3}}[/tex]

[tex]tan 60^{\circ} = \sqrt{3}[/tex]

solving

[tex]\sqrt{3} = \frac{a}{4\sqrt{3}}[/tex]

[tex]a = \sqrt{3}\times 4 \sqrt{3}[/tex]

As

[tex]\sqrt{3}\times \sqrt{3} = 3[/tex]

put in the above

a = 4 × 3

a = 12 units

The  length of the altitude of the equilateral triangle is 12 units .

Option (F) is correct .

Hence proved

Answer:

F. 12

Step-by-step explanation:

We have been given an image of a triangle and we are asked to find the length of the altitude of our given triangle.

Since we know that altitude of an equilateral triangle splits it into two 30-60-90 triangle.

We will use Pythagoras theorem to solve for the altitude of our given triangle.

[tex]\text{Leg}^2+\text{Leg}^2=\text{Hypotenuse}^2[/tex]

Upon substituting our given values in above formula we will get,

[tex](4\sqrt{3})^2+a^2=(8\sqrt{3})^2[/tex]

[tex]16*3+a^2=64*3[/tex]

[tex]48+a^2=192[/tex]

[tex]48-48+a^2=192-48[/tex]

[tex]a^2=144[/tex]

Upon taking square root of both sides we will get,

[tex]a=\sqrt{144}[/tex]

[tex]a=12[/tex]

Therefore, the length of the altitude of our given equilateral triangle is 12 units and option F is the correct choice.

ANSWER:

To construct a triangle given one side and the angle at each end of it with compass and straightedge or ruler. It works by first copying the line segment to form one side of the triangle, then copy the two angles on to each end of it to complete the triangle.  Follow these steps to draw the recreate a triangle using a protector, a string and angle, angle side congruence theorem.

STEP-BY-STEP EXPLANATION:

Step 1: You will need to choose two angle and one length for both triangle. Noted that the two angle and one length must be the same for both triangle.

Step 2: Draw a long line on a paper

Step 3: Use a ruler to measure the length ,and then make a mark on the string.

Step 4: Use the string to mark on the long line.

Step 5: Use a protractor to measure

Step 6: Use AAS Congruence Theorem to prove that both triangle are congruent.

After converting 0.4 kilograms to ounces using the appropriate conversion factors, it is determined that 0.4 kilograms is slightly greater than 14 ounces.

To determine which is greater, 14 ounces or 0.4 kilograms, we first need to use the appropriate conversion factors to compare the two measurements in the same unit. Since 1 pound is equivalent to 0.4536 kilograms, and 1 pound is also equal to 16 ounces, we need to convert one of the measures so that both are in either ounces or kilograms for a direct comparison.

Converting 0.4 kilograms to ounces:

Now we have both weights in ounces, and we can see that 14.11 ounces is slightly more than 14 ounces. Therefore, 0.4 kilograms is greater than 14 ounces.

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D: The cost of natural gas is $0.83 per unit.

Answer:

Option d is correct

The cost of natural gas (y) = $0.83 per unit

Step-by-step explanation:

Given the system of equation:

750x + 17y = 141.61                 ......[1]

300x + 30y =75.90                 ......[2]

where x is the cost of electricity and y is the costs of natural gas.

Multiply equation [1] by 30 and equation [2] by 75 we get;

[tex]30 \cdot (750x + 17y) = 30 \cdot 141.61[/tex]

Simplify:

22500x + 510y = 4248.3        ......[3]

[tex]75 \cdot (300x + 30y) = 75 \cdot 75.90[/tex]

Simplify:

22500x + 2250y = 5692.5      .......[4]

Subtract equation [3] from  [4]  we get;

1740 y = 1444.2

Divide by 15 both sides, we get

116 y = 96.28

Since, Katrina solves the above system using linear equation and arrives at the equation :

116 y = 96.28

Division property of equality states that  divide the same number to both sides of an equation:

Divide by 116 to both sides of an equation, to solve for y;

[tex]\frac{116 y}{116} =\frac{96.28}{116}[/tex]

Simplify:

y = 0.83 where y represents the cost of natural gas

Therefore, the cost of natural gas y ,is, $ 0.83 per unit.

Answer: 0.3 is the mode

Step-by-step explanation: The mode is what number shows up the most and 0.3 shows up 3 times.

0 shows up twice

a. the total change in the size of the tumor during the first 6 months is 64 - 1 = 63 mm³.

b. the average rate of change in the size of the tumor during the first 6 months is 10.5 mm³ per month.

c. the estimated rate at which the tumor is growing at t = 6 is approximately 44.352 mm³ per month.

a) Total change in the size of the tumor during the first 6 months:

To find the total change in the size of the tumor, we need to subtract the initial size from the size after 6 months. Since [tex]\( S = 2^t \)[/tex], we can calculate the size at t = 6 by plugging in t = 6 into the equation. Then, we can subtract the initial size from this value.

[tex]\[ S(6) = 2^6 = 64 \][/tex]

The initial size is given by [tex]\( S(0) = 2^0 = 1 \).[/tex]

So, the total change in the size of the tumor during the first 6 months is 64 - 1 = 63 mm³.

b) Average rate of change in the size of the tumor during the first 6 months:

The average rate of change is given by the total change divided by the number of months. We already found the total change (63 mm³), and the number of months is 6.

[tex]\[ \text{Average rate of change} = \frac{\text{Total change}}{\text{Number of months}} = \frac{63}{6} = 10.5 \][/tex]

So, the average rate of change in the size of the tumor during the first 6 months is 10.5 mm³ per month.

c) Estimate the rate at which the tumor is growing at t = 6:

To estimate the rate at which the tumor is growing at t = 6, we can use calculus to find the derivative of the function [tex]\( S(t) = 2^t \)[/tex] with respect to t. The derivative represents the rate of change of S with respect to t at any given time.

[tex]\[ \frac{dS}{dt} = \frac{d}{dt} (2^t) = (\ln 2) \cdot (2^t) \][/tex]

Now, plug in t = 6 to find the rate of change at that specific time.

[tex]\[ \frac{dS}{dt}\bigg|_{t=6} = (\ln 2) \cdot (2^6) \][/tex]

[tex]\[ \frac{dS}{dt}\bigg|_{t=6} = (\ln 2) \cdot 64 \][/tex]

Since [tex]\( \ln 2 \approx 0.693 \)[/tex], we have:

[tex]\[ \frac{dS}{dt}\bigg|_{t=6} \approx 0.693 \cdot 64 \approx 44.352 \][/tex]

So, the estimated rate at which the tumor is growing at t = 6 is approximately 44.352 mm³ per month.

Answer:

A

Step-by-step explanation:

Answer:

$55

Step-by-step explanation: