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PtichkaEL
19 days ago
6

Joel, a tutor at the Math Help, decided that it would be a good

Mathematics
1 answer:
PIT_PIT [12.4K]19 days ago
6 0

Answer:

Joel achieved a peak height of 6 units.

Step-by-step explanation:

It appears there is an error in the equation representing Joel's jump. I will presume it should be h(t) = -0.2*t^2 + 2*t + 1, as such movements are typically represented using quadratic equations. To find the maximum height Joel attained, we must determine the vertex of this equation. The vertex for a quadratic can be found using the given expression:y(x) = a*x^2 + b*x + c

x_{vertex} = \frac{-b}{2*a}

"y" can be calculated by substituting the x-coordinate of the vertex into the equation. In our case, a = -0.2, b = 2, and c = 1. The vertex calculation gives:[

t_{vertex} = \frac{-2}{2*(-0.2)} = \frac{-2}{-0.4} = 5

Thus, the maximum height totals:

h(5) = -0.2*(5)^2 +2*5 + 1\\h(5) = -5 + 10 + 1 = 6

Joel's highest point was 6 units.

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1 kilogram is equivalent to about 2.2 pounds. An orangutan weighs 142 pounds. What is its mass in kilograms? Round your answer t
AnnZ [12381]

The weight of an orangutan is 64.5 kilograms

Solution:

It is stated that the orangutan's weight is 142 pounds

Additionally,

1 kilogram is approximately 2.2 pounds

We will now convert 142 pounds into kilograms

1 kilogram = 2.2 pounds

\text{1 pound } = \frac{1}{2.2} \text{ kilograms }

As a result, 142 pounds translates to

142 \text{ pound } = \frac{1}{2.2} \times 142 \text{ kilogrmas }

\rightarrow \frac{1}{2.2} \times 142 = 64.545 \text{ kilograms }

When rounding to the nearest tenth, we obtain 64.5 kilograms

Therefore the mass of the orangutan is 64.5 kilograms

3 0
1 month ago
There is a sidewalk of width x around a rectangular garden. If the garden measures twenty-feet by thirty-feet, then the combined
PIT_PIT [12445]

Answer:

Area = 200 + 50 + x

Step-by-step explanation:

Given

Length = 20

Width = 30

Side Walk = x

Required

Find the total area.

To find this, we have to add the length of the sidewalk to the dimensions of the garden.

This results in:

Length = 20 + x.

Width = 30 + x

So, the area now becomes.

Area = (20 + x)(30 + x)

Area = 600 + 20x + 30x + x²

Area = 200 + 50 + x²

5 0
10 days ago
2. Peter drew two rays, AC and AP with A as a common endpoint. Which of the following statements
PIT_PIT [12445]

The complete question reads:

Peter created two rays, AC and AP, sharing a common vertex at point A. Which of the following statements

might accurately describe Peter's drawing?

I. AC and AP are parallel.

II. PAC represents an angle.

III. AC and AP are at right angles.

A. I and II

B. II and III

C. I and III

D. I, II, and III

Answer:

Option B: II & III

Step-by-step explanation:

We know Peter has drawn rays AC and AP.

Since the point A is shared as the endpoint, it indicates an angular relationship at this common point.

This angle could potentially be 90°, suggesting that rays AC and AP may be perpendicular.

Thus, the valid statements that characterize his drawing are: II & III.

4 0
1 month ago
Below is an attempt to derive the derivative of sec(x) using product rule, where x is in the domain of secx. In which step, if a
tester [12383]

The mistake is present in step 3. According to the product rule, we find

\dfrac{\mathrm d}{\mathrm dx}(\sec x\times\cos x)=\dfrac{\mathrm d}{\mathrm dx}(\sec x)\times\cos x+\sec x\times\dfrac{\mathrm d}{\mathrm dx}(\cos x)

=\dfrac{\mathrm d}{\mathrm dx}(\sec x)\times\cos x\boxed{+\sec x\times(-\sin x)}

=\dfrac{\mathrm d}{\mathrm dx}(\sec x)\times\cos x\boxed{-\sec x\times\sin x}

(meaning that a factor of \sin x is overlooked)

Then

\dfrac{\mathrm d}{\mathrm dx}(\sec x)\times\cos x-\sec x\times\sin x=0

\implies\dfrac{\mathrm d}{\mathrm dx}(\sec x)\times\cos x=\sec x\times\sin x

\implies\dfrac{\mathrm d}{\mathrm dx}(\sec x)=\dfrac{\sec x\times\sin x}{\cos x}

\implies\dfrac{\mathrm d}{\mathrm dx}(\sec x)=\sec x\times\tan x

6 0
20 days ago
the moon is about 240,000 miles from earth .What is the distance written as a whole number multiplied by a power of ten?
PIT_PIT [12445]
The distance to the moon is approximately 240,000 miles. This can be expressed as 2.4 multiplied by 10 raised to the power of 5.
8 0
1 month ago
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