35x^7+15x^5-10x^3/5x^3

The probability of randomly choosing a blue marble, not replacing it, and then choosing a green marble is,

⇒ 2 / 19

The term probability refers to the likelihood of an event occurring. Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.

Given that;

A jar contains 5 blue marbles, 7 yellow marbles, and 8 green marbles.

Hence, Number of marbles = 5 + 7 + 8 = 20

Thus, the probability of randomly choosing a blue marble is,

⇒ 5 / 20

⇒ 1 / 4

And, the probability of randomly choosing a green marble is,

⇒ 8 / 19

Thus, the probability of randomly choosing a blue marble, not replacing it, and then choosing a green marble is,

⇒ 1/4 × 8/19

⇒ 2 / 19

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Using the equation 't = d/v', where 'd' is distance and 'v' is velocity, the time 't' for a dragonfly to fly 50 feet at 25 feet per second is calculated to be 2 seconds.

To find the time it takes for a dragonfly to fly a distance of 50 feet at a speed of 25 feet per second, we can use the formula d = vt, where d is the distance, v is the velocity (speed), and t is the time. To solve for time (t), we rearrange the formula to t = d/v.

So, plugging in the values:

We get:

t = d/v = 50 feet / 25 feet/second = 2 seconds

The dragonfly takes 2 seconds to fly 50 feet at a speed of 25 feet per second.

Two pairs of fractions that meet the criteria are 3/4 and 1/4, and 5/6 and 1/3, as the difference of each pair reduces to 1/2.

Finding two fractions with a difference of 1/2 where neither denominator is equal to 2 requires a bit of trial and error. An example of this could be the two fractions 3/4 and 1/4. If you subtract 1/4 from 3/4, you get 2/4, which reduces to 1/2. This meets the requirements of the question as neither denominator is equal to 2 and the difference of the two fractions is 1/2.

For another example, consider the fractions 5/6 and 1/3. Subtract 1/3 from 5/6, and your result is 3/6, which likewise reduces to 1/2. Again, this answer meets the criteria set by the question.

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The perimeter of the right triangle WXY is 60 units.

To find the perimeter of a right triangle WXY, we need to know the lengths of the two legs. Given that the radius of the circle is 4 units and WY is 20 units, we can determine that the hypotenuse, XY, is equal to 20 units. So the perimeter of the triangle is the sum of the lengths of its three sides: 20 + 20 + 20 = 60 units.

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We calculate the perimeter of the right triangle by first finding the lengths of the legs using the Pythagorean theorem and the given radius, then summing these lengths with the hypotenuse.

The subject of this question is Geometry, specifically involving the calculation of the perimeter of a right triangle. In a right triangle, the lengths of the three sides follow the Pythagorean theorem, written as [tex]a^2 + b^2 = c^2[/tex], where c is the hypotenuse and a and b are the lengths of the other two sides.

Here, the length of the hypotenuse WY is given as 20 units. In a right-angled triangle, where a circle is inscribed, the radius of the circle is equal to half the sum of the two legs of the triangle (radius = (a+b)/2). In this case, the radius is given as 4 units. Thus, the total of the two legs a and b is 8 units.

The Pythagorean theorem can now be used to solve for a and b, knowing that [tex]a^2 + b^2 = 20^2[/tex] and a + b = 8; then to find the perimeter, sum the hypotenuse and the lengths of the legs, i.e. Perimeter = a + b + c.

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Answer:

false

Step-by-step explanation:

fr ong

Answer: 18 - apex

Step-by-step explanation:

The distance between the first base and third base in the given problem is 127.28 feet.

Distance is a measure of how far apart two objects are. It's a scalar quantity, which means that it has magnitude but no direction. The distance between two points can be gauged in various ways, including using a ruler, a tape measure, or a laser rangefinder. The primary difference between the distance and the displacement is that the latter has the direction but not the former.

As evident from the figure,

the running track is a square.

So, all the sides are equal to each other in length.

So, the distance between the first base and the second base is equal to the distance between that of the second base and the third base which is equal to 90 feet.

Let us name the square ABCD.

Join vertex A and vertex D.

The triangle ABC is the right-angled isosceles triangle.

So, by Pythagorean Theorem,

[tex]\sqrt{AC^2}[/tex] = [tex]\sqrt{AB^2+BC^2}[/tex]

          = [tex]\sqrt{90^2 + 90^2}[/tex]

          = [tex]\sqrt{8100+8100}[/tex]

          = [tex]\sqrt{16200}[/tex]

          = 127.28 (approximately)

So, the first base and third base are 127.28 feet apart.

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[tex] |\Omega|=4\cdot12=48\\
|A|=3\cdot3.14\cdot2^2=37.68\\\\
P(A)=\dfrac{37.68}{48}\approx0.79 [/tex]

Answer: 77.9

Step-by-step explanation: .

There is 4 roll of coins of nickel and 3 rolls of coins of dimes.

Given that,

Joanna has a total of 7 rolls of coins worth $23.

She has rolls of dimes worth $5 each and rolls of nickel worth $2.

We have to determine,

How many rolls of each kind does she have?

According to the question,

Let the number of coins of dimes be x

And the number of coins of nickel be y.

Then,

The number of coins of dimes + Number of coins of nickel = Total number of coins.

[tex]\rm x+ y =7[/tex]

And the total price of the coins is equal to the price of each coin of dimes and nickel.

[tex]\rm 5x+2y=23[/tex]

Solving both the equation,

From equation 1,

[tex]\rm x+ y = 7\\\\x = 7-y[/tex]

Substitute the value of x in equation 2,

[tex]\rm 5x+2y = 23\\\\5(7-y) +2y = 23\\\\35-5y+2y=23\\\\-3y = 23-35\\\\-3y = -12\\\\y = \dfrac{-12}{-3}\\\\y = 4[/tex]

Substitute the value of y in equation 1,

[tex]\rm x+y = 7\\\\x+4 = 7\\\\x = 7-4\\\\x =3[/tex]

Hence, there is 4 roll of coins of nickel and 3 rolls of coins of dimes.

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Set of sides that will make right triangle is

Set 4 : 12 ft, 16 ft, 20 ft

Pythagoras theorem states that in a right angled triangle square of the hypotenuse is equal to the sum of the square of perpendicular and square of base. The longest side is the hypotenuse.

Given measurements

Set 1:

4cm , 5cm , 6cm

By Pythagoras theorem

4² + 5² = 6²

16 + 25 = 36

41 ≠ 36

Pythagoras theorem is not satisfied,

this set of sides will not make right triangle

Set 2

8 in., 12 in., 20 in.

By Pythagoras theorem

8² + 12² = 20²

64 + 144 = 400

208 ≠ 400

Pythagoras theorem is not satisfied,

this set of sides will not make right triangle

Set 3

10 mm, 20 mm, 30 mm

By Pythagoras theorem

10² + 20² = 30²

100 + 400 = 900

500 ≠ 900

Pythagoras theorem is not satisfied

this set of sides will not make right triangle

Set 4

12 ft, 16 ft, 20 ft

By Pythagoras theorem

12² + 16² = 20²

144 + 256 = 400

400 = 400

Pythagoras theorem is satisfied

this set of sides will make right triangle.

Hence, Set 4 of sides 12 ft, 16 ft, 20 ft will form a right triangle.

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The Weight on Venus is 3.64 pounds

The unitary technique involves first determining the value of a single unit, followed by the value of the necessary number of units.

For example, Let's say Ram spends 36 Rs. for a dozen (12) bananas.

12 bananas will set you back 36 Rs. 1 banana costs 36 x 12 = 3 Rupees.

As a result, one banana costs three rupees. Let's say we need to calculate the price of 15 bananas.

This may be done as follows: 15 bananas cost 3 rupees each; 15 units cost 45 rupees.

Given:

They need to find out how much a rock that weights 4 pounds on earth would weight on Venus.

So, the Weight on Venus

= 4 x 0.91

= 4x 0.9+ 4 x 0.01

= 3.6+ 0.04

= 3.64

Hence, the Weight on Venus is 3.64 pounds

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Answer:31.88

Step-by-step explanation:

a. The height of the ball when it leaves the child's hand is 3 feet, as [tex]\(x = 0.[/tex]

b. The maximum height of the ball is 195 feet, calculated by finding the vertex of the parabolic function.

c. The distance from the child where the ball strikes the ground is approximately 47.24 feet, using the quadratic formula.

Given the equation representing the height y of a ball as a function of the horizontal distance x :

[tex]\[y = \frac{1}{16} x^2 + 4x + 3\][/tex]

Part (a): Height of the ball when it leaves the child's hand

The ball leaves the child's hand when x = 0.

Substituting x = 0 into the equation:

[tex]\[y = \frac{1}{16} \cdot 0^2 + 4 \cdot 0 + 3 = 3\][/tex]  

Thus, the height of the ball when it leaves the child's hand is 3 feet.

Part (b): Maximum height of the ball

The given function represents a parabola, and since [tex]\(a = \frac{1}{16}[/tex] is positive, it opens upward.

To find the maximum height, we need the x-coordinate of the vertex, which can be found using:

[tex]\[x = -\frac{b}{2a} = -\frac{4}{2 \cdot \frac{1}{16}} = -\frac{4}{\frac{1}{8}} = -32\][/tex]

However, the correct derivation is:

[tex]\[x = -\frac{4}{2 \cdot \frac{1}{16}} = -\frac{4}{\frac{1}{8}} = -32 \cdot 8 = -64\][/tex]

But the correct value for x should be:

[tex]\[x = -\frac{4}{2 \times \frac{1}{16}} = -4 \times 8 = -32[/tex]

Considering this, let's correct the previous explanation and use the correct approach:

1. Find the x-coordinate of the vertex:

[tex]\[x = -\frac{4}{2 \times \frac{1}{16}} = -8 = 32\][/tex]

With the x-coordinate of the vertex, find the maximum height (the y-coordinate of the vertex):

[tex]\[y = \frac{1}{16} \cdot 32^2 + 4 \cdot 32 + 3 = \frac{1024}{16} + 128 + 3 = 64 + 128 + 3 = 195\][/tex]

Thus, the maximum height is 195 feet

Part (c): How far from the child does the ball strike the ground?

To find out when the ball strikes the ground, we need to set \(y = 0\) and solve for \(x\):

[tex]\[0 = \frac{1}{16} x^2 + 4x + 3\][/tex]

Using the quadratic formula, where[tex]\(a = \frac{1}{16}\), \(b = 4\), and \(c = 3\):[/tex]

[tex]\[x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot \frac{1}{16} \cdot 3}}{2 \cdot \frac{1}{16}} = \frac{-4 \pm \sqrt{16 - \frac{12}{16}}}{\frac{1}{8}} \][/tex]

To find the x-coordinates where the ball strikes the ground, you can set (y = 0) and solve for (x):

[tex]\[0 = \frac{1}{16} x^2 + 4x + 3\][/tex]

Using the quadratic formula, where [tex]\(a = \frac{1}{16}\), \(b = 4\), and \(c = 3\) :[/tex]

[tex]\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\][/tex]

Substitute the given values of a ,b , and c into the quadratic formula:

Calculate [tex]\(b^2 - 4ac\) :[/tex]

  - [tex]\(b^2 = 4^2 = 16\),[/tex]

  - [tex]\(4 \cdot \frac{1}{16} \cdot 3 = \frac{12}{16} = \frac{3}{4}\),[/tex]

  - So [tex]\(b^2 - 4ac = 16 - \frac{3}{4} = \frac{64}{4} - \frac{3}{4} = \frac{61}{4}\).[/tex]

Calculate the roots :

  - The value for [tex]\(a\) is \(\frac{1}{16}\), so \(2a = \frac{1}{8}\),[/tex]

  - Now we can find \(x\):

   [tex]\[ x = \frac{-4 \pm \sqrt{16 - \frac{3}{4}}}{\frac{1}{8}} = \frac{-4 \pm \sqrt{61/4}}{1/8} = -32 \pm 2 \sqrt{61}. \][/tex]

Simplifying the roots :

  - The roots for x are:

   [tex]\[ x_1 = -32 + 2 \sqrt{61}, \, x_2 = -32 - 2 \sqrt{61}. \][/tex]

However, we consider only the positive root, as it represents the distance from the point where the ball is thrown to where it lands.

Thus, the value of [tex]\(x_1\)[/tex] is:

[tex]\[x = -32 + 2 \sqrt{61} \approx -32 + 15.6 = -16 + 16 \cdot \sqrt{61} \approx 47.24\][/tex]

Thus, the correct answer with [tex]\(x\approx 47.24\).[/tex]

The complete question is : The height y (in feet) of a ball thrown by a child is

[tex]y = (1)/(16) {x}^{(2)}+ 4x + 3[/tex]

where x is the horizontal distance in feet from the point at which the ball is thrown

a).how high is the ball when it leaves the child's hand ?

b).what is the maximum high of the ball ?

c).how far from the child does the ball strike the ground ?

The answer provides step-by-step explanations for finding the initial height, maximum height, and horizontal distance at which the ball strikes the ground, given the equation of the ball's height as a function of the horizontal distance.

To find the height of the ball when it leaves the child's hand (a), we need to determine the value of y when x is 0. Substituting x = 0 into the equation, we get:

y = -1/14(0)² + 2(0) + 3

y = 3 feet

Therefore, the ball is 3 feet high when it leaves the child's hand.

let's find the maximum height of the ball (b), we need to find the vertex of the quadratic equation. The vertex can be found using the formula x = -b/2a, where a, b, and c are the coefficients of the quadratic equation. In this case, a = -1/14, b = 2, and c = 3. Substituting these values into the formula, we get:

x = -2 / (2 * (-1/14))

x = 7 feet

Substituting x = 7 into the equation, we can find the maximum height:

y = -1/14(7)² + 2(7) + 3

y = 10.5 feet

Therefore, the maximum height of the ball is 10.5 feet.

let's find how far from the child the ball strikes the ground (c), we need to find the x-coordinate when y = 0. Setting y = 0 in the equation, we get:

0 = -1/14x² + 2x + 3

Solving this quadratic equation,

hence we find two possible values for x: x = -7 and x = 28. Since the ball is thrown in a positive direction, the ball strikes the ground 28 feet away from the child.

Answer: –18 feet

Step-by-step explanation:

Given: Garrett begins 18 feet above water level to inspect the lighting, descends 32 feet to inspect the base of the bridge underwater, then descends another 4 feet to check on the concrete anchoring the base.

Then using integers, the expression which represents Garrett’s position with respect to the water level is given by :-

[tex]+18-32-4=-18[/tex]

Hence, Garrett’s position with respect to the water level =-18 feet.

Answer:

And.Each Trip will have to take 41.667kilograms

Step-by-step explanation:

1/6×1000÷4=41.6666667