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2 years ago

11

There is space for 120 students to go on a field trip. Currently, 74 students have signed up. Write and solve an inequality to f

ind how many more students can sign up for the field trip.
Please help

1 answer:

sukhopar [10]2 years ago

5 0

Answer:

46

Step-by-step explanation:

$120 - 74 = 46$

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One leg of a right triangular piece of land has a length of 24 yards. They hypotenuse has a length of 74 yards. The other leg ha

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Solutions

To solve this problem we have to use the Pythagorean theorem. You can only use the Pythagorean theorem in Right Triangles. The longest side of the triangle is called the "hypotenuse". C² is the longest side so it is the hypotenuse . To calculate c² we have to do α² + β² = c².

Given

One leg of a right triangular piece of land has a length of 24 yards. They hypotenuse has a length of 74 yards. The other leg has a length of 10x yards.

First leg (24 yards) would be α

Second leg would be β

Hypotenuse (74 yards) would be c

Now we have points α β c.

a² (24) + β² ( x ) = c² (74)

Calculations

c² = α² + β²

74² = 24²+ β²

<span>5476 = 576 + </span>β²

5476 - 576 = β²
<span> </span>
<span>4900 = </span>β²

→√4900
<span> </span>
β<span> = 70 yards
</span>
<span>70 = 10x
</span>
<span>x = 70</span>÷<span>10 = 7 yards
</span>
The second leg = 7 yards

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3 years ago

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Step-by-step explanation:

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Write an equation for a linear function whose graph has the given characteristics.

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Answer:

y-1=1/4(x-8)

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3 years ago

A standard deck of playing cards has 52 cards: 13 spades, 13 clubs, 13 hearts, and 13 diamonds. What is the probability of drawi

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13/ 52 = 1/4 chances.

There are 13 spades and 52 cards. You divide the spades and cards in the deck to find your chances of drawing one.

You have a 25% chance of getting a spade.

Now you put it back. Since you put it back, you still have the same number of cards and the same number of spades. So when you divide 13/52 again, the numbers are still the same. 1/4. 25% chance.

So, you still have a 25% chance the second time around.

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3 years ago

A golfball is hit from the ground with an initial velocity of 200 ft/sec. The horizontal distance that the golfball will travel,

algol13

Answer:

The golfball launched with an initial velocity of 200ft/s will travel the maximum possible distance which is 1250 ft when it is hit at an angle of $\pi/4$ .

Step-by-step explanation:

The formula from the maximum distance of a projectile with initial height h=0, is:

$d(\theta)=\frac{v_i^2sin(2\theta)}{g}$

Where $v_i$ is the initial velocity.

In the closed interval method, the first step is to find the values of the function in the critical points in the interval which is $[0, \pi/2]$ . The critical points of the function are those who make $d'(\theta)=0$ :

$d(\theta)=\frac{v_i^2\sin(2\theta)}{g}\\d'(\theta)=\frac{v_i^2\cos(2\theta)}{g}*(2)\\d'(\theta)=\frac{2v_i^2\cos(2\theta)}{g}$

$d'(\theta)=0\\\frac{2v_i^2\cos(2\theta)}{g}=0\\\cos(2\theta)=0\\2\theta=\pi/2,3\pi/2,5\pi/2,...\\\theta=\pi/4,3\pi/4,5\pi/4,...$

The critical value inside the interval is $\pi/4$ .

$d(\theta)=\frac{v_i^2sin(2\theta)}{g}\\d(\pi/4)=\frac{v_i^2sin(2(\pi/4))}{g}\\d(\pi/4)=\frac{v_i^2sin(\pi/2)}{g}\\d(\pi/4)=\frac{v_i^2(1)}{g}\\d(\pi/4)=\frac{(200)^2}{32}\\d(\pi/4)=\frac{40000}{32}\\d(\pi/4)=1250ft$

The second step is to find the values of the function at the endpoints of the interval:

$d(\theta)=\frac{v_i^2sin(2\theta)}{g}\\\theta=0\\d(0)=\frac{v_i^2sin(2(0))}{g}\\d(0)=\frac{v_i^2(0)}{g}=0ft\\\theta=\pi/2\\d(\pi/2)=\frac{v_i^2sin(2(\pi/2))}{g}\\d(\pi/2)=\frac{v_i^2sin(\pi)}{g}\\d(\pi/2)=\frac{v_i^2(0)}{g}=0ft$

The biggest value of f is gived by $\pi/4$ , therefore $\pi/4$ is the absolute maximum.

In the context of the problem, the golfball launched with an initial velocity of 200ft/s will travel the maximum possible distance which is 1250 ft when it is hit at an angle of $\pi/4$ .

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3 years ago