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ololo11 [35]
3 years ago
8

A ball is hit into the air with an initial velocity of 100 ft/s. Its height, h(t), after t seconds is given by the function h(t)

= -16t² + 100t + 5. What is the ball's maximum height?
Mathematics
1 answer:
Lana71 [14]3 years ago
8 0

at maximum. dt/dh=0

if you differentiate you will get

-32t=-100

t=3.125

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Function h(x) = x + 4<br> If your input was 2, what is your output?
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h(2)=6

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Jessica wants to plan where she will place furniture in her room, so she has drawn a scale model of her room in order to begin p
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To solve this problem you must apply the proccedure shown below:
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5 0
3 years ago
Anyone can help me solve this equation using cross multiplying
Natali5045456 [20]

9514 1404 393

Answer:

  x = 1 or 5

Step-by-step explanation:

The notion of "cross-multiplying" is the idea that the numerator on the left is multiplied by the denominator on the right, and the numerator on the right is multiplied by the denominator on the left. This looks like ...

  \displaystyle \frac{x-1}{7}=\frac{2x-2}{3x-1}\ \longrightarrow\ (x-1)(3x-1)=(7)(2x-2)

Then the solution proceeds by eliminating parentheses, and solving the resulting quadratic equation.

  3x^2-4x+1=14x-14\\\\3x^2-18x+15=0\qquad\text{subtract $14x-14$}\\\\x^2-6x+5=0 \qquad\text{divide by 3}\\\\(x-1)(x-5)=0\qquad\text{factor}\\\\x\in\{1,5\}

_____

<em>Comment on "cross multiply"</em>

Like a lot of instructions in Algebra courses, the idea of "cross multiply" describes <em>what the result looks like</em>. It doesn't adequately describe how you get there. The <em>one and only rule</em> in solving Algebra problems is "<em>whatever is done to one side of the equation must also be done to the other side of the equation</em>." If you multiply one side by one thing and the other side by a different thing, you are violating this rule.

What looks like "cross multiply" is really "<em>multiply by the product of the denominators</em> and cancel like terms from numerator and denominator." Here's what that looks like with the intermediate steps added.

  \displaystyle \frac{x-1}{7}=\frac{2x-2}{3x-1}\\\\\frac{x-1}{7}\times7(3x-1)=\frac{2x-2}{3x-1}\times7(3x-1)\\\\(x-1)(3x-1)=(2x-2)(7)\qquad\textit{looks like}\text{ cross multiply}

8 0
2 years ago
Write a situation for 15 x - 20 is less than or equal to 130 and solve
ankoles [38]
15x-20<130
15x<130+20
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X=150/15
x<10
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3 years ago
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