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kolbaska11 [484]
2 years ago
11

How do I write names for 0.550

Mathematics
1 answer:
mojhsa [17]2 years ago
3 0

Answer:

  five hundred fifty thousandths

Step-by-step explanation:

Recognize the place value of the rightmost digit, and write the name of the number that ends in that place.

The third place to the right of the decimal point is the <em>thousandths</em> place. The number between that and the decimal point is 550, <em>five hundred fifty</em>.

  five hundred fifty thousandths

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Describe how (2 ^ 3)(2 ^ - 4) can be simplified . Multiply the bases and add the exponents. Then find the reciprocal and change
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Answer:

Add the exponents and keep the same base Then find the reciprocal and change the sign of the exponent .

Step-by-step explanation:

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Answer:

a) -16t^{2}+64t\geq40

b)[2-\frac{\sqrt{6}}{2} , 2+\frac{\sqrt{6}}{2}]

(see attached picture for the number line)

This interval means that from about 0.78s to about 3.22s the ball will not be visible because it will be higher than 40 feet high.

c) h_{max}=64ft

d) t=2s

Step-by-step explanation:

a)

Basically the ball will disappear when it is higher than 40 ft, so we can build the inequality like this:

-16t^{2}+64t\geq40

this is because the function represents it's height, so when the function is greater than or equal to 40, then the ball will disappear.

b) to solve the inequality, first we need to turn the ≥ symbol to an = symbol and solve for t, so we get:

-16t^{2}+64t =40

we can move the 40 to the other side of the equation so we get:

-16t^{2}+64t-40=0

we can factor the equation so it's easier to solve, so we get:

-8(2t^{2}-8t+5) =0

which simplifies to:

2t^{2}-8t+5=0

and we can use the quadratic formula to solve this equation, so we get:

t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}

so we get:

t=\frac{-(-8)\pm\sqrt{(-8)^{2}-4(2)(5)}}{2(2)}

which yields:

2\pm \frac{\sqrt{6}}{2}

and next we need to test the posible intervals to see which one makes the inequality true, the possible intervals are:

(-\infty, 2-\frac{\sqrt{6}}{2}]  for a test value of 0

[2-\frac{\sqrt{6}}{2},2+\frac{\sqrt{6}}{2}] for a test value of 1

[2+\frac{\sqrt{6}}{2},\infty) for a test value of 4

so next we test each of the values in the original inequality and see if the inequality is true:

(-\infty, 2-\frac{\sqrt{6}}{2}] for a test value of 0

-16(0)^{2}+64(0)\geq40

0\geq40 false

so this is not an answer.

[2-\frac{\sqrt{6}}{2},2+\frac{\sqrt{6}}{2}] for a test value of 1

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t=-\frac{64}{2(-16)}

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h=-16(2)^{2}+64(2)

which yields:

h=64 ft

d) We found the answer for d on the previous part where we got that t=2s

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