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Virty [35]
3 years ago
15

3. A driver's path when diving off of a platform is given by h(t)=-5t^2+10t+20, where h(t) is the distance above water, in feet,

and t is the time from the beginning of the dive, in seconds.
3a. How high is the diving platform?
3b. After how many seconds is the diver 25 feet above water?
3c. When does the diver enter the water?
Mathematics
1 answer:
agasfer [191]3 years ago
5 0

Answer:

3a.=20ft  3b.=1 second 3c.=3.25seconds

Step-by-step explanation:

Hello, Courtney!

h(t)=-5t^2+10t+20

20 would be the height so 3.a would be 20ft

3b.  25=-5t^2+10t+20

-5t^2+10t-5 then divide by greatest factor the answer would 1 second

3c.    0=-5t^2+10t+20 divide by greatest factor and would get 3.25 and -1.25 but can't have negative so the answer is 3.25.

I hope this helps

best you know who!

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What is -x/3 = 9. Solve for x
vlada-n [284]

Note: I just noticed that you only have one negative sign. If you mean to put -9, change the sign to positive. If not, leave the answer.

Note the equal sign. what you do to one side, you do to the other. Isolate the x. First, multiply 3 to both sides

-x/3(3) = 9(3)

-x = 9(3)

-x = 27

Next, to isolate the x, divide -1 from both sides

-x/-1 = 27/-1

x = 27/-1

x = -27

-27 is your answer for x.

hope this helps

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The perimeter of a rectangular garden is 322 feet. If the width of the garden is 72 feet, what is the length?
zhannawk [14.2K]
The length is gonna be 89 ft 
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If logb2=x and logb3=y, evaluate the following in terms of x and y:
Alja [10]

log_b{162} = x + 4y\\\\log_b324 = 2x+4y\\\\log_b\frac{8}{9} = 3x-2y\\\\\frac{log_b27}{log_b16} = 3y-4x

<em><u>Solution:</u></em>

Given that,

log_b2 = x\\\\log_b3 = y --------(i)

<em><u>Use the following log rules</u></em>

Rule 1: log_b(ac) = log_ba + log_bc

Rule 2: log_b\frac{a}{c} = log_ba - log_bc

Rule 3: log_ba^c = clog_ba

a) log_b{162}

Break 162 down to primes:

162 = 2^1 \times 3^4

log_b{162} =log_b 2^1. 3^4\\\\By\ rule\ 1\\\\ log_b{162} = log_b 2^1 +log_b 3^4\\\\By\ rule\ 3\\\\1log_b2 + 4log_b3\\\\1x+4y\\\\x+4y

Thus we get,

log_b162 = x + 4y

Next

b) log_b 324

Break 324 down to primes:

324 = 2^2 \times 3^4

log_b324 = log_b 2^2.3^4\\\\By\ rule\ 1\\\\log_b324 = log_b2^2 + log_b3^4\\\\By\ rule\ 3\\\\log_b324 = 2log_b2 + 4log_b3\\\\From\ (i)\\\\log_b324 = 2x + 4y

Next

c) log_b\frac{8}{9}

By rule 2

log_b\frac{8}{9} = log_b8 - log_b9\\\\log_b\frac{8}{9} = log_b 2^3 - log_b3^2\\\\By\ rule\ 3\\\\log_b\frac{8}{9} =  3 log_b2 - 2log_b3\\\\From\ (i)\\\\log_b\frac{8}{9} =  3x - 2y

Next

d) \frac{log_b27}{log_b16}

By rule 2

\frac{log_b27}{log_b16} = log_b27 - log_b16\\\\ \frac{log_b27}{log_b16} = log_b3^3 - log_b2^4\\\\By\ rule\ 2\\\\ \frac{log_b27}{log_b16} = 3log_b3 - 4log_b2 \\\\From\ (i)\\\\\frac{log_b27}{log_b16} =  3y - 4x

Thus the given are evaluated in terms of x and y

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