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hichkok12 [17]
3 years ago
12

I know the answers but i need the work. plz help

Mathematics
1 answer:
inysia [295]3 years ago
5 0
For number 3.

You add 8 to both sides as you need to separate the variable

4x-8=32
+8 +8

4x=32

now divide 32 by 4

x=8

for number 4

you use pemdas to multiply the parenthisis

2(x-5)=-20

(2•x)-(2•5)

2x-10=-20

then add 10 to both sides

2x-10=-20
+10 +10

2x=10

divide 10 by two

x=5


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If log2 5 = k, determine an expression for log32 5 in terms of k.
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Answer:

log_3_2(5)=\frac{1}{5} k

Step-by-step explanation:

Let's start by using change of base property:

log_b(x)=\frac{log_a(x)}{log_a(b)}

So, for log_2(5)

log_2(5)=k=\frac{log(5)}{log(2)}\hspace{10}(1)

Now, using change of base for log_3_2(5)

log_3_2(5)=\frac{log(5)}{log(32)}

You can express 32 as:

2^5

Using reduction of power property:

log_z(x^y)=ylog_z(x)

log(32)=log(2^5)=5log(2)

Therefore:

log_3_2(5)=\frac{log(5)}{5*log(2)}=\frac{1}{5} \frac{log(5)}{log(2)}\hspace{10}(2)

As you can see the only difference between (1) and (2) is the coefficient \frac{1}{5} :

So:

\frac{log(5)}{log(2)} =k\\

log_3_2(5)=\frac{1}{5} \frac{log(5)}{log(2)} =\frac{1}{5} k

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3 years ago
A ball is thrown into the air with an upward
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To find the 'maximum height' we will need to take the derivative of h(t) = –16t² + 32t + 6 then set it equal to zero, then solve for t. this t will be the time at which the ball reaches it's maximum height.
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I need help asap :):)/
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Answer:

35

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