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salantis [7]
3 years ago
13

PLEASE HELP !!

Mathematics
2 answers:
Pavlova-9 [17]3 years ago
6 0
Log 10  100 =2
Basically, we have to know how to solve for x in this equation:
10^x = 100

For log3 = 9 we have to solve for x in this equation:
3^x = 9
Therefore, x = 2


maw [93]3 years ago
3 0
Log base 3 (9) can be rewritten as 3^x = 9. Of course 3^2 = 9 so x = 2, which is B above.
You might be interested in
Nadine can send or receive a text message for $0.15 or get an unlimited number for $5.00. Write and solve an inequality to deter
34kurt

Answer:

0.15*t < 5.00

33 texts

Step-by-step explanation:

given : $0.15 per text; unlimited cost is $5.00

# of texts = t

0.15*t < 5.00

t < 33.33; since you cannot make 33.33 texts i will round down to 33 texts.

Basically you can send or receive up to 33 texts and it will be cheaper than $5.00

0.15(33) = $4.95

at 34 texts, 0.15(34) = $5.10

8 0
3 years ago
Order the fractions from least to greatest: 2/3,5/6,29/36,8/9. Briefly explain your method.
elena-14-01-66 [18.8K]

Answer: 2/3, 29/36, 5/6, 8/9

Step-by-step explanation:

You find the common denominator you could use for all fractions. In this instance, you can use 36 for the common denominator.

2/3: 36/3=12, 2x12=24 therefore 2/3 becomes 24/36.

29/36: you leave this one alone, it has already been simplified.

5/6: 36/6=6, 6x5=30 therefore 5/6 becomes 30/36.

8/9: 36/9=4, 8x4=32 therefore 8/9 becomes 32/26.

4 0
3 years ago
Define $g$ by $g(x)=5x-4$. If $g(x)=f^{-1}(x)-3$ and $f^{-1}(x)$ is the inverse of the function $f(x)=ax+b$, find $5a+5b$.
Slav-nsk [51]

Answer:

<h2>2</h2>

Step-by-step explanation:

Given g(x)=5x-4 and g(x)=f^{-1}(x)-3

Substituting g(x) =  5x-4 into the second equation we have;

5x-4 =  f^{-1}(x)-3

f^{-1}(x) = 5x-4+3

f^{-1}(x) = 5x-1

To get f(x), let us first make y to be equal  f^{-1}(x)

y = 5x-1

expressing x in terms of y to get f(x), we have;

5x = y+1

x = y/5+1/5

replacing y with x, we will have;

y = x/5 + 1/5

F(x) = x/5 + 1/5

Comparing x/5 + 1/5 with ax+b, a = 1/5 and b = 1/5

5a + 5b = 5(1/5)+ 5(1/5)

5a+5b = 1+1

5a+5b = 2

8 0
3 years ago
Which of the following function types exhibit the end behavior f(x)--&gt;0 as x --&gt; -infinity?
ipn [44]
Let's consider the functions one by one.

i) y=x^n,

x --> -infinity means that x is a very very small negative number. To model it in our mind let's think of -10^{15}. This number to an even power becomes 10^{30}, 10^{60}... etc.

Indeed, we can see that the smaller the x, the greater the value x^n. In fact, as x --> -infinity,  f(x)-->+infinity.

ii)
y=x, 

this clearly means that the behaviors of x and y are identical.

As x --> -infinity, y --> -infinity as well.


iii) y=|x|,

We can think of x --> -infinity, again, as a very very small number, like -10^{15}. For this value of x, y is |-10^{15}|, that is 10^{15}.

Indeed, the smaller the x, the greater the y. As in the function in part (i), as x --> -infinity,  f(x)-->+infinity.

iv) 

y=1/x

consider the values of y for the following values of x:

for x=-10, -100, -1000, y is respectively -0.1, -0.01, -0.001.

We can see that the smaller the x, the closer y gets to 0.

Thus, <span>f(x)-->0 as x --> -infinity
</span>
v) n'th root of x in not defined for negative x, when n is even.

vi) y=b^x, b>0

Here note that b cannot be equal to 1, otherwise the function is not exponential.

Let b=5, consider the values of y for the following values of x:

for x=-10, -100, -1000, y is respectively \displaystyle{ \frac{1}{5^{10}},  \frac{1}{5^{100}}, \frac{1}{5^{1000}}.

That is, the smaller the value of x, the closer y gets to 0.

Thus, <span>f(x)-->0 as x --> -infinity</span>
5 0
3 years ago
If x varies directly as y, and x=12 when y=3, find x when y=7
Stella [2.4K]

Answer:

28

Step-by-step explanation:

x=12 and y=3

x=4 and y =1

x=28 and y=7

3 0
3 years ago
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