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GenaCL600 [577]

3 years ago

10

I'll pay someone to do my math <3

1 answer:

ss7ja [257]3 years ago

8 0

I would like to be paid, then xD

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The equation 3s equals 21 can be used to find the number of students s and each van on a field trip how many students are in eac

zubka84 [21]

3s=21
s=7
7 students are in each van.

6 0

3 years ago

Determine the domain and range of the following

Tasya [4]

Answer:

Domain: (-3,-2,0,1,2,3)

Range: (9,4,1,0,1,4,9)

Step-by-step explanation:

5 0

3 years ago

Find the linear approximation of the function g(x) = 3 1 + x at a = 0. g(x) ≈ Correct: Your answer is correct. Use it to approxi

EleoNora [17]

Answer:

$3.296x+3$

2.835

3.330

Step-by-step explanation:

$g(x) = 3^{1+x}$

Let the linear approximation be L(x) at a = 0. This is given by

$L(x) = g(a) + g'(a)(x-a)$

$g'(x)$ is the derivative of $g(x)$ . To find this, we use the form

If $f(x) = a^x$ , then $f'(x) =a^x\ln a$

By doing this and applying chain rule for the power (which is a function of x), we have

$g'(x) = 3^{1+x}\ln 3$

Then

$g'(0) = 3^{1+0}\ln 3 = 3.296$

Also $g(0) = 3^{1+0} = 3$

Hence $L(x) = 3+(x-0)\times3.296 = 3.296x + 3$

For $3^{0.95}$ , $1+x = 0.95$ and $x=-0.05$

Using this in $L(x)$ ,

$3^{0.95} = 3.296(-0.05) + 3 = 2.835$

For $3^{1.1}$ , $1+x = 1.1$ and $x=0.1$

Using this in $L(x)$ ,

$3^{1.1} = 3.296(0.1) + 3 = 3.330$

7 0

3 years ago

The temperature in Bloomington on Sunday was 32.6°C. On Monday, the temperature changed by -8.25°C. What was the temperature on

Afina-wow [57]

Subtract 8.25 from 32.6 (32.6-8.25) = 24.35

8 0

3 years ago

What values must x have so that log (x) is greater than 2? justify your answer

kotegsom [21]

Answer:

The value of x must be greater than 100.

Step-by-step explanation:

We need to find the value x for which the value of log(x) is greater than than 2.

$log(x)>2$

According to the property of logarithm, if a is a real number.

$log(x)=a \Rightarrow x=10^a$

Using the above property of logarithm, the given inequality can be rewritten as

$x>10^2$

$x>100$

We conclude that the value of log(x) is greater than 2 for real values of x which are greater than 100.

Therefore, the value of x must be greater than 100.

3 0

3 years ago