What is the spread of Cassandra’s grades?

Mathematics

2 answers:

Ksju [112]3 years ago

5 0

Answer:i THINK 79, i dont know fully, but i think 79

OLEGan [10]3 years ago

3 0

79 sorry if i am wrong :)

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Help me ASAP please

Harrizon [31]

Answer:

$\large\boxed{7^{-8}}$

Step-by-step explanation:

$7^{-1}\cdot7^{-7}\qquad\text{use}\ a^n\cdot a^m=a^{n+m}\\\\=7^{-1+(-7)}=7^{-8}$

6 0

3 years ago

I really dont know how to start this Calculus 1 question!

hichkok12 [17]

According to the given plot, $f'(1)=2$ . Then the linear approximation to $f(x)$ at $x=1$ is

$L(x)=f(1)+f'(1)(x-1)=6+2(x-1)=2x+4$

Then

$f(0.95)\approx L(0.95)=5.9$

$f(1.05)\approx L(1.05)=6.1$

On the interval [0, 1], the plot of $f'(x)$ is positive, so $f(x)$ is an increasing function here. But we can see that $f'(x)$ is approaching 0. This means tangent lines to $f(x)$ have a positive slope, but the slopes are approaching 0 and are thus becoming less steep. This in turn means the tangent lines lie above the curve, so the approximations are greater than the actual values of $f(0.95)$ and $f(1.05)$ .