There are 12 donuts in one dozen donuts. Which equation could be used to find the total (T) number of donuts (d) in 6 dozen?

Which graph represents the inequality y>3−x?

nekit [7.7K]

Answer:

B

Step-by-step explanation:

The first step is to draw the line y = 3 - x to see what the line itself looks like. Is it going from left to right as in A and C is going up as you go from right to left as in B and D? The graph on the left (below) gives you the answer. It is going up as you progress from right to left.

The next step is to answer which is it: B or D.

y has to be above the line. it is B. See the graph below on the right.

4 0

3 years ago

I need help with this because I'm not very good at understanding this.

HACTEHA [7]

Answer:

the initial value is 2

Step-by-step explanation:

"y=mx+b" is the equation

"b" is the initial value, and there is no "b" value

4 0

3 years ago

Simplify the expression:<br> 3(3 + 7b)

cupoosta [38]

Answer:

The answer is 9 + 21b

Step-by-step explanation:

Simplify the expression.

~Hoped this helped~

4 0

3 years ago

Read 2 more answers

20 Points! Which ordered pair is the best estimate for the solution of the system of equations?

yan [13]

Answer:

The best estimate for the solution is the ordered pair $(-6.5,-3.5)$

Step-by-step explanation:

we have

$y=\frac{3}{2}x+6$ ------> equation A

$y=\frac{1}{4}x-2$ ------> equation B

we know that

using a graphing tool, the solution of the system of equations is the intersection point both graphs

The intersection point is $(-6.4,-3.6)$

therefore

The best estimate for the solution is the ordered pair $(-6.5,-3.5)$

3 0

3 years ago

How many nonzero terms of the Maclaurin series for ln(1 x) do you need to use to estimate ln(1.4) to within 0.001?

Vilka [71]

Answer:

The estimate of In(1.4) is the first five non-zero terms.

Step-by-step explanation:

From the given information:

We are to find the estimate of In(1 . 4) within 0.001 by applying the function of the Maclaurin series for f(x) = In (1 + x)

So, by the application of Maclurin Series which can be expressed as:

$f(x) = f(0) + \dfrac{xf'(0)}{1!}+ \dfrac{x^2 f"(0)}{2!}+ \dfrac{x^3f'(0)}{3!}+... \ \ \ \ \ --- (1)$

Let examine f(x) = In(1+x), then find its derivatives;

f(x) = In(1+x)

$f'(x) = \dfrac{1}{1+x}$

f'(0) $= \dfrac{1}{1+0}=1$

f ' ' (x) $= \dfrac{1}{(1+x)^2}$

f ' ' (x) $= \dfrac{1}{(1+0)^2}=-1$

f ' ' '(x) $= \dfrac{2}{(1+x)^3}$

f ' ' '(x) $= \dfrac{2}{(1+0)^3} = 2$

f ' ' ' '(x) $= \dfrac{6}{(1+x)^4}$

f ' ' ' '(x) $= \dfrac{6}{(1+0)^4}=-6$

f ' ' ' ' ' (x) $= \dfrac{24}{(1+x)^5} = 24$

f ' ' ' ' ' (x) $= \dfrac{24}{(1+0)^5} = 24$

Now, the next process is to substitute the above values back into equation (1)

$f(x) = f(0) + \dfrac{xf'(0)}{1!}+ \dfrac{x^2f' \ '(0)}{2!}+\dfrac{x^3f \ '\ '\ '(0)}{3!}+\dfrac{x^4f '\ '\ ' \ ' \(0)}{4!}+\dfrac{x^5f' \ ' \ ' \ ' \ '0)}{5!}+ ...$

$In(1+x) = o + \dfrac{x(1)}{1!}+ \dfrac{x^2(-1)}{2!}+ \dfrac{x^3(2)}{3!}+ \dfrac{x^4(-6)}{4!}+ \dfrac{x^5(24)}{5!}+ ...$

$In (1+x) = x - \dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}- \dfrac{x^6}{6}+...$

To estimate the value of In(1.4), let's replace x with 0.4

$In (1+x) = x - \dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}- \dfrac{x^6}{6}+...$

$In (1+0.4) = 0.4 - \dfrac{0.4^2}{2}+\dfrac{0.4^3}{3}-\dfrac{0.4^4}{4}+\dfrac{0.4^5}{5}- \dfrac{0.4^6}{6}+...$

Therefore, from the above calculations, we will realize that the value of $\dfrac{0.4^5}{5}= 0.002048$ as well as $\dfrac{0.4^6}{6}= 0.00068267$ which are less than 0.001

Hence, the estimate of In(1.4) to the term is $\dfrac{0.4^5}{5}$ is said to be enough to justify our claim.

∴

The estimate of In(1.4) is the first five non-zero terms.

8 0

3 years ago