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Nataliya [291]
3 years ago
9

How would you best describe the two students outside the circles?

Mathematics
1 answer:
ruslelena [56]3 years ago
5 0

Answer:

A. They are boys who are not in sixth grade and don't wear glasses.

Explanation:

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A blueprint shows the mater bedroom to be 5 inches long and 3 inches wide. The scale on the blueprint is 2 in : 11 ft. What is t
Anna35 [415]

Answer:

453.75 sq. feet

Step-by-step explanation:

In this question, we are going to calculate the area of a room given the blue print measurement and the actual measurements with the answer being in the actual measurement.

To correctly do this, it is best if the measurements are converted to the actual.

The measurements we’ll be trying to convert is 5 inches by 3 inches with a scale of 2in to 11ft ( same as 1 in to 5.5 ft)

The 5 inches blueprint measurement will be 5 * 5.5 = 27.5 ft

The 3 inches blueprint measurement will be 3 * 5.5 = 16.5 ft

The area in sq. feet will be length * width = 27.5 * 16.5 = 453.75 sq feet

7 0
3 years ago
Identify the phrase that cannot be described by the same absolute value as the other three. Explain your reasoning.
r-ruslan [8.4K]
The answer would probably be 18 degrees below normal because the absolute values of the rest are 8 not 18 which is the absolute value of 18 degrees below normal.
6 0
4 years ago
Derivative of tan(2x+3) using first principle
kodGreya [7K]
f(x)=\tan(2x+3)

The derivative is given by the limit

f'(x)=\displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}h

You have

\displaystyle\lim_{h\to0}\frac{\tan(2(x+h)+3)-\tan(2x+3)}h
\displaystyle\lim_{h\to0}\frac{\tan((2x+3)+2h)-\tan(2x+3)}h

Use the angle sum identity for tangent. I don't remember it off the top of my head, but I do remember the ones for (co)sine.

\tan(a+b)=\dfrac{\sin(a+b)}{\cos(a+b)}=\dfrac{\sin a\cos b+\cos a\sin b}{\cos a\cos b-\sin a\sin b}=\dfrac{\tan a+\tan b}{1-\tan a\tan b}

By this identity, you have

\tan((2x+3)+2h)=\dfrac{\tan(2x+3)+\tan2h}{1-\tan(2x+3)\tan2h}

So in the limit you get

\displaystyle\lim_{h\to0}\frac{\dfrac{\tan(2x+3)+\tan2h}{1-\tan(2x+3)\tan2h}-\tan(2x+3)}h
\displaystyle\lim_{h\to0}\frac{\tan(2x+3)+\tan2h-\tan(2x+3)(1-\tan(2x+3)\tan2h)}{h(1-\tan(2x+3)\tan2h)}
\displaystyle\lim_{h\to0}\frac{\tan2h+\tan^2(2x+3)\tan2h}{h(1-\tan(2x+3)\tan2h)}
\displaystyle\lim_{h\to0}\frac{\tan2h}h\times\lim_{h\to0}\frac{1+\tan^2(2x+3)}{1-\tan(2x+3)\tan2h}
\displaystyle\frac12\lim_{h\to0}\frac1{\cos2h}\times\lim_{h\to0}\frac{\sin2h}{2h}\times\lim_{h\to0}\frac{\sec^2(2x+3)}{1-\tan(2x+3)\tan2h}

The first two limits are both 1, and the single term in the last limit approaches 0 as h\to0, so you're left with

f'(x)=\dfrac12\sec^2(2x+3)

which agrees with the result you get from applying the chain rule.
7 0
3 years ago
What is 4/5 x 1 1/6=
andrew11 [14]
4/5 x 1 1/6 
<span>= 4/5 x (6 x 1 + 1)/6 </span>
<span>= 4/5 x 7/6 </span>
<span>= (4 x 7)/(5 x 6) </span>
<span>= 28/30 </span>
<span>= (28 ÷ 2)/(30 ÷ 2) </span>
<span>= 14/15</span>
6 0
3 years ago
Read 2 more answers
Which statement is true of the function f(x) = ? Select three options.
Marianna [84]

Answer:

The function has a domain of all real numbers.

The function has a range of {y|–∞ < y <∞ }.

The function is a reflection of y=∛x

Step-by-step explanation:

Given:

f(x)=-∛x

domain is set of all values that x can take for which the function is defined, so

for above function domain= set of all real numbers

range is set of values that corresponds to the set of values of domain, so for given f(x) range={y|–∞ < y <∞ } set of real numbers

Now f(x)=-∛x hence its reflection will be

-f(x)=-(-∛x)

y=∛x !

3 0
3 years ago
Read 2 more answers
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