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Rufina [12.5K]
3 years ago
13

Which of the tables represents a function?

Mathematics
1 answer:
True [87]3 years ago
8 0
Table B. The same input can't go to different outputs
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Help me with this please and thank you!! :)
Ann [662]

Answer:

Step-by-step explanation:

  • A \cap C denotes the set of elements in both A and C, which is \{20, 24, 28\}.
  • B^{C} denotes the complement of set B, which is the set of all elements that are in the universal set that are not in set B. In this case, this set is \{3, 4, 5, 6, 8, 11, 13, 16, 17, 19, 20, 22, 23, 24, 25, 26, 27, 28\}
7 0
1 year ago
10. 10. If AB bisects CAF, and m&lt;EAF = 72°,<br>then the m&lt;BAF =​
never [62]

The measure of ∠BAF is 54°.

Solution:

DF and CE are intersecting lines.

m∠EAF = 72° and AB bisects ∠CAF.

∠EAF and ∠DAC are vertically opposite angles.

Vertical angle theorem:

<em>If two lines are intersecting, then vertically opposite angles are congruent.</em>

∠DAC ≅ ∠EAF

m∠DAC = 72°

<em>Sum of the adjacent angles in a straight line = 180°</em>

m∠DAE + m∠EAF = 180°

m∠DAE + 72° = 180°

Subtract 72° from both sides.

m∠DAE = 108°

∠CAF and ∠DAE are vertically opposite angles.

⇒ m∠CAF = m∠DAE

⇒ m∠CAF = 108°

AB bisects ∠CAF means ∠CAB = ∠BAF

m∠CAB + m∠BAF = 108°

m∠BAF + m∠BAF = 108°

2 m∠BAF = 108°

Divide by 2 on both sides, we get

m∠BAF = 54°

Hence the measure of ∠BAF is 54°.

4 0
3 years ago
Derivative, by first principle<br><img src="https://tex.z-dn.net/?f=%20%5Ctan%28%20%5Csqrt%7Bx%20%7D%20%29%20" id="TexFormula1"
vampirchik [111]
\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}h

Employ a standard trick used in proving the chain rule:

\dfrac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\cdot\dfrac{\sqrt{x+h}-\sqrt x}h

The limit of a product is the product of limits, i.e. we can write

\displaystyle\left(\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\right)\cdot\left(\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt x}h\right)

The rightmost limit is an exercise in differentiating \sqrt x using the definition, which you probably already know is \dfrac1{2\sqrt x}.

For the leftmost limit, we make a substitution y=\sqrt x. Now, if we make a slight change to x by adding a small number h, this propagates a similar small change in y that we'll call h', so that we can set y+h'=\sqrt{x+h}. Then as h\to0, we see that it's also the case that h'\to0 (since we fix y=\sqrt x). So we can write the remaining limit as

\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan\sqrt x}{\sqrt{x+h}-\sqrt x}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{y+h'-y}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{h'}

which in turn is the derivative of \tan y, another limit you probably already know how to compute. We'd end up with \sec^2y, or \sec^2\sqrt x.

So we find that

\dfrac{\mathrm d\tan\sqrt x}{\mathrm dx}=\dfrac{\sec^2\sqrt x}{2\sqrt x}
7 0
3 years ago
Solve this equation! Please Help!<br><br> √3 x 66.15/4.41<br><br> (SHOW YOUR WORK)
finlep [7]
Hi Jessica,
<span>Remember PEMDAS (Parenthesis, Exponents, Multiplication & Division, Addition & Subtraction).

√3 x 66.15/4.41 {Exponents/Cube & Square Roots First}

1.73 x 66.15 ÷ 4.41 {Multiplication}

114.4395 ÷ 4.41 {Division}

25.95 {Final Answer}

Cheers,
Izzy</span>
7 0
3 years ago
Find the non-permissible replacement for: 4/5x
Alik [6]
Hello : 
answer <span>B) 0
because : 4/(5×0)  no exist</span>
8 0
3 years ago
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