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3 years ago

I have a math final tomorrow and I don’t know how to do this. Someone please help!

Mathematics

1 answer:

Serga [27]3 years ago

5 0

135+105=240
Divide by 2
120
Then 180-120=60

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SOMEONE HELP! I know the answer I just need to show the work.

gayaneshka [121]

okay so

117° + angle a = 180° (supplementary angle)

angle a = 180 - 117

= 63°

then angle a + angle b + 81° = 180° ( angle sum property)

so... 63° + angle b + 81° = 180

144 + angle b = 180°

angle b = 180 - 144

= 36°

hope this helps.

6 0

3 years ago

Please helppp me!!!!!

Scrat [10]

Answer:

Step 2 contains error in the given problem.

Step-by-step explanation:

Given expression is:

$\frac{1}{2}x-3=\frac{1}{3}x+6$

Step 1: identifying the LCM.

The LCM identified is 6.

This step is correct.

In the next step, we multiply the LCM with each term of the equation.

Step 2:

$\frac{6}{1}(\frac{1}{2}x) -3 =\frac{6}{1}(\frac{1}{3}x)+6$

However,

In the given solution, the LCM is not multiplied with each term.

Hence,

Step 2 contains error in the given problem.

7 0

3 years ago

Please answer all of them! SHOW WORK!!

antiseptic1488 [7]

Wait I can help fm dm

8 0

2 years ago

juilo is paid 1.4 times his normal hourly rate for each hour he works over 30 hours in a week. Last week he worked 35hours and e

Licemer1 [7]

Let x be Juilo's normal hourly rate.

so for the 35 hr week we have:-

30x + 5*1.4x = 436.60

37x = 436.60

x = 436.60 / 37 = $11.80

His normal hourly rate = $11.80

So for a 30 hr week he will earn 11.80 * 30 = $354

So the amount $436.60 is a reasonable figure for a 35 hr week.

4 0

3 years ago

Find the indicated limit, if it exists.

kondor19780726 [428]

Answer:

d) The limit does not exist

General Formulas and Concepts:

Calculus

Limits

Right-Side Limit: $\displaystyle \lim_{x \to c^+} f(x)$
Left-Side Limit: $\displaystyle \lim_{x \to c^-} f(x)$

Limit Rule [Variable Direct Substitution]: $\displaystyle \lim_{x \to c} x = c$

Limit Property [Addition/Subtraction]: $\displaystyle \lim_{x \to c} [f(x) \pm g(x)] = \lim_{x \to c} f(x) \pm \lim_{x \to c} g(x)$

Step-by-step explanation:

*Note:

In order for a limit to exist, the right-side and left-side limits must equal each other.

Step 1: Define

Identify

$\displaystyle f(x) = \left\{\begin{array}{ccc}5 - x,\ x < 5\\8,\ x = 5\\x + 3,\ x > 5\end{array}$

Step 2: Find Right-Side Limit

Substitute in function [Limit]: $\displaystyle \lim_{x \to 5^+} 5 - x$
Evaluate limit [Limit Rule - Variable Direct Substitution]: $\displaystyle \lim_{x \to 5^+} 5 - x = 5 - 5 = 0$

Step 3: Find Left-Side Limit

Substitute in function [Limit]: $\displaystyle \lim_{x \to 5^-} x + 3$
Evaluate limit [Limit Rule - Variable Direct Substitution]: $\displaystyle \lim_{x \to 5^+} x + 3 = 5 + 3 = 8$

∴ Since $\displaystyle \lim_{x \to 5^+} f(x) \neq \lim_{x \to 5^-} f(x)$ , then $\displaystyle \lim_{x \to 5} f(x) = DNE$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Limits

5 0

2 years ago