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2 years ago

10

FIND THE MISSING SIDE LENGTHS!!!

1 answer:

velikii [3]2 years ago

7 0

Answer:

Step-by-step explanation:

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Find the indicated limit, if it exists.

kondor19780726 [428]

Answer:

d) The limit does not exist

General Formulas and Concepts:

<u>Calculus</u>

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.

<u>Step 1: Define</u>

<em>Identify</em>

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

<u>Step 2: Find Right-Side Limit</u>

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

<u>Step 3: Find Left-Side Limit</u>

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

Point A is the midpoint of side XZ and point B is the

Ludmilka [50]

Answer:

4 units

Step-by-step explanation:

just took a quiz and got it right

6 0

2 years ago

Read 2 more answers

out of some students appeared in an examination 80% passed in mathematics 75% passed in english and 5% failed in both subjects i

Advocard [28]

Answer:

5000 students appeared in the examination.

Step-by-step explanation:

We solve this question using Venn probabilities.

I am going to say that:

Event A: Passed in Mathematics

Event B: Passed in English.

5% failed in both subjects

This means that 100 - 5 = 95% pass in at least one, which means that $P(A \cup B) = 0.95$

80% passed in mathematics 75% passed in english

This means that $P(A) = 0.8, P(B) = 0.75$

Proportion who passed in both:

$P(A \cap B) = P(A) + P(B) - P(A \cup B)$

Considering the values we have for this problem

$P(A \cap B) = P(A) + P(B) - P(A \cup B) = 0.8 + 0.75 - 0.95 = 0.6$

3000 of them were passed both subjects how many students appeared in the examination?

3000 is 60% of the total t. So

$0.6t = 3000$

$t = \frac{3000}{0.6}$

$t = 5000$

5000 students appeared in the examination.

4 0

2 years ago

Can anyone help me plot this?<br><br>The Equation is y= 45 - x

asambeis [7]

The first point could be (5, 40)

The second could be (25, 20)

That's just assuming the number of minutes are the x-coordinates...

Hope this helps!

3 0

3 years ago

Find the equation of the circle with center at the origin that contains the point<br> (-8,6)

BigorU [14]

X^2+y^2=R^2
64+36=R^2
R=10
x^2+y^2=100

4 0

3 years ago