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2 years ago

I'm confused can you guys please help me ASAP.

Mathematics

1 answer:

Darina [25.2K]2 years ago

3 0

Answer:

Step-by-step explanation:

I think I mean to me it makes since

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100 POINTS AND BRAINLIEST

zysi [14]

Answer:

8 square units and $\frac{40}{3}$ square units

Step-by-step explanation:

The area of the triangle ABC is 24 square units.

1. Triangles ABC and FBG are similar with scale factor $\frac{1}{3},$ then

$\dfrac{A_{\triangle FBG}}{A_{\triangle ABC}}=\dfrac{1}{9}\Rightarrow A_{\triangle FBG}=\dfrac{1}{9}\cdot 24=\dfrac{8}{3}\ un^2.$

2. Triangles ABC and DBE are similar with scale factor $\frac{2}{3},$ then

$\dfrac{A_{\triangle DBE}}{A_{\triangle ABC}}=\dfrac{4}{9}\Rightarrow A_{\triangle DBE}=\dfrac{4}{9}\cdot 24=\dfrac{32}{3}\ un^2.$

3. Thus, the area of the quadrilateral DFGE is

$A_{DFGE}=A_{\triangle DBE}-A_{\triangle FBG}=\dfrac{32}{3}-\dfrac{8}{3}=8\ un^2.$

and the area of the quadrilateral ADEC is

$A_{ADEC}=A_{\triangle ABC}-A_{\triangle DBE}=24-\dfrac{32}{3}=\dfrac{40}{3}\ un^2.$

4 0

3 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=%5Clim_%7Bx%5Cto%20%5C%200%7D%20%5Cfrac%7B%5Csqrt%7Bcos2x%7D-%5Csqrt%5B3%5D%7Bcos3x%7D%20%7D%7

salantis [7]

Answer:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{1}{2}$

General Formulas and Concepts:

Calculus

Limits

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

L'Hopital's Rule

Differentiation

Derivatives
Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

We are given the limit:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)}$

When we directly plug in x = 0, we see that we would have an indeterminate form:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{0}{0}$

This tells us we need to use L'Hoptial's Rule. Let's differentiate the limit:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)}$

Plugging in x = 0 again, we would get:

$\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \frac{0}{0}$

Since we reached another indeterminate form, let's apply L'Hoptial's Rule again:

$\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)}$

Substitute in x = 0 once more:

$\displaystyle \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)} = \frac{1}{2}$

And we have our final answer.

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

Unit: Limits

6 0

3 years ago

Huxian's father gives her $120 as her monthly allowance. a) If she spends an average of 5 dollars a day, find the amount on mone

abruzzese [7]

Answer:

3 days= $105, 6 days= $90, 10 days= $70

Step-by-step explanation:

You can rewrite this equation as y=120-5x

x is the amount of days passed

plug in 3 for x and you get 15. 120-15=105

plug in 6 for x and you get 30. 120-30=90

plug in 10 for x you get 50. 120-50=70

Hope this helps!

4 0

2 years ago

Find the secant of angle Z A.4/5 B.4/3 C.3/5 D.5/3

Alex73 [517]

Answer:

D 5/3

Step-by-step explanation:

Secθ= Hypotenuse/Adjacent

Secθ=10/6

Simplify: 10/6 -->5/3

Ans: D 5/3

remember your six trigonometric functions:

Sinθ=Opp/Hyp -->Cscθ=Adj/Hyp

Cosθ=Adj/Hyp -->Secθ=Hyp/Adj

Tanθ=Opp/Adj-->Cotθ=Opp?Adj

It is basically the opposite of each other, don't forget when you are at the last step you simplify! never forget. Also working with square roots will be a bit tricky when converting to Csc/ Sec/ Cot. Just noting you that Square roots don't belong in denominators!

3 0

3 years ago

The figure shows two right triangles, each with its longest side on the same line. R 4 How long is XY? Type the answer in the bo

Leokris [45]

Answer: 6

Step-by-step explanation:

Because corresponding sides of similar triangles are proportional,

$\frac{XY}{3}=\frac{4}{2}\\ \\ \frac{XY}{3}=2\\ \\ XY=\boxed{6}$

8 0

2 years ago