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vladimir1956 [14]

3 years ago

15

In the straightedge and compass construction of the equilateral triangle below, which of the following reasons can you use to pr

ove that AC ≅ BC?

2 answers:

Kaylis [27]3 years ago

5 0

Answer:

reason A

Step-by-step explanation:

AB and AC both are the radii of the first circle

AB=AC

also AB and BC both are the radii of the second circle

hence

AB=BC

By using transitive property

AB=AC and AB=BC

so AC=BC

Therefore reason A "AB and AC are the radii of the same circle and AB and BC are the radii of the same circle so AB=AC and AB=BC and AC=BC" !

romanna [79]3 years ago

4 0

Answer: A and D are both correct

Step-by-step explanation:

just took this test

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The Big Falcon Rocket (BFR or Starship) from Space X can carry approximately 220,000 pounds. If they only carried $100 bills, ho

kifflom [539]

Answer:

$9,979,032,100 or $9.979 billion

Step-by-step explanation:

The approximate weight of a $100 bill is 1 gram.

All of the calculations bellow assume that the volume of the bills would not be an issue and only concerns weight.

1 pound is equivalent to approximately 453.59237 grams.

The weight in grams that the Big Falcon Rocket can carry is:

$W= 220,000*453.59237=99,790,321.4\ g$

Since each bill weighs 1 gram, the number of bills it could carry, rounded to nearest whole bill is 99,790,321. The total amount it could carry is:

$A=99,790,321*\$100=\$9,979,032,100$

4 0

3 years ago

Help plz!!!!! will be gratefully appreciated

Morgarella [4.7K]

(A) is your answer

In this case, only rotational symmetry can work, with a turn of 180°

hope this helps

8 0

3 years ago

the square of a certain whole number n is n^2. if 60 is a factor of n^2, it's possible that ? is not a factor of n^2. a)16 b)25

DanielleElmas [232]

If 60 is a factor of n², then √60 is a factor of n. However, n is a whole number, so its factors are whole numbers.

Simplify √60:
$\sqrt{60}=\sqrt{4 \times 15}=2\sqrt{15}$

If 2√15 is a factor of a whole number n, then √15 must be another factor to make it a whole number.

$2\sqrt{15} \times \sqrt{15}=2 \times 15=2 \times 3 \times 5$

If 60 is a factor of n², then 2, 3, and 5 must be factors of n. The factors of n² are the squares of the factors of n, so 2, 2, 3, 3, 5, and 5 must be factors of n².

Now, if 2, 2, 3, 3, 5, and 5 are factors of n², then:
* 5×5=25 must also be its factor
* 2×2×3×3=36 must also be its factor
* 2×2×5×5=100 must also be its factor

Only 16 may not be a factor of n². The answer is A.

4 0

3 years ago

Andrew spent $80 on travel last month.

xz_007 [3.2K]

195/30 * 80 = $520 spent on household bills last month

8 0

2 years ago

Evaluate the following integral (Calculus 2) Please show step by step explanation!

barxatty [35]

Answer:

$\dfrac{1}{2} \left(25 \arcsin \left(\dfrac{x}{5}\right) -x\sqrt{25-x^2}\right) + \text{C}$

Step-by-step explanation:

<u>Fundamental Theorem of Calculus</u>

$\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))$

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

Given indefinite integral:

$\displaystyle \int \dfrac{x^2}{\sqrt{25-x^2}}\:\:\text{d}x$

Rewrite 25 as 5²:

$\implies \displaystyle \int \dfrac{x^2}{\sqrt{5^2-x^2}}\:\:\text{d}x$

<u>Integration by substitution</u>

<u />

$\boxed{\textsf{For }\sqrt{a^2-x^2} \textsf{ use the substitution }x=a \sin \theta}$

$\textsf{Let }x=5 \sin \theta$

$\begin{aligned}\implies \sqrt{5^2-x^2} & =\sqrt{5^2-(5 \sin \theta)^2}\\ & = \sqrt{25-25 \sin^2 \theta}\\ & = \sqrt{25(1-\sin^2 \theta)}\\ & = \sqrt{25 \cos^2 \theta}\\ & = 5 \cos \theta\end{aligned}$

Find the derivative of x and rewrite it so that dx is on its own:

$\implies \dfrac{\text{d}x}{\text{d}\theta}=5 \cos \theta$

$\implies \text{d}x=5 \cos \theta\:\:\text{d}\theta$

<u>Substitute</u> everything into the original integral:

$\begin{aligned}\displaystyle \int \dfrac{x^2}{\sqrt{5^2-x^2}}\:\:\text{d}x & = \int \dfrac{25 \sin^2 \theta}{5 \cos \theta}\:\:5 \cos \theta\:\:\text{d}\theta \\\\ & = \int 25 \sin^2 \theta\end{aligned}$

Take out the constant:

$\implies \displaystyle 25 \int \sin^2 \theta\:\:\text{d}\theta$

$\textsf{Use the trigonometric identity}: \quad \cos (2 \theta)=1 - 2 \sin^2 \theta$

$\implies \displaystyle 25 \int \dfrac{1}{2}(1-\cos 2 \theta)\:\:\text{d}\theta$

$\implies \displaystyle \dfrac{25}{2} \int (1-\cos 2 \theta)\:\:\text{d}\theta$

$\boxed{\begin{minipage}{5 cm}\underline{Integrating $\cos kx$}\\\\$\displaystyle \int \cos kx\:\text{d}x=\dfrac{1}{k} \sin kx\:\:(+\text{C})$\end{minipage}}$

$\begin{aligned} \implies \displaystyle \dfrac{25}{2} \int (1-\cos 2 \theta)\:\:\text{d}\theta & =\dfrac{25}{2}\left[\theta-\dfrac{1}{2} \sin 2\theta \right]\:+\text{C}\\\\ & = \dfrac{25}{2} \theta-\dfrac{25}{4}\sin 2\theta + \text{C}\end{aligned}$

$\textsf{Use the trigonometric identity}: \quad \sin (2 \theta)= 2 \sin \theta \cos \theta$

$\implies \dfrac{25}{2} \theta-\dfrac{25}{4}(2 \sin \theta \cos \theta) + \text{C}$

$\implies \dfrac{25}{2} \theta-\dfrac{25}{2}\sin \theta \cos \theta + \text{C}$

$\implies \dfrac{25}{2} \theta-\dfrac{5}{2}\sin \theta \cdot 5 \cos \theta + \text{C}$

$\textsf{Substitute back in } \sin \theta=\dfrac{x}{5} \textsf{ and }5 \cos \theta = \sqrt{25-x^2}:$

$\implies \dfrac{25}{2} \theta-\dfrac{5}{2}\cdot \dfrac{x}{5} \cdot \sqrt{25-x^2} + \text{C}$

$\implies \dfrac{25}{2} \theta-\dfrac{1}{2}x\sqrt{25-x^2} + \text{C}$

$\textsf{Substitute back in } \theta=\arcsin \left(\dfrac{x}{5}\right) :$

$\implies \dfrac{25}{2} \arcsin \left(\dfrac{x}{5}\right) -\dfrac{1}{2}x\sqrt{25-x^2} + \text{C}$

Take out the common factor 1/2:

$\implies \dfrac{1}{2} \left(25 \arcsin \left(\dfrac{x}{5}\right) -x\sqrt{25-x^2}\right) + \text{C}$

Learn more about integration by trigonometric substitution here:

brainly.com/question/28157322

6 0

2 years ago