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1 year ago

14

Find the area of AAOB 9√3 un² 3√3 un² 4.5√3 un

1 answer:

butalik [34]1 year ago

7 0

Answer:

Step-by-step explanation:

The area of the equilateral triangle is $\frac{\sqrt{3}}{4}(6)^{2}=\boxed{9\sqrt{3}}$

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3⁵.2⁴.2¹.3⁶.(3²)⁴ and down (2³)².3⁶<br>

BARSIC [14]

Answer:

$\dfrac{3^{13}}{2}$

Step-by-step explanation:

Given expression:

$\dfrac{3^5 \cdot 2^4 \cdot 2^1 \cdot 3^6 \cdot (3^2)^4}{(2^3)^2 \cdot 3^6}$

$\textsf{Apply exponent rule} \quad (a^b)^c=a^{bc}:$

$\implies \dfrac{3^5 \cdot 2^4 \cdot 2^1 \cdot 3^6 \cdot 3^8}{2^6 \cdot 3^6}$

$\textsf{Apply exponent rule} \quad a^b \cdot a^c=a^{b+c}:$

$\implies \dfrac{3^{(5+6+8)} \cdot 2^{(4+1)}}{2^6 \cdot 3^6}$

$\implies \dfrac{3^{19} \cdot 2^{5}}{2^6 \cdot 3^6}$

$\implies \dfrac{3^{19} \cdot 2^{5}}{3^6 \cdot 2^6}$

$\textsf{Apply exponent rule} \quad \dfrac{a^b}{a^c}=a^{b-c}:$

$\implies 3^{(19-6)} \cdot 2^{(5-6)}$

$\implies 3^{13} \cdot 2^{-1}$

$\textsf{Apply exponent rule} \quad a^{-n}=\dfrac{1}{a^n}:$

$\implies \dfrac{3^{13}}{2^1}$

$\textsf{Apply exponent rule} \quad a^1=a:$

$\implies \dfrac{3^{13}}{2}$

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1 year ago

AssAssessment captures blank growth

jekas [21]

What do you mean by this

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3 years ago

On a scale drawing with a scale of 1 to 25 feet, the height of the building is 8 inches on a drawing. How tall is the actual bui

Wewaii [24]

Answer:

200 feet

Step-by-step explanation:

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3 years ago

Read 2 more answers

21 is 42% of what number

Step2247 [10]

21 is 42% of 50 , So the correct answer is 50

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3 years ago

Determine if triangle ABC with coordinates A (0, 2), B (2, 5), and C (−1, 7) is an isosceles triangle. Use evidence to support y

kirill115 [55]

Answer:

The triangle ABC is an isosceles right triangle

Step-by-step explanation:

we have

The coordinates of triangle ABC are

A (0, 2), B (2, 5), and C (−1, 7)

we know that

An isosceles triangle has two equal sides and two equal internal angles

The formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

step 1

Find the distance AB

substitute in the formula

$d=\sqrt{(5-2)^{2}+(2-0)^{2}}$

$d=\sqrt{(3)^{2}+(2)^{2}}$

$dAB=\sqrt{13}\ units$

step 2

Find the distance BC

substitute in the formula

$d=\sqrt{(7-5)^{2}+(-1-2)^{2}}$

$d=\sqrt{(2)^{2}+(-3)^{2}}$

$dBC=\sqrt{13}\ units$

step 3

Find the distance AC

substitute in the formula

$d=\sqrt{(7-2)^{2}+(-1-0)^{2}}$

$d=\sqrt{(5)^{2}+(-1)^{2}}$

$dAC=\sqrt{26}\ units$

step 4

Compare the length sides

$dAB=\sqrt{13}\ units$

$dBC=\sqrt{13}\ units$

$dAC=\sqrt{26}\ units$

$dAB=dBC$

therefore

Is an isosceles triangle

Applying the Pythagoras Theorem

$(AC)^{2} =(AB)^{2}+(BC)^{2}$

substitute

$(\sqrt{26})^{2} =(\sqrt{13})^{2}+(\sqrt{13})^{2}$

$26=13+13$

$26=26$ -----> is true

therefore

Is an isosceles right triangle

8 0

3 years ago