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3 years ago

EASY QUICK POINTS!!!! Find the altitude of an equilateral triangle if each side of the triangle has the length of 6 meters!!!

Mathematics

1 answer:

KonstantinChe [14]3 years ago

7 0

Answer:

5.20 to the nearest hundredth.

Step-by-step explanation:

The altitude line bisects the base of the triangle to form 2 right angled triangles with hypotenuse = 6 and the base = 3.

So, by the Pythagoras theorem the altitude is calculated as follows:

6^2 = 3^2 + x^2

x^2 = 36 - 9 = 27

x = √27

= 5.20 to the nearest hundredth.

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lord [1]

Answer:

680≤400+8h

280≤8h

35≤h

Step-by-step explanation:

400 is the constant, and h is your variable. Solve the inequality for h, number of hours worked. Briana must work at least 35 hours.

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3 years ago

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klasskru [66]

Answer:

See explanation

Step-by-step explanation:

The question is incomplete as the required trapezoid is not given. SO, I will answer using genera rules.

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3 years ago

NO LINKS!!! Find the arc measure and arc length of AB. Then find the area of the sector ABQ.

Norma-Jean [14]

Answer:

<u>Arc Measure</u>: equal to the measure of its corresponding central angle.

<u>Formulas</u>

$\textsf{Arc length}=2 \pi r\left(\dfrac{\theta}{360^{\circ}}\right)$

$\textsf{Area of a sector of a circle}=\left(\dfrac{\theta}{360^{\circ}}\right) \pi r^2$

$\textsf{(where r is the radius and the angle }\theta \textsf{ is measured in degrees)}$

<h3><u>Question 39</u></h3>

Given:

r = 7 in
$\theta$ = 90°

Substitute the given values into the formulas:

Arc AB = 90°

$\textsf{Arc length of AB}=2 \pi (7) \left(\dfrac{90^{\circ}}{360^{\circ}}\right)=3.5 \pi=11.00\:\sf in\:(2\:d.p.)$

$\textsf{Area of the sector AQB}=\left(\dfrac{90^{\circ}}{360^{\circ}}\right) \pi (7)^2=\dfrac{49}{4} \pi=38.48\:\sf in^2\:(2\:d.p.)$

<h3><u>Question 40</u></h3>

Given:

r = 6 ft
$\theta$ = 120°

Substitute the given values into the formulas:

Arc AB = 120°

$\textsf{Arc length of AB}=2 \pi (6) \left(\dfrac{120^{\circ}}{360^{\circ}}\right)=4\pi=12.57\:\sf ft\:(2\:d.p.)$

$\textsf{Area of the sector AQB}=\left(\dfrac{120^{\circ}}{360^{\circ}}\right) \pi (6)^2=12 \pi=37.70\:\sf ft^2\:(2\:d.p.)$

<h3><u>Question 41</u></h3>

Given:

r = 12 cm
$\theta$ = 45°

Substitute the given values into the formulas:

Arc AB = 45°

$\textsf{Arc length of AB}=2 \pi (12) \left(\dfrac{45^{\circ}}{360^{\circ}}\right)=3 \pi=9.42\:\sf cm\:(2\:d.p.)$

$\textsf{Area of the sector AQB}=\left(\dfrac{45^{\circ}}{360^{\circ}}\right) \pi (12)^2=18 \pi=56.55\:\sf cm^2\:(2\:d.p.)$

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2 years ago

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Strike441 [17]

Answer:

Step-by-step explanation:

(x÷5)*4=80÷(5*4)

(x÷5)*4 = 80÷20

(x÷5)*4 = 4

x÷5 = 4÷4

$\frac{x}{5}=1\\\\x=1*5\\\\x=5$

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lina2011 [118]

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