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3 years ago

PLS HELP!! GIVING OUT BRAINLIEST!

Mathematics

1 answer:

Makovka662 [10]3 years ago

4 0

Answer:

path A: 2.5 mph

path B: 2 mph

path C: 3 mph

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620,000 round off by ten thousand

Softa [21]

620,000 is already rounded by ten thousand

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3 years ago

Ed wants to bicycle at least 75 miles this week. The inequality 11 + 4b > 75 can be used to find the average number of miles

jeyben [28]

Answer: Ed should bike at least more than 16 miles

Step-by-step explanation:

11 + 4b > 75

4b > 64

b > 16

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3 years ago

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1.The temperature was 80∘F and then fell 20∘F. 2.The temperature was -13∘F and then rose 9∘F. 3.The temperature was -5∘F and the

deff fn [24]

If your looking for the ending temperatures on each one here they are in Celsius and Fahrenheit!

1. 60°F or 15.6°C

2. -4°F or -20°C

3. -13°F or -26°C

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4 years ago

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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°

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

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

8 0

2 years ago

Y=-2 y=3x+1 in graph

Airida [17]

Answer:

The solution of the system is x=-1, y=-2

Step-by-step explanation:

we have

$y=-2$ ----> equation A

$y=3x+1$ ----> equation B

Solve the system by graphing

Remember that the solution of the system is the intersection point both lines

using a graphing tool

The intersection point is (-1,-2)

see the attached figure

therefore

The solution of the system is x=-1, y=-2

3 0

3 years ago