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3 years ago

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Mathematics

1 answer:

olchik [2.2K]3 years ago

4 0

When you express the equation of a line in the form $y=mx+q$ , the coefficient m is exactly the slope of the line.

So, let's manipulate the equation to fulfill our goal:

$12 = 4x-6y \iff 6y = 4x-12 \iff y = \dfrac{4}{6}x - 2 \iff y = \dfrac{2}{3}x-2$

So, the slope is 2/3

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86 divide by 3 using area model,

IgorLugansk [536]

Answer:

86/3=28.67

Step-by-step explanation:

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3 years ago

Please help with geometry!!!!

AlladinOne [14]

1) Parallelogram ABCD 1) Given

2) AD || BC; DC || AB 2) Definition of a parallelogram (opposite sides are ||)

3) <A + <B = 180 3) Same-side interior angles are supplementary

<B + <C = 180

4) <A and <B are supplementary 4) Definition of supplementary angles

<B and <C are supplementary

Hope this helps

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4 years ago

this week leah three as much as she did last week. During both weeks leah earned $72. how much did she earn last week?

Nikitich [7]

she earned $36 last week.

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3 years ago

WILL GIVE A BRAINLIEST IF UR RIGHT PLEASE HELP!!!

stellarik [79]

For (8, 16): 3(8) - 8 = 24 - 8 = 16 [solution]
For (3, 4): 3(3) - 8 = 9 - 8 = 1 [not a solution]
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I hope I helped! :)

6 0

4 years ago

Read 2 more answers

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