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3 years ago

?? please help a dude out

Mathematics

1 answer:

finlep [7]3 years ago

7 0

Tis so easyAnswer:

1:> 2:< 3>

Step-by-step explanation:

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Identify the terms, like terms, coefficients, and constants in each expression.

astra-53 [7]

Answer:

2. Like terms: 5a 4. Lt: 4y, and y

coefficient: 2,-7 coefficient: 4, -3

Constant: 2,-7 constant: 4

3. like terms: 3h, 2h, and 6h can you follow these and do the remaining??

if no, I'll help u

coefficient: 3,2,6

constant: 9

5 0

3 years ago

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Find the distance of PQ<br> P(0,4) and Q(10,-6)

solong [7]

Answer:

PQ ≈ 14.14 units

Step-by-step explanation:

calculate the distance d using the distance formula

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

with (x₁, y₁ ) = P (0, 4 ) and (x₂, y₂ ) = Q (10, - 6 )

PQ = $\sqrt{(10-0)^2+(-6-4)^2}$

= $\sqrt{10^2+(-10)^2}$

= $\sqrt{100+100}$

= $\sqrt{200}$

≈ 14.14 units ( to 2 dec. places )

4 0

2 years ago

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Consider the equation 5/3v + 4 + 1/3v = 8. What is the resulting equation after the first step in the solution?

BigorU [14]

You would combing like terms. In this case, it is 5/3v and 1/3v. So you would add them together to get 6/3v,or just 2 because 6 divided by 3 is 2. So now you have 2v+4=8.

8 0

4 years ago

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How do i solve it and round to the nearest tenth of a percent

Zepler [3.9K]

Divide 42 by 89
now the answer is 0.4719... but to get a percent multiply by 100
the answer is now 47.19... so to round to the nearest tenth your answer is 47.2%

6 0

4 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°

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

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

8 0

2 years ago