Liam volunteers to help plan the event. First month he volunteered for x hours. The next month he volunteered for 2 times as many hours as he had the first month. If he volunteered for a total of 45 hours,how many hours did he volunteer during the first month? Write an equation to represent the situation and solve

Answers

Answer 1

Answer:

0/3 that’s what my teacher said on this test

Step-by-step explanation:

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6x-4y=54
-9x+2y=-6
substitution

Answers

$\left \{ {{6x-4y=54} \atop {-9x+2y=-6\ \ | *2}} \right. \n\n \left \{ {{6x-4y=54} \atop {-18x+4y=-12}} \right. \n+-----\naddition\ method\n\n-12x=42\ \ \ | divide\ by\ -12\nx=-3.5\n\n2y=-6+9x\ny=(-6+9x)/(2)=(-6+31.5)/(2)=(25.5)/(2)=12.75\n\n \left \{ {{y=12.75} \atop {x=-3.5}} \right.$

-9x + 2y = -6 and that's how u solve it and u can thank me wants u get it right lol

in an over-fished area, the catch of a certain type of fish is decreasing at an average rate of 8% per year. If this decline persists, how long will it take for the catch to reach half of the amount before the decline???

Answers

n = n0(1 - 0.08)^t
= n0(0.92)^t

Putting n = n0 / 2:
1 / 2 = 0.92^t
t = log(1 / 2) / log(0.92)
= 8.31 yr.

Five elevenths times a number is 55

Answers

5/11x = 55
11/5 * 5/11x = 55 * 11/5
x = 121

Final answer:

The number is found by setting up an algebraic equation and solving for x. The solution to the equation (5/11)x = 55 is x = 121.

Explanation:

The problem states: Five elevenths times a number is 55. This is a typical algebra problem. The first step to solve it is to set up the equation: (5/11)x = 55. In order to isolate x to have it by itself on one side of the equation, you divide 55 by 5/11. When we divide a number by a fraction, we multiply it by the reciprocal of that fraction. So, we have x = 55 * (11/5) = 121.

Learn more about Algebra here:

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Need help with this question

Answers

a₁ = -8 This is the first term in an arithmetic sequence.
d = 3, the common difference

a(n) = a₁ + d(n-1)

a₇ = -8 + 3(7-1)
a₇ = -8 + 3(6)
a₇ = -8 + 18
a₇ = 10

To check:
a₁ = -8
a₂ = -8 + 3 = -5
a₃ = -5 + 3 = -2
a₄ = -2 + 3 = 1
a₅ = 1 + 3 = 4
a₆ = 4 + 3 = 7
a₇ = 7 + 3 = 10

Last year , a large trucking company delivered 5.5*10^5 tons goods with a average value of $23,000 per ton . What was the total value of the goods delivered ? Write your answer in scientific notation . Please , I have to do this before bed time . Can you help me ?
Thank you :)

Answers

5.5*10^5 = 550,000 tons

$23,000 per ton

23,000 x 550,000 = $12,650,000,000 (total value)

Now to convert to scientific notation.

1.265 x 10^10

Hope that was useful :)

How to find the perimeter and the area of a rectangle on a coordinate plane using the distance formula? : A(3,8) B(5,4) C(-4,-1) D(-6,3) Round to the nearest tenth if necessary.

Answers

I used some site to plot the points.

First, let's recall the formulas for Perimeter and Area of a Rectangle.
Perimeter = 2(l+w)
Area = l×w
Also, the distance formula is
D = $\sqrt{( x_(2)-x_(1))^2+{(y_(2)-y_(1))^2}$

Now, we need to determine l and w.
So, the length is the distance of AD or BC
and the width is the distance of DC or AB
(I'll just use the sides that I've labelled for ease)

So first to determine the length, we need to calculate the distance of AD
Points are A(3,8) and D(-6,3)
$x_(1) =3, x_(2) =8, y_(1) =6, y_(2) =-3$
$D_(AD)= \sqrt{( 6-3)^2+{(-3-8)^2}$
$D_(AD)= √((3)^2+(-11)^2)$
$D_(AD)= √(9+121)$
$D_(AD)= √(130) ≈ 11.40$
So the length of the rectangle is 11.40 units.

Now, the width!
So first to determine the width, we need to calculate the distance of DC
Points are D(-6,3) and C(-4,-1)
$x_(1) =6, x_(2) =-4, y_(1) =-3, y_(2) =-1$
$D_(DC)= \sqrt{(-4-6)^2+{(-1- -3)^2}$
$<span>D_(DC)= \sqrt{(-4-6)^2+{(-1+3)^2}$
$D_(DC)= √((-10)^2+(2)^2)$
$D_(AD)= √(100+4)$
$D_(AD)= √(104) ≈ 10.20$
So the width of the rectangle is ≈10.20 units.

Let's now solve for Perimeter and Area using
l = 11.40
w = 10.20

Perimeter = 2(l+w)
Perimter = 2(11.40+10.20)
Perimter = 2(21.60)
Perimter = 43.2 units

Area = l×w
Area = (11.40)(10.20)
Area = 116.28
Area = 116.3 units² (rounding)

In conclusion, given points A(3,8) B(5,4) C(-4,-1) D(-6,3), the Perimeter is 43.2 units and the Area is ≈116.3 units² using the distance formula.

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