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Rama09 [41]
3 years ago
15

Can someone double check my answers and tell me the correct ones

Mathematics
1 answer:
Alex73 [517]3 years ago
5 0

Answer:

all are right ma'am ☺️☺️☺️☺️

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X=(y)/(y-3) make y the subject
Liula [17]

Answer:

\displaystyle y=\frac{3x}{x-1}

Step-by-step explanation:

<u>Solving Equations</u>

We have the expression:

\displaystyle x=\frac{y}{y-3}

And it's required to make y the subject.

It can be done by isolating y on the left side of the equation and the rest of the expression on the right side.

The first step is removing the denominator by multiplying both sides by y-3:

\displaystyle (y-3)x=(y-3)\frac{y}{y-3}

Simplifying the right side:

\displaystyle (y-3)x=y

Remove the parentheses:

yx-3x=y

Move the y's to the left side and the rest of the expression to the right side:

yx-y=3x

Factor out y:

y(x-1)=3x

Divide by x-1:

\mathbf{\displaystyle y=\frac{3x}{x-1}}

7 0
3 years ago
A football is thrown from the top of the stands, 50 feet above the ground at an initial velocity of 62 ft/sec and at an angle of
Anvisha [2.4K]

a. i. The parametric equation for the horizontal movement is x = 43.84t

ii. The parametric equation for the vertical movement is y = 50 + 43.84t

b. the location of the football at its maximum height relative to the starting point is (60.1 ft, 60.1 ft)

<h2>a. Parametric equations</h2>

A parametric equation is an equation that defines a set of quantities a functions of one or more independent variables called parameters.

<h3>i. Parametric equation for the horizontal movement</h3>

The parametric equation for the horizontal movement is x = 43.84t

Since

  • the angle of elevation is Ф = 45° and
  • the initial velocity, v = 62 ft/s,

the horizontal component of the velocity is v' = vcosФ.

So, the horizontal distance the football moves in time, t is x = vcosФt

= vtcosФ

= 62tcos45°

= 62t × 0.7071

= 43.84t

So, the parametric equation for the horizontal movement is x = 43.84t

<h3>ii Parametric equation for the vertical movement</h3>

The parametric equation for the vertical movement is y = 50 + 43.84t

Also, since

  • the angle of elevation is Ф = 45° and
  • the initial velocity, v = 62 ft/s,

the vertical component of the velocity is v" = vsinФ.

Since the football is initially at a height of h = 50 feet, the vertical distance the football moves in time, t relative to the ground is y = 50 + vsinФt

= 50 + vtcosФ

= 50 + 62tsin45°

= 50 + 62t × 0.7071

= 50 + 43.84t

<h3>b. Location of football at maximum height relative to starting point</h3>

The location of the football at its maximum height relative to the starting point is (60.1 ft, 60.1 ft)

Since the football reaches maximum height at t = 1.37 s

The x coordinate of its location at maximum height is gotten by substituting t = 1.37 into x = 48.84t

So, x = 43.84t

x = 43.84 × 1.37

x = 60.0608

x ≅ 60.1 ft

The y coordinate of the football's location at maximum height relative to the ground is y = 50 + 48.84t

The y coordinate of the football's location at maximum height relative to the starting point is y - 50 = 48.84t

So,  y - 50 = 48.84t

y - 50 = 43.84 × 1.37

y - 50 = 60.0608

y - 50 ≅ 60.1 ft

So, the location of the football at its maximum height relative to the starting point is (60.1 ft, 60.1 ft)

Learn ore about parametric equations here:

brainly.com/question/8674159

5 0
2 years ago
Arnold deposited $732.19 in a savings account that earns 2.7% simple interest. What is Arnold's account balance after seven year
liq [111]

Answer:

Step-by-step explanation:

$870.57

use equation total = Principal(1 + rate*time)

3 0
3 years ago
What is the quotient of 2 ÷ 2/3?​
Neko [114]

Answer:

The answer is 3! I believe.

Step-by-step explanation:

3 0
2 years ago
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Two factors of 24 that add up to 11
Vadim26 [7]
Simple....

8*3=24

8+3=11

Thus, your answer.
8 0
3 years ago
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