Isaac bought 20 quarts of water.

At A Local Restaurant, The Health Inspector Visits Every 7 Days, And The Fire Inspector Visits Every 12 Days.

Xelga [282]

PART A :OK so here is the table for the health inspector; all you are really doing is 7+7+7, and so on:

PART A: health inspector: 7,14,21,28,35,42,49,56,63,70,77,84..........

PART A:same thing for the fire inspector, except this time your adding 12+12+12, and so on:

PART A: fire inspector: 12,24,36,48,60,72,84,96.......

PART B: and to figure out what day they will both come is you have to find the LCM (least common multiple)

PART B: which in this case is Day 84

hopes this helps!!!!!

8 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

22. Find the perimeter and area.

zmey [24]

Answer:

Part 22) The area is $A=15a^3b^6\ units^2($ and the perimeter is $P=10a^2b^4+6ab^2\ unit$

Part 24) The area is $A=16m^3n\ units^2$ and the perimeter is $P=24mn\ units$

Part 26) The area is equal to $A=9\pi x^6y^{2}\ units^2$

Step-by-step explanation:

Part 22) Find the perimeter and area

step 1

The area of a rectangle is equal to

$A=LW$

we have

$L=5(a^2)(b^4)\ units$

$W=3(a)(b^2)\ units$

Remember that

When multiply exponents with the same base, adds the exponents and maintain the base

substitute in the formula

$A=(5(a^2)(b^4))(3(a)(b^2))$

$A=15a^3b^6\ units^2$

step 2

The perimeter of a rectangle is equal to

$P=2(L+W)$

we have

$L=5(a^2)(b^4)\ units$

$W=3(a)(b^2)\ units$

substitute in the formula

$P=2(5(a^2)(b^4)+3(a)(b^2))$

$P=10a^2b^4+6ab^2\ unit$

Part 24) Find the perimeter and area

step 1

The area of triangle is equal to

$A=\frac{1}{2}bh$

where

$b=8mn\ units$

$h=4m^2\ units$

Remember that

When multiply exponents with the same base, adds the exponents and maintain the base

substitute the given values

$A=\frac{1}{2}(8mn)(4m^2)$

$A=16m^3n\ units^2$

step 2

Find the perimeter

I will assume that is an equilateral triangle (has three equal length sides)

The perimeter of an equilateral triangle is

$P=3b$

where

$b=8mn\ units$

substitute

$P=3(8mn)$

$P=24mn\ units$

Part 26) Find the area

The area of a circle is equal to

$A=\pi r^{2}$

where

$r=3x^3y\ units$

Remember the property

$(a^{m})^{n}=a^{m*n}$

substitute in the formula the given value

$A=\pi (3x^3y)^{2}$

$A=9\pi x^6y^{2}\ units^2$

6 0

3 years ago

PLEASE PLEASE HELP what is f(1)

Dafna1 [17]

Answer:

5

Step-by-step explanation:

f(1) means what is the value when x=1. y is at 5.

4 0

3 years ago

Read 2 more answers

Create a real life example using either temperature or money to compare -3 and -5

lana66690 [7]

The temperature when I went to the bus stop in the morning felt like it was -3°, sadly it was only -5° only two degrees warmer, which means it is still freezing

7 0

3 years ago

Read 2 more answers