MATHEMATICS
HIGH SCHOOL

Answer: The independent variable is the hours and the dependent variable is the calories Kim burns in h amount of hours

HIGH SCHOOL

Which is an example of an algebraic expression? A. 2k + 25 B. z – 8 = 44 C. 3 + 7 = 10 D. 76 – 10

B. would be a great example of an algebraic expression

Please mark as brainliest answer

Please mark as brainliest answer

Options a and b are the algebraic expressions

HIGH SCHOOL

A rectangular piece of cardboard, whose area is 216 square centimeters, is made into an open box by cutting a 2-centimeter square from each corner and turning up the sides. if the box is to have a volume of 224 cubic centimeters, what should the original dimensions of the cardboard start with

Let the length of the original cardboard be x and width be y.

Area=xy=216

thus y=216/x

after cutting 2 cm from the edges, the dimensions will be:

(x-4) cm by (216/x-4) by 2 cm

thus the volume will be:

V=L×W×H

V=(216/x-4)×(x-4)×2=224

thus solving for x we get:

x=12 or x=18

Hence:

length=18cm width=12 cm

Area=xy=216

thus y=216/x

after cutting 2 cm from the edges, the dimensions will be:

(x-4) cm by (216/x-4) by 2 cm

thus the volume will be:

V=L×W×H

V=(216/x-4)×(x-4)×2=224

thus solving for x we get:

x=12 or x=18

Hence:

length=18cm width=12 cm

Answer:

Length: 18 cm

Width: 12 cm

Step-by-step explanation:

A better explanation for this is the following:

First we know that the area of the card is 216 cm². This can be expressed as:

A = L * W so

216 = L * W (1)

Now, let's change the variables here, we will denote the Length as "x" and the width as "y". Then (1) can be rewritten as:

216 = xy (2)

Now, we know that in order to make the open box, it was cut from each corner of the card, 2 cm², and the volume of the box is 224 cm³. According to this, we know that the volume of the box is:

V = L' * W' * H (3)

The Height of the box would be the 2 cm that were cut and L' and W' would be x and y. However, as the Length and Width has been cut, then the expression for both of them is the following

For the Length:

L' = x - 4

For the Width:

W'= y - 4

Replacing in expression (3):

224 = 2 * (x-4) * (y-4)

112 = (x-4)(y-4) (4)

Now, in (2) we can solve either x or y to make a new expression and then, do the same in (4), and thus, we can actually solve for one of the dimensions. In this case, we will solve for y first, so let's solve for y in (2) and (4):

216 = xy

y = 216/x (5)

Solving now for y, from (4):

112/x-4 = y - 4

y = (112/x-4) + 4 (6)

So now, all we have to do is equal (5) and (6), and in that way we can find the value of x:

216/x = (112/x-4) + 4

216/x = 112 + 4(x-4) / (x-4)

216(x-4) = x(112 + 4x - 16)

216x - 864 = 112x + 4x² - 16x

4x² - 120x + 864 = 0 (7)

From here, we can either do the general expression and solve for x, or we can just factorize (7) and get the 2 values of x at once. In this case let's use the general expression. Although is longer, but we will get the correct result using this method so:

4x² - 120x + 864 = 0 (Divide by 4 all terms)

x² - 30x + 216 = 0 (8)

the general equation:

x = -b ±√b² - 4ac / 2a

From (8), we know that a = 1, b = -30, c = 216. Replacing:

x = 30 ± √(-30)² - 4*1*216 / 2

x = 30 ±√900 - 864 / 2

x = 30 ± 6 / 2

x1 = 30 + 6 / 2 = 18

x2 = 30 - 6 / 2 = 12

So the values for the Length and width are:

L = 18 cm

W = 12 cm

If you put this numbers into equations (2) and (4):

216/18 = 12

(18-4)(12-4) = 112

MIDDLE SCHOOL

Amos has 93 cents two of his coins are quarters what is the largest number of nickels he might have

Amos can have 8 nickels, and he will have 3 cents left

93 - 50 = 43

5 goes into 43 eight times

5x8 = 40

Hope this helps!

93 - 50 = 43

5 goes into 43 eight times

5x8 = 40

Hope this helps!

2 are quarters

quarters are worth 25 cents each

2*25=50

50 cents are quarters

93-50=43 cents that aren't quarters

now if we want max number of nickles

how many times can 5 go into 43

43/5=8 and a remainder of 3

so 8 times

max number is 8 nickles

quarters are worth 25 cents each

2*25=50

50 cents are quarters

93-50=43 cents that aren't quarters

now if we want max number of nickles

how many times can 5 go into 43

43/5=8 and a remainder of 3

so 8 times

max number is 8 nickles

MIDDLE SCHOOL

the perimeter of a triangle is 34 units. its width is 6.5 units. write an equation to determine the length (l) of the rectangle

Perimeter = length of the 3 sides of the triangle ie 34units

Since width we are given 6.5 units

We have to find the remaining 2 sides:

Formula of Triangle for Perimeter

P = a + b + c (sum of sides)

simplify: P= a + 2b (its an isosceles)

34 = 6.5 + 2b

34 - 6.5 = 2b

27.5 = 2b

Só b = 13.75 units each length of triangle

Relationship between triangle and rectangle

= rectangle has four right angles, so a triangle has an angle of half since it bisects the edges (ie 45*)

Use Trig Ratios to find L

SOHCAHTOA

we want adjacent ( L)

13.75 sin 45

L = 9.72

Thus your equation is

Equation used to find Triangle × Trig ratios

P = a + 2b

sin 45

Where L is length of rectangle

a is width of triangle

L =(P - a +2b)× sin 45

Since width we are given 6.5 units

We have to find the remaining 2 sides:

Formula of Triangle for Perimeter

P = a + b + c (sum of sides)

simplify: P= a + 2b (its an isosceles)

34 = 6.5 + 2b

34 - 6.5 = 2b

27.5 = 2b

Só b = 13.75 units each length of triangle

Relationship between triangle and rectangle

= rectangle has four right angles, so a triangle has an angle of half since it bisects the edges (ie 45*)

Use Trig Ratios to find L

SOHCAHTOA

we want adjacent ( L)

13.75 sin 45

L = 9.72

Thus your equation is

Equation used to find Triangle × Trig ratios

P = a + 2b

sin 45

Where L is length of rectangle

a is width of triangle

L =(P - a +2b)× sin 45