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3 years ago

At the same rate, how many miles did she bike in the next 15 minutes?

Mathematics

1 answer:

mamaluj [8]3 years ago

8 0

Distance = speed* time
D= constant rate *1/4 miles/hour

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If mc022-1.j pg and mc022-2.j pg, what is the value of (f – g)(144)? –84 –60 0 48

Mashutka [201]

Answer:

Step-by-step explanation:

f(x) = √(x) + 12

g(x) = 2√(x)

(f-g)(x) = √(x) + 12 - 2√(x)

(f-g)(x) = 12 - √(x)

if x = 144

(f-g)(144) = 12 - √(144) = 12 - 12 = 0

5 0

3 years ago

Read 2 more answers

Write the slope-intercept form of the equation of the line through the given points.

Lubov Fominskaja [6]

Answer:

y= -3x + 11

Step-by-step explanation:

Slope formula: y^2- y^1 divide by x^2- x^1

y^2 is -1

y^1 is 5

x^2 is 4

x^1 is 2

-1 - 5 = -6

4 - 2 = 2

-6/2 = - 3

Y-intercept: Using the slope and either of the two points, replace the y, m, and x in y= mx+b.

Using (2, 5)

5= -3(2) + b

5= -6 + b

Add 6 to both sides in order to eliminate -6.

-6 + 6 = 0

5 + 6 = 11

11 = b

Combine your slope-intercept formula:

y= 3x + 11

6 0

3 years ago

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Pls Guy, I Need The Answer To This Question For A Good Grade In My Class :,(

kozerog [31]

From top to bottom
31
13
-41
-68
-104

3 0

2 years ago

An open box with a square base is constructed so that it has a volume of 64in^3. The base needs to be made out of a material tha

Ivahew [28]

The volume of a box is the amount of space in the box

The dimensions that minimize the cost of the box is 4 in by 4 in by 4 in

<h3>How to determine the dimensions that minimize the cost</h3>

The dimensions of the box are:

Width = x

Depth = y

So, the volume (V) is:

$V = x^2y$

The volume is given as 64 cubic inches.

So, we have:

$x^2y = 64$

Make y the subject

$y = \frac{64}{x^2}$

The surface area of the box is calculated as:

$A =x^2 + 4xy$

The cost is:

$C = 2x^2 + 4xy$ --- the base is twice as expensive as the sides

Substitute $y = \frac{64}{x^2}$

$C = 2x^2 + 4x * \frac{64}{x^2}$

$C =2x^2 + \frac{256}{x}$

Differentiate

$C' =4x - \frac{256}{x^2}$

Set to 0

$4x - \frac{256}{x^2} = 0$

Multiply through by x^2

$4x^3 - 256= 0$

Divide through by 4

$x^3 - 64= 0$

Add 64 to both sides

$x^3 = 64$

Take the cube roots of both sides

$x = 4$

Recall that:

$y = \frac{64}{x^2}$

So, we have:

$y = \frac{64}{4^2}$

$y = 4$

Hence, the dimensions that minimize the cost of the box is 4 in by 4 in by 4 in