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1 year ago

The temperature of a substance is -29.9 F. The temperature decreases by 8.7 and then again by 17.8"

Mathematics

1 answer:

earnstyle [38]1 year ago

6 0

Answer:

3.4

Step-by-step explanation:

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If a b and m∠2 = 71°, what is m∠1?

Nikolay [14]

Answer:

71°

Step-by-step explanation:

In order to be able to solve this problem, we must assume the relation between lines "a" and "b" is that they are parallel.

If a║b, then ∠1 and ∠2 are corresponding, hence congruent.

m∠1 = m∠2 = 71°

8 0

2 years ago

Which equations below represent lines that are perpendicular to the line that contains (1, -2) and (3, 4)? (CHECK PICS FOR ANSWE

Gala2k [10]

First, let's find the slope of the line from the points given.

m = (4 - - 2) / (3 - 1)

m = 6 / 2

m = 3

Secondly, we know that a line perpendicular to the original must have a slope that is the opposite reciprocal of the original. For the given points, the opposite reciprocal slope would be -1/3.

Now, we can put all of the equations below into slope intercept form and find the ones that have a slope of -1/3.

Equation 1: Correct

y = -1/3x - 5

Equation 2: Incorrect

y = 3x - 3

Equation 3: Incorrect

y - 2 = 3(x + 1)

y - 2 = 3x + 1

y = 3x + 2

Equation 4: Correct

x + 3y = 9

3y = -x + 9

y = -1/3x + 3

Equation 5: Incorrect

3x + y = -5

y = -3x - 5

Hope this helps!! :)

8 0

2 years ago

What is the equation of the line that passes through the point (-5,2 ) and has the slope of -2/5

valina [46]

Answer:

y = -2/5x + 2

Step-by-step explanation:

8 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