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2 years ago

Determine the standard form of an equation that passes through (2,2) and (-6,4).

1 answer:

7 0

Slope = (4 - 2)/(-6 - 2)= 2/-8 = -1/4

y = mx + b
2 = -1/4(2) + b
2 = -1/2 + b
b = 5/2

equation
y = -1/4x + 5/2
4y = -x + 10
x + 4y = 10

answer
A) x+4y=10

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The length of a rectangle is 24 units. can the perimeter x of the rectangle be 60 units when its width y is 11 units? (1 point)

Sergio [31]

Okay, well we start out with the equation P=66, where P is perimeter. You should create equations using variables to explain each piece of information you are given. Follow the equations below and see if you can understand how to do another one like this. In this problem, l is length and w is width.

P = 66 The perimeter is equal to 66

l = 3 + w The length of one side is 3 more than the width

2l + 2w = 66 A rectangle's perimeter is calculated by adding the lengths and widths

2(3 + w) + 2w = 66 Use what you know about length from step 2 to replace the variable in step 3

6 + 2w + 2w = 66 Multiply

6 + 4w = 66 Add like terms

4w = 60 Subtract

w = 15 Divide

l = 3 + w Remember step 2?

l = 3 + 15 Replace the variable using your value for w

l = 18 Add

And you're done! Always check your work. It helps to create a picture of a rectangle while you're doing these problems as well. As you get used to these problems more and more, you can show more or less work than I've shown, but try to stay true to what the teacher asks of you. Good luck!

8 0

2 years ago

Read 2 more answers

Which expression is equivalent to square root of 8x^7y^6? assume x is greater than or equal to 0

laila [671]

Answer:

$2x^3y^3\sqrt{2x}$

Step-by-step explanation:

= $\sqrt{8x^7y^6}$

= $\sqrt{2^2 *2*x^6*x*y^6$

= $2x^3y^3\sqrt{2x}$

6 0

2 years ago

While visiting Yosemite National Forrest, Joe approximated the angle of elevation to the top of a hill to be 40 degrees. After w

NARA [144]

Draw a diagram to illustrate the problem as shown in the figure below.

Let h = the height of the hill.
At position A, the angle of elevation is 40°, and the horizontal distance to the foot of the hill is x.
By definition,
tan(40°) = h/x
h = x tan40 = 0.8391x (1)

At position B, Joe is (x - 450) ft from the foot of the hill. His angle of elevation is 40 + 18 = 58°.
By definition,
tan(58°) = h/(x - 450)
h = (x - 450) tan(58°) = 1.6003(x-450)
h = 1.6003x - 720.135 (2)

Equate (1) and (2).
1.6003x - 720.135 = 0.8391x
0.7612x = 720.135
x = 946.0523

From (1), obtain
h = 0.8391*946.0523 = 793.8 ft

Answer: The height of the hill is approximately 794 ft (nearest integer)

8 0

2 years ago

Maya collected data about the number of ice cubes and milliliters of juice in several glasses of juice and organized the data in

ziro4ka [17]

It would be c I believe,

Because you can tell by what you wrote

6 0

2 years ago

Read 2 more answers

A doctor is measuring the average height of male students at a large college. The doctor measures the heights, in inches, of a s

amid [387]

Answer:

The correct conclusion is:

"The doctor can be 95% confident that the mean height of male students at the college is between 63.5 inches and 74.4 inches."

Step-by-step explanation:

A doctor is measuring the average height of male students at a large college.

The doctor measures the heights, in inches, of a sample of 40 male students from the baseball team.

Using this data, the doctor calculates the 95% confidence interval (63.5, 74.4).

The following conclusions is valid:

"The doctor can be 95% confident that the mean height of male students at the college is between 63.5 inches and 74.4 inches."

Since we know that the confidence interval represents an interval that we can guarantee that the target variable will be within this interval for a given confidence level.

For the given case, the confidence level is 95% and the corresponding confidence interval is (63.5, 74.4) which represents the true mean of heights for male students at the college where the doctor measured heights.

Therefore, it is valid to conclude that the doctor is 95% confident that the mean height of male students at the college is within the interval of (63.5, 74.4).

6 0

2 years ago