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2 years ago

Find the midpoint of the line segment.

Mathematics

1 answer:

Dmitry [639]2 years ago

7 0

Answer:

(2, -1)

Step-by-step explanation:

$\left(\frac{0+4}{2}, \frac{0-2}{2} \right)=(2, -1)$

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Someone please help with this question please

natima [27]

Step-by-step explanation:

Since Angle DAE = Angle BCE, lines AD amd BC are parallel (by Z-angles).

This means that Angle ADE = Angle CBE (by Z-angles).

We have 2 congruent angles and 1 congruent side (AD = BC, given).

By ASA congruence, triangles AED and CEB are congruent.

5 0

3 years ago

Read 2 more answers

Consider the parent graph f(x) = x2 . Then, g(x) is a translation according the rule: y = (x + 3)2 + 4. Find the graph g(x).

krok68 [10]

For this case we have the following parent function:
f (x) = x ^ 2
We apply the following function transformation:
Horizontal translations
Suppose that h> 0
To graph y = f (x + h), move the graph of h units to the left.
We have then:
y = (x + 3) ^ 2
Vertical translations
Suppose that k> 0
To graph y = f (x) + k, move the graph of k units up.
We have then:
y = (x + 3) ^ 2 + 4
Answer:
The graph g (x) is the graph f (x) 3 units to the left and 4 units up:
y = (x + 3) ^ 2 + 4

3 0

3 years ago

PLEASE HELP I'LL GIVE YOU 15 POINTS

ELEN [110]

Answer:

I cant see whats on the page

Step-by-step explanation:

4 0

3 years ago

Lines a and b are parallel. When they are cut by transversals, they form a triangle Parallel lines a and b are cut by transversa

lord [1]

Answer:

Therefore, the angle 3 have 28 degrees.

Step-by-step explanation:

We know that the lines a and b are parallel. When they are cut by transversals, they form a triangle. Parallel lines a and b are cut by transversals s and t to form a triangle. Angle 1 is 90 degrees, angle 2 is 62 degrees.

We know that the number of degrees in a triangle equals 180.

We get:

$90+62+x=180\\\\152+x=180\\\\x=180-152\\\\x=28\\$

Therefore, the angle 3 have 28 degrees.

6 0

4 years ago

Read 2 more answers

According to the National Postsecondary Student Aid Study conducted by the U.S. Department of Education in 2008, 62% of graduate

Ronch [10]

The standard error for the distribution of sampling proportions will be 0.0686.

<h3>What is standard error for the distribution of sampling proportions?</h3>

In mathematics, the difference between a data set and the populace's true average is known as primary data deviation from the mean.

According to the National Postsecondary Student Aid Study conducted by the U.S.

Department of Education in 2008, 62% of graduates from public universities had student loans.

We randomly select 50 students at a time.

Then the standard error for the distribution of sampling proportions will be

$\rm Standard\ error = \sqrt{\dfrac{p(1-p)}{n}}\\\\Standard\ error = \sqrt{\dfrac{0.62(1-0.62)}{50}}\\\\Standard\ error = 0.0686$

More about the standard error for the distribution link is given below.

brainly.com/question/14524236

#SPJ1

6 0

2 years ago