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e-lub [12.9K]
3 years ago
15

This is not a question but who here is in RSM Middle school?

Mathematics
1 answer:
zysi [14]3 years ago
8 0
Not me but I do go to somewhere is spring where are y’all?
You might be interested in
In a right triangle, the measure of one acute angle is 12 more than twice the measure of the other acute angle. Find the measure
soldi70 [24.7K]

Given:

In a right triangle, the measure of one acute angle is 12 more than twice the measure of the other acute angle.

To find:

The measures of the 2 acute angles of the triangle.

Solution:

Let x be the measure of one acute angle. Then the measure of another acute is (2x+12).

According to the angle sum property, the sum of all interior angles of a triangle is 180 degrees. So,

x+(2x+12)+90=180

3x+102=180

3x=180-102

3x=78

Divide both sides by 3.

x=\dfrac{78}{3}

x=26

The measure of one acute angle is 26 degrees. So, the measure of another acute angle is:

2x+12=2(26)+12

2x+12=52+12

2x+12=64

Therefore, the measures of two acute angles are 26° and 64° respectively.

8 0
3 years ago
What are the next three terms in the pattern 2, 6, 18, 54, ...?
castortr0y [4]

Answer:

OPTION B: 162, 486, 1458

Step-by-step explanation:

The given sequence is 2, 6, 18, 54, . . .

It is a geometric sequence and the common difference is 3.

The general form of a geometric sequence is: a, ar, ar², ar³, . . .

Here a = 2 and r = 3.

$ n^{th} $ term of a Geometric progression is $ ar^{n - 1} $.

Note that the fourth term is 54.

i.e., $ ar^3 = 54 $

$ \implies ar^4 = ar^3 . r = 54 . 3 = 162 $.

Similarly, $ ar^6 = 162 \times 3 = 486 $.

Also, $ ar^7 = ar^6 . r = 486 \times 3 = 1458 $.

Hence, OPTION B is the answer.

8 0
3 years ago
Read 2 more answers
4.20. According to a report released by the National Center for Health Statistics, 51% of U.S. households has only cell phones (
Sergeu [11.5K]

Answer:

The probability that the household has only cell phones and has high-speed Internet is 0.408

Step-by-step explanation:

Let A be the event that represents U.S. households has only cell phones

Let B be the event that represents U.S. households have high-speed Internet.

We are given that 51% of U.S. households has only cell phones

P(A)=0.51

We are given that 70% of the U.S. households have high-speed Internet.

P(B)=0.7

We are given that U.S. households having only cell phones, 80% have high-speed Internet. A U.S household is randomly selected.

P(B|A)=0.8

\frac{P(A\capB)}{P(A)}=0.8\\P(A\capB)=0.8 \times P(A)\\P(A\capB)=0.8 \times 0.51\\P(A\capB)=0.408

Hence the probability that the household has only cell phones and has high-speed Internet is 0.408

7 0
3 years ago
The sum of the measures of the two angles of a linear pair must be...?
Rina8888 [55]

The answer is 180 degrees

5 0
3 years ago
Country A has a population of 514,482 and Country B has a population of 523,407. How would you compare these populations
prohojiy [21]
You can subtract the populations to see by how many more people/citizens does country B have
8 0
3 years ago
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