Work out the area of this circle. Take to be 3.142 and give your answer to 2 decimal places. 11.2 cm

How to solve this triangle ABC? i did super wrong all :( who can help me please?

Alex73 [517]

It looks like you had some of the numbers correct, but you just labeled it and rounded it incorrectly, and the teacher gave you super partial credit for it.

For my work, I'm assuming that side A is opposite of angle A and side B is opposite of angle B, etc...

5 0

3 years ago

Triangle m is similar to triangle n. Triangle m has two angles with measures of 32°and 93°. Which two angles measures could be i

inysia [295]

Given that triangle m and n are similar, then the implication is the ratio of the corresponding sides are the same and the corresponding angles are equal. This implies that if the two angles of triangle m measure 32° and 93°, then the possible size for the two angles in triangle n will be 32° and 93°.

6 0

3 years ago

Can whole numbers be rational numbers

yKpoI14uk [10]

Use this diagram, it might help.

From the inside, whatever that layer is, it can be any of the layers outside of it.

All whole numbers can be a rational number, but not all rational numbers be a whole number.

Kind of like all squares can be rectangles, but not all rectangles can be a square.

The whole number has all the attributes of a rational number, but not all rational numbers have all of the attributes of a whole number.

6 0

3 years ago

Can some one just tell me if these are right? If not can I have the answers?

-Dominant- [34]

Yes you have them correct

6 0

3 years ago

Which values are in the solution set of the compound inequality? Check all that apply. 4(x + 3) ≤ 0 or x + 1 > 3 –6 –3 0 3 8

Colt1911 [192]

we have

$4(x + 3) \leq 0$ -------> inequality 1

or

$x + 1 > 3$ -------> inequality 2

we know that

In this system of inequalities, for a value to be the solution of the system, it is enough that it satisfies at least one of the two inequalities.

let's check each of the values

case 1) x=-6

Substitute the value of x=-6 in the inequality 1

$4(-6 + 3) \leq 0$

$4(-3) \leq 0$

$-12 \leq 0$ -------> is ok

The value of x=-6 is a solution of the compound inequality-----> It is not necessary to check the second inequality, because the first one satisfies

case 2) x=-3

Substitute the value of x=-3 in the inequality 1

$4(-3 + 3) \leq 0$

$4(0) \leq 0$

$0 \leq 0$ -------> is ok

The value of x=-3 is a solution of the compound inequality-----> It is not necessary to check the second inequality, because the first one satisfies

case 3) x=0

Substitute the value of x=0 in the inequality 1

$4(0 + 3) \leq 0$