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baherus [9]
3 years ago
10

Insert parenthesis () to make the following problem true: 3+6-2x4=19

Mathematics
1 answer:
tia_tia [17]3 years ago
5 0

Answer:

3+(6-2)*4=19

Step-by-step explanation:

Due to PEMDAS, it would first be required to do "6-2", which is 4.

Then, the 4 in the parenthesis is multiplied by the 4 on the outside, making 16.

Finally, 3 would be added to 16, making 19.

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An accountant earns 6% simple interest. You want to find the interest earned on $200 after 8 months.
tamaranim1 [39]

Answer:

The interest is $12

Step-by-step explanation:

You just put the percent in a decimal form(6%-0.06)Then you multiply 200 with 0.06. And the answer you get is the interest

Hope that helps :)

5 0
3 years ago
Three point zero one divided by seven
vivado [14]
The answer is zero point four three

You could also write this answer as 0.43
7 0
3 years ago
Number 7 and 8 plz answer quickly
Flauer [41]

(7) m∠A = 52°

(8) m∠B = 117°

Solution:

(7) Let us first define the supplementary and complementary angles.

Supplementary angles: Two angles are said to be supplementary angles if their sum is add up to 180°

Complementary angles: Two angles are said to be complementary angles if their sum is add up to 90°

Given supplement of 142° = 180° – 142°

                                           = 38°

Complement of ∠A = Supplement of 142°

⇒ Complement of ∠A = 38°

Measure of ∠A = 90° – 38°

                          = 52°

Hence m∠A = 52°.

(8) Given complement of 27° = 90° – 27°

                                                 = 63°

Supplement of ∠B = Complement of 27°

⇒ Supplement of ∠B = 63°

Measure of ∠B = 180° – 63°

                          = 117°

Hence m∠B = 117°.

3 0
3 years ago
Sixty percent of Company A’s employees are considered top performers. Fifty five percent of Company A’s employees are in a rigor
pickupchik [31]
Not exactly. there are still 40 percent of the employees that are not top performers. for all you know those 40 percent could make up the majority of the employees in the training program<span />
5 0
3 years ago
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g Let G be a not necessarily abelian group with normal subgroups H and K such that H contains K (i.e., K ✂ G, H ✂ G, K ≤ H) and
allsm [11]

Answer:

Lets a,b be elements of G. since G/K is abelian, then there exists k ∈ K such that ab * k = ba (because the class of ab, [ab]_K is equal to [ba]_K, thus ab and ba are equal or you can obtain one from the other by multiplying by an element of K.

Since K is a subgroup of H, then k ∈ H. This means that you can obtain ba from ab by multiplying by an element of H, k. Thus, [ab]_H = [ba]_H . Since a and b were generic elements of H, then H/G is abelian.

4 0
3 years ago
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