All categories

4 years ago

How do you factor the expression 20k+50

2 answers:

5 0

Find the greates common factor between the two numbers being added which is 10, and divide them. Next put the two quotients in separate parentheses and multiply each one by 10 separatelyas shown:

10(2k) + 10(5)

Aliun [14]4 years ago

4 0

Answer is 10(2k+5)

-----------------------------------

Work Shown:

20k+50 = 10*2k + 10*5
20k+50 = 10(2k+5)

Note how we can distribute the 10 back in to check our work
10(2k+5) = 10(2k)+10(5) = 20k+50
so that confirms we have the right answer

Another thing to notice is that 10 is the largest factor that we can pull out of 20k and 50. The value 10 is the GCF (greatest common factor) of 20 and 50.

You might be interested in

Which do I click for the answer?

beks73 [17]

You Should click :

B) The row categories are "Wears Glasses" and "Do Not Wear Glasses"

C) Two teachers do not wear glasses

E) Total of 6 teachers were polled

<u>Explanation:</u>

In a class there are 129 students, among that 32 wear glasses, 97 don't wear glasses.
There are 6 teachers in those 4 wear glasses and 2 do not wear glasses.
Totally 135 present. In those 36 wear and 99 don't wear glasses.

7 0

3 years ago

1. Prove or give a counterexample for the following statements: a) If ff: AA → BB is an injective function and bb ∈ BB, then |ff

Fantom [35]

Answer:

a) False. A = {1}, B = {1,2} f: A ⇒ B, f(1) = 1

b) True

c) True

d) B = {1}, A = N, f: N ⇒ {1}, f(x) = 1

Step-by-step explanation:

a) lets use A = {1}, B = {1,2} f: A ⇒ B, f(1) = 1. Here f is injective but 2 is an element of b and |f−¹({b})| = 0., not 1. This statement is False.

b) This is True. If A were finite, then it can only be bijective with another finite set with equal cardinal, therefore, B should be finite (and with equal cardinal). If A were not finite but countable, then there should exist a bijection g: N ⇒ A, where N is the set of natural numbers. Note that f o g : N ⇒ B is a bijection because it is composition of bijections. This, B should be countable. This statement is True.

c) This is true, if f were surjective, then for every element of B there should exist an element a in A such that f(a) = b. This means that f−¹({b}) has positive cardinal for each element b from B. since f⁻¹(b) ∩ f⁻¹(b') = ∅ for different elements b and b' (because an element of A cant return two different values with f). Therefore, each element of B can be assigned to a subset of A (f⁻¹(b)), with cardinal at least 1, this means that |B| ≤ |A|, and as a consequence, B is finite.

b) This is false, B = {1} is finite, A = N is infinite, however if f: N ⇒ {1}, f(x) = 1 for any natural number x, then f is surjective despite A not being finite.

4 0

4 years ago

In tests for intoxication, blood alcohol levels are expressed as percentages by volume. A blood alcohol level of 0.080% by volum

GaryK [48]

Answer:

The required volume is 5 ml.

Step-by-step explanation:

Given : In tests for intoxication, blood alcohol levels are expressed as percentages by volume. A blood alcohol level of 0.080% by volume means 0.080 mL of ethanol per 100 mL of blood and is considered proof of intoxication. If a person's total blood volume is 5.0 L.

To find : What volume of alcohol in the blood gives a blood alcohol level of 0.100 % by volume?

Solution :

According to question we use the formula,

$\text{Present Volume}=\dfrac{\text{Vol}_{\text{Alcohol}}}{\text{Vol}_{\text{total}}}\times 100\%$

The percentage of a blood alcohol level is 0.100%

The total volume of the blood is V=5L=5000 ml.

Substitute the value in the formula,

$0.100=\dfrac{\text{Vol}_{\text{Alcohol}}}{5000}\times 100$

$\frac{0.100}{100}=\dfrac{\text{Vol}_{\text{Alcohol}}}{5000}$

$\text{Vol}_{\text{Alcohol}}=\frac{0.100\times 5000}{100}$

$\text{Vol}_{\text{Alcohol}}=5$

Therefore, the required volume is 5 ml.

8 0

4 years ago

Best and correct answer gets brainliest!!!

dmitriy555 [2]

59 degrees!
because you set up your problem: 3c+2+2c-7=90
then combine all like terms: 5c-5=90
Add 5 to both sides: 5c=95
then divide both sides by 5: c=19
Plug in your answer: 3(19)+2=59

8 0

3 years ago

Please help <br> What’s the answer

saul85 [17]

What is the question you are looking for?

8 0

3 years ago