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Vadim26 [7]
4 years ago
12

Tell whether the given value is a solution of the equation.

Mathematics
1 answer:
Anvisha [2.4K]4 years ago
3 0

Answer: a. y=11 is a solution

b. z=8 is not a solution

Step-by-step explanation:

a. if y+6=17, then y=17-6. 17-6=11, so y=11, making it a solution.

b. if 7z=42, then y=42/7. 42/7 is 6, not 8, so z=8 is not a solution.

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A group of 5 men and 5 women are applying for a job at a local company. Each of the 10 job candidates has the same chance of rec
Zielflug [23.3K]

Answer: 1/4

Explanation: Because there is a 1/2 chance that a female will get the first job, and a half chance that a lady will get a second job. So (1/2)*(1/2)=(1/4)

7 0
3 years ago
Use a calculator to approximate the measures of ∠A, ∠B, and ∠C to the nearest tenth of a degree. sin A = 0.96, tan B = 0.53, cos
ivolga24 [154]
To find each of the angles we must make the following clearances:
 For ∠A
 sin A = 0.96
 A = Asin (0.96)
 A = 73.7 degrees
 For ∠B
 tan B = 0.53
 B = Atan (0.53)
 B = 27.9 degrees
 For ∠C
 cos C = 0.44
 C = Acos (0.44)
 C = 63.9 degrees
 Answer:
 approximate measures of ∠A, ∠B, and ∠C are:
 A = 73.7 degrees
 B = 27.9 degrees
 C = 63.9 degrees
3 0
3 years ago
Mrs. Rodger got a weekly raise of $145. If she gets paid every other week, write an integer describing how the raise will affect
m_a_m_a [10]
It will affect her paycheck by it going up and she gets more money
4 0
3 years ago
Read 2 more answers
Melissa makes necklaces and sells them online. She charges $88 per necklace. Her monthly expenses are $3,745. How many necklaces
Dovator [93]
She will need to sell 18 necklaces a month to reach 1650
4 0
4 years ago
Read 2 more answers
Given a function I and a subset A of its domain, let I(A) represent the range of lover the set A; that is, I(A) = {I(x) : x E A}
Ede4ka [16]

Step-by-step explanation:

(a)

Using the definition given from the problem

f(A) = \{x^2  \, : \, x \in [0,2]\} = [0,4]\\f(B) = \{x^2  \, : \, x \in [1,4]\} = [1,16]\\f(A) \cap f(B) = [1,4]  = f(A \cap B)\\

Therefore it is true for intersection. Now for union, we have that

A \cup B = [0,4]\\f(A\cup B ) = [0,16]\\f(A) = [0,4]\\f(B)= [1,16]\\f(A) \cup f(B) = [0,16]

Therefore, for this case, it would be true that f(A\cup B) = f(A)\cup f(B).

(b)

1 is not a set.

(c)

To begin with  

A\cap B \subset A,B

Therefore

g(A\cap B) \subset g(A) \cap g(B)

Now, given an element of g(A) \cap g(B) it will belong to both sets, therefore it also belongs to g(A\cap B), and you would have that

g(A)\cap g(B) \subset  g(A \cap B), therefore  g(A)\cap g(B)  =  g(A \cap B).

(d)

To begin with A,B  \subset A \cup B, therefore

g(A) \cup g(b) \subset g(A\cup B)

7 0
4 years ago
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