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KATRIN_1 [288]
4 years ago
14

Which question is biased?

Mathematics
2 answers:
Anni [7]4 years ago
8 0
B. Don't you think Latin class is a waste of time?
because the person who is asking the question has bias against Latin class, or they would've said 'Do you' instead of 'Don't you'
I'm 75% sure..
pishuonlain [190]4 years ago
4 0
I think the answer is B. because it says "waste of time"
You might be interested in
Identify The domain of the rational function, g(x)= 1/x+3
8_murik_8 [283]

Answer:

The domain is (-∞ , -3) ∪ (-3, ∞) ⇒ D

Step-by-step explanation:

<em>The domain of the rational fraction is t</em><em>he values of x which make the fraction defined</em><em>. That means </em><em>the domain does not contain the values of x which make the denominator equal to 0</em><em>.</em>

∵ g(x) = \frac{1}{x+3}

∴ The denominator = x + 3

→ Equate the denominator by 0

∵ x + 3 = 0

→ Subtract 3 from both sides

∴ x + 3 - 3 = 0 - 3

∴ x = -3

→ That means the domain can not have -3 because it makes the denominator

   equal to 0

∴ The domain is all values of real numbers except x = -3

∴ The domain = {x : x ∈ R, x ≠ -3}

∴ The domain = (-∞ , -3) ∪ (-3, ∞)

4 0
3 years ago
(5-3)to the power of 2?
Kipish [7]
You would subtract 3 from 5 getting 2 and multiply 2 by itself getting the Answer: 4

Work: (5-3)= 2^ of 2 or 2*2 = 4

4 0
3 years ago
5000 jums for 7 hours
Sunny_sXe [5.5K]

Answer:

do you mean jumps? and if so thts alot of exersize i dont understand how im solving

Step-by-step explanation:

5 0
3 years ago
Find the area, in square inches, of the following composite figure.
Mama L [17]
I think the answer it’s 15 in not 100% sure tho
8 0
3 years ago
PLEASE HELP!!!!!!!<br>please
Nataly_w [17]

Answer:

see explanation

Step-by-step explanation:

To find the zeros let h(t) = 0, that is

t² + 4t + 3 = 0 ← in standard form

(t + 3)(t + 1) = 0 ← in factored form

Equate each factor to zero and solve for t

t + 3 = 0 ⇒ t = - 3 ← smaller t

t + 1 = 0 ⇒ t = - 1 ← larger t

(2)

given a parabola in standard form : ax² + bx + c ( a ≠ 0)

Then the x- coordinate of the vertex is

x_{vertex} = - \frac{b}{2a}

h(t) = t² + 4t + 3 ← is in standard form

with a = 1 and b = 4, thus

x_{vertex} = - \frac{4}{2} = - 2

Substitute t = - 2 into h(t) for y- coordinate

h(- 2) = (- 2)² + 4(- 2) + 3 = 4 - 8 + 3 = - 1

Vertex = (- 2, - 1 )

4 0
3 years ago
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