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goldfiish [28.3K]
4 years ago
13

What is the range for the function ƒ(x) = 7?

Mathematics
1 answer:
Citrus2011 [14]4 years ago
4 0
Find the domain by finding where the function is define on the graph. the rang is the set of values that correspond with the domain. 

Domain: {x/x<R}

Range:7 
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Simplify 10/16 to lowest terms and find an equivalent fraction that has a denominator of 32?
goblinko [34]

Answer:

The lowest fraction for \frac{10}{16} is \frac{5}{8} and a equivalent fraction with denominator of 32 is \frac{20}{32}.

Step-by-step explanation:

First, to have the fraction to the lowest terms we will have to divide the numerator and denominator by 2, this is:

\frac{10/2}{16/2}=\frac{5}{8}

As can be seen the expression couldn't be simplified more because there is no a common number that could divide the numerator and denominator.

Second, to have an equivalent fraction with 32 in the denominator we will have to multiply the denominator by 2 and of course the numerator, this is:

\frac{10*2}{16*2}=\frac{20}{32}

The lowest fraction for \frac{10}{16} is \frac{5}{8} and a equivalent fraction with denominator of 32 is \frac{20}{32}.

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Name a pair of overlapping congruent triangles in each diagram. State whether the triangles are congruent by SSS, SAS, ASA, AAS,
MariettaO [177]

Answer:

ΔPTS≅ΔRTA by AAS axiom of congruency

Step-by-step explanation:

Consider ΔPQA and ΔRQS

∠PQA=∠RQS     (Vertically Opposite Angles)

∠QAP=∠QSR     (Complementary of two equal angles, ∠RAT and∠PST)

Due to angle sum property of a triangle, we come to the conclusion that

∠APQ=∠SRQ

Consider ΔPTS and ΔRTA

TA=TS  (Given)

∠RAT=∠PST(Given)

∠APQ=∠SRQ (Proved above)

Therefore, ΔPTS≅ΔRTA by AAS axiom of congruency.

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here \\ \angle 6 = 75 \degree \\ we \: know \: that \\ \angle6 + \angle 8 = 180 \degree \\ \implies 75 \degree \: + \angle 8 = 180 \degree \\ \implies \angle8 = 180 \degree - 75 \degree \\ \implies \angle8 = 105 \degree \\ since \: line \: l \parallel \: line \: m \\ \angle 8 = \angle3 \: (corresponding \: angles) \\ therefore \: \angle3 = 105 \degree
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