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AveGali [126]
3 years ago
11

What is the greatest common factor (gcf) of the numerator and denominator in the rational expression below? 7x+14/2x^2+x-6

Mathematics
1 answer:
Mashutka [201]3 years ago
4 0
The answer is C. The numerator 7x+14 can be factored as 7(x+2). The denominator 2x^2+2x-6 can be factored as (x+2)(2x-3). The greatest common divisor of 7(x+2) and (x+2)(2x-3) is obviously x+2, which is contained in both of the two expressions.
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In a class, every student knows French or German (or both). 15 students know French, and 17 students know German.
vazorg [7]
If you apply the or both
Only 1 of the students would need to know the "or both", therefore maximizing the remaining amount of students you can put in.

Gerald, let's call him, knows French AND German, so there's only one less student that knows french and german. Gerald is 1 student.

MAXIMUM:
There are now 14 monolinguistic French speakers and 16 monolinguistic German's, 30 students + Gerald=31.

Minimum:
As a bonus, the minimum is 15 students knowing french AND German and only 2 monolinguistic German speakers, so 17.
8 0
3 years ago
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3. The cost of a ticket twill be no more than $26.
goblinko [34]
Any number 1-26 should work. the expression would me x<$26.
7 0
3 years ago
At sunrise, the temperature was -12 °F. The temperature increased 19 °F each hour for 3 hours. What was the temperature 3 hours
yawa3891 [41]
-12 degree F + 19 + 19 + 19 = 49 degree F
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4 0
3 years ago
3a-3/a+6=9/6 what is a
Gekata [30.6K]

Answer:

a=1/2

Step-by-step explanation:

8 0
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Please solve these questions for me. i am having a difficult time understanding.
s2008m [1.1K]

Answer:

1) AD=BC(corresponding parts of congruent triangles)

2)The value of x and y are 65 ° and 77.5° respectively

Step-by-step explanation:

1)

Given : AD||BC

AC bisects BD

So, AE=EC and BE=ED

We need to prove AD = BC

In ΔAED and ΔBEC

AE=EC (Given)

\angle AED = \angel BEC ( Vertically opposite angles)

BE=ED (Given)

So, ΔAED ≅ ΔBEC (By SAS)

So, AD=BC(corresponding parts of congruent triangles)

Hence Proved

2)

Refer the attached figure

\angle ABC = 90^{\circ}

In ΔDBC

BC=DC (Given)

So,\angle CDB=\angle DBC(Opposite angles of equal sides are equal)

So,\angle CDB=\angle DBC=x

So,\angle CDB+\angle DBC+\angle BCD = 180^{\circ} (Angle sum property)

x+x+50=180

2x+50=180

2x=130

x=65

So,\angle CDB=\angle DBC=x = 65^{\circ}

Now,

\angle ABC = 90^{\circ}\\\angle ABC=\angle ABD+\angle DBC=\angle ABD+x=90

So,\angle ABD=90-x=90-65=25^{\circ}

In ΔABD

AB = BD (Given)

So,\angle BAD=\angle BDA(Opposite angles of equal sides are equal)

So,\angle BAD=\angle BDA=y

So,\angle BAD+\angle BDA+\angle ABD = 180^{\circ}(Angle Sum property)

y+y+25=180

2y=180-25

2y=155

y=77.5

So, The value of x and y are 65 ° and 77.5° respectively

8 0
3 years ago
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