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Yuki888 [10]
4 years ago
10

What is the GCF of 96.5 and 64x2? 32x 3222 32,3 32,5

Mathematics
2 answers:
erica [24]4 years ago
5 0
The correct answer is 32.5
Keith_Richards [23]4 years ago
4 0
Ator
Find the greatest common factor, common factors, and all factors for a set of numbers, to use when simplifying or reducing fractions
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Find the area of trapezium where area=5cm base=6cm height=8cm
Jlenok [28]

Answer:

44cm2

Step-by-step explanation:

[8×(5+6)]/2

=(8×11)/2

=88/2

=44cm2

6 0
2 years ago
I need help with this
andre [41]

Answer:

  f(x)=\dfrac{(x+3)(x-1)^2}{(x+4)(x+3)(x-2)^2}

Step-by-step explanation:

Each vertical asymptote corresponds to a zero in the denominator. When the function does not change sign from one side of the asymptote to the other, the factor has even degree. The vertical asymptote at x=-4 corresponds to a denominator factor of (x+4). The one at x=2 corresponds to a denominator factor of (x-2)², because the function does not change sign there.

__

Each zero corresponds to a numerator factor that is zero at that point. Again, if the sign doesn't change either side of that zero, then the factor has even multiplicity. The zero at x=1 corresponds to a numerator factor of (x-1)².

__

Each "hole" in the function corresponds to numerator and denominator factors that are equal and both zero at that point. The hole at x=-3 corresponds to numerator and denominator factors of (x-3).

__

Taken altogether, these factors give us the function ...

  \boxed{f(x)=\dfrac{(x+3)(x-1)^2}{(x+4)(x+3)(x-2)^2}}

8 0
3 years ago
Please help this is on one of my finals <br> Find the sum <br> 8/14 + 7/14 + 2/14
mylen [45]
The answer would be 1 3/14
6 0
3 years ago
Which of the following values are zeroes of x(x+5)(x−3)? Select three that apply.
max2010maxim [7]
A,C, and E are all correct answers
3 0
3 years ago
Read 2 more answers
Given: △ABC, m∠A=120°, AB=AC=1<br> Find: The radius of circumscribed circle
Citrus2011 [14]

In Δ ABC, ∠A=120°, AB=AC=1

To draw a circumscribed circle Draw perpendicular bisectors of any of two sides.The point where these bisectors meet is the center of the circle.Mark the center as O.

Then join OA, OB, and OC.

Taking any one OA,OB and OC as radius draw the circumcircle.

Now, from O Draw OM⊥AB and ON⊥AC.

As chord AB and AC are equal,So OM and ON will also be equal.

The reason being that equal chords are equidistant from the center.

AM=MB=1/2 and AN=NC=1/2  [ perpendicular from the center to the chord bisects the chord.]

In Δ OMA and ΔONA

OM=ON [proved above]

OA is common.

MA=NA=1/2  [proved above]

ΔOMA≅ ONA [SSS]

∴ ∠OAN =∠OAM=60° [ CPCT]

In Δ OAN

\cos60=\frac {AN}{OA}

\frac{1}{2}=\frac{\frac{1}{2}}{OA}

OA=1

∴ OA=OB=OC=1, which is the radius of given Circumscribed circle.





4 0
3 years ago
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