Mathematics, 19.04.2021 21:10 tarangorogelio
David found and factored out the GCF of the polynomial 80b4 – 32b2c3 + 48b4c. His work is below.
GFC of 80, 32, and 48: 16
GCF of b4, b2, and b4: b2
GCF of c3 and c: c
GCF of the polynomial: 16b2c
Rewrite as a product of the GCF:
16b2c(5b2) – 16b2c(2c2) + 16b2c(3b2)
Factor out GCF: 16b2c(5b2 – 2c2 + 3b2)
Which statements are true about David’s work? Check all that apply.
The GCF of the coefficients is correct.
The GCF of the variable b should be b4 instead of b2.
The variable c is not common to all terms, so a power of c should not have been factored out.
The expression in step 5 is equivalent to the given polynomial.
In step 6, David applied the distributive property.
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`if you have a set of parallel lines a and b cut by transversal x, if angle 1 = 167 degrees, what is angle 4 and why. also what is the measure of angle 5 and why? you need to use the names of the angle relationship and explain. ''.
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David found and factored out the GCF of the polynomial 80b4 – 32b2c3 + 48b4c. His work is below.
GF...
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