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lapo4ka [179]
3 years ago
8

Solve the equation. 3(k−6) = 15 k = ____?

Mathematics
2 answers:
Anvisha [2.4K]3 years ago
6 0
This is the answer..........

const2013 [10]3 years ago
5 0
Distribute
3k-18=15
Get k by itself, add 18 to both sides
3k=33
Divide by 3 to both sides to make k by itself
k=11
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Please hurry for quiz
fenix001 [56]
The third answer is correct
5 0
3 years ago
How do you solve for b in this expression
Leona [35]
Remember you can do anything to an equation as long as you do it to both sides


8-(3+b)=b-9

remember the distributive property
a(b+c)=ab+ac

distribute that invisible -1
-(3+b)=-1(3+b)=-1(3)-1(b)=-3-b

8-3-b=b-9
add like terms
5-b=b-9
add b to both sides
5=2b-9
add 9 to both sides
14=2b
divide both sides by 2
7=b
b=7
7 0
3 years ago
A candy store makes 6 1/2 pounds of fudge in 1/4 hr how many pounds of fudge can the store make in 1 hr
ohaa [14]

Answer:

26 lbs

Step-by-step explanation:

there are four 1/4 hours in an hour  1/4 x4=1 hour

6 1/2 x4= 26 lbs in an hour

5 0
3 years ago
The measure of an angle is 56°. What is the measure of its supplementary angle?
allsm [11]

Answer:

Step-by-step explanation: Supplemenatry equal 180 degrees so if you know one find the other

6 0
3 years ago
PLEASE HURRY!!!
mylen [45]

Step-by-step answer:

Given:

Three roots of a fifth degree polynomial function f(x) are –2, 2, and 4 + i. Which statement describes the number and nature of all roots for this function?

A. f(x) has two real roots and one imaginary root.

B. f(x) has three real roots.

C. f(x) has five real roots.

D. f(x) has three real roots and two imaginary roots.

Solution:

We know from the fundamental theorem of algebra that every non-constant single variable polynomial with real coefficients of degree N has exactly N roots, multiplicities included.

Therefore there are five roots of in polynomial function f(x).

Of the five roots, we are already given that 2 of them are real, one of them is complex.

Since for a polynomial with real coefficients, each complex root (4+i), has its conjugate (4-i) as another root, so there is a minimum of 2 real and 2 complex roots.

The remaining (fifth) root must therefore be real, since we cannot have an odd number of complex roots.  This gives a total of three real and two complex roots.

8 0
3 years ago
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