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ale4655 [162]
3 years ago
11

Factor this trinomial.b^2-6b-7?your answer should consist of two binomials.

Mathematics
2 answers:
ss7ja [257]3 years ago
6 0

Short Answer: (b - 7)(b + 1)

The answer will consist of 2 factors each of which is a binomial. Please don't take that comment personally. It is just on Brainly we are told to be accurate.

This factors by sight. You do not need the quadratic formula (but you could use it).

(b - 7)(b + 1) are the two factors that you need. When you remove the brackets, you get b^2 - 6b- 7 back.

Stolb23 [73]3 years ago
4 0

b^2-6b-7=b^2+b-7b-7=b(b+1)-7(b+1)\\\\=\boxed{(b+1)(b-7)}

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The Little League baseball club in town holds open registration. During one weekend they registered 432 baseball players. If 12
nekit [7.7K]

Answer:

36

Step-by-step explanation:

having 432 players and one team has 12 players => the number of team is 432/12=36 teams

3 0
3 years ago
Read 2 more answers
Solve (x - 2 < 5) U (x + 7 > 6).
dolphi86 [110]

Answer:

A) {x | -1 < x < 7}

Step-by-step explanation:

Given in the question two inequalities,

Inequality 1

(x - 2 < 5)

 x < 5 + 2

 x < 7

x is smaller than 7

Inequality 2

(x + 7 > 6)

 x > 6 - 7

 x > -1

x is greater then -1

When (x - 2 < 5) U (x + 7 > 6).

              (x < 7)   U  (x > -1)    

                    -1 <  x  < 7

When combined, x is greater than -1 but smaller than 7

8 0
4 years ago
Lowering powers write in terms of first power of cosine. Cos^6
yaroslaw [1]

The main identity you need is the double angle one for cosine:

\cos^2x=\dfrac{1+\cos2x}2

We get

\cos^6x=(\cos^2x)^3=\left(\dfrac{1+\cos2x}2\right)^3=\dfrac{(1+\cos2x)^3}8

Expand the numerator to apply the identity again:

\cos^6x=\dfrac{1+3\cos2x+3\cos^22x+\cos^32x}8

\cos^6x=\dfrac{1+3\cos2x+3\left(\frac{1+\cos2(2x)}2\right)+\cos2x\left(\frac{1+\cos2(2x)}2\right)}8

\cos^6x=\dfrac{1+3\cos2x+\frac32+\frac32\cos4x+\frac12\cos2x(1+\cos4x)}8

\cos^6x=\dfrac5{16}+\dfrac7{16}\cos2x+\dfrac3{16}\cos4x+\dfrac1{16}\cos2x\cos4x

Finally, make use of the product identity for cosine:

\cos2x\cos4x=\dfrac{\cos6x+\cos2x}2

so that ultimately,

\cos^6x=\dfrac5{16}+\dfrac7{16}\cos2x+\dfrac3{16}\cos4x+\dfrac1{32}\cos2x+\dfrac1{32}\cos6x

\cos^6x=\dfrac5{16}+\dfrac{15}{32}\cos2x+\dfrac3{16}\cos4x+\dfrac1{32}\cos6x

4 0
4 years ago
I need help with this question ​
satela [25.4K]

Answer:

all you need to do is say if you agree or disagree and put when you think that! :)

7 0
3 years ago
Read 2 more answers
I need help with this please ​
Andrei [34K]

Answer:

x=12 and y=1

to find the x value, you multiply the y value by 2.

6×2=12

to find y, you divide x by 2.

2/2=1

3 0
3 years ago
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