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Ahat [919]
3 years ago
9

How do you do C and B

Mathematics
1 answer:
Hoochie [10]3 years ago
4 0
"the argument" of a function, is simply "the expression being evaluated by that function".

\bf a)\qquad (x+2)^3\impliedby \stackrel{function}{(\qquad )^3}\qquad \stackrel{argument}{x+2}
\\\\\\
b)\qquad f(2x-1)\impliedby \stackrel{function}{f(\qquad )}\qquad \stackrel{argument}{2x-1}
\\\\\\
c)\qquad \sqrt{x^2+4}\qquad \stackrel{function}{\sqrt{\qquad }}\qquad \stackrel{argument}{x^2+4}
\\\\\\
d)\qquad sin(5x)\impliedby \stackrel{function}{sin(\qquad )}\qquad \stackrel{argument}{5x}
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CAN SOMEONE PLEASE HELP ME WITH THIS QUESTION AND PLEASE SHOW WORK. ASAP
kaheart [24]

Answer:

x = 20

Step-by-step explanation:

15/9 = x/12

180 = 9x

x = 20

7 0
3 years ago
Expand the following using either the Binomial Theorem or Pascal’s Triangle. You must show your work for credit. (x - 3)^5
Rudiy27
I used bionomial expansion up to and including the term x^{3}

(-3 + x)^{5}

-3^{5} (1 +  -\frac{1}{3}x)  ^{5}

-243 (1 +  -\frac{1}{3}x)  ^{5}

n = 5
x = - \frac{1}{3}x


Binomial expansion:

1 + nx +  \frac{n(n-1) x^{2} }{2!} + \frac{n(n-1)(n-2) x^{3} }{3!}


Therefore:

1 + (5*-  \frac{1}{3}x) +  \frac{5(4) (- \frac{1}{3}x)^{2}  }{2} + \frac{5(4)(3) (- \frac{1}{3}x)^{3}  }{6}

=

1 -  \frac{5}{3} x +  \frac{10}{9} x^{2} -  \frac{10}{27}  x^{3} 



Multiply by -243

=

-243 + 405x - 270x^{2} +  90  x^{3} 

3 0
3 years ago
Read 2 more answers
An eighth-grade teacher wants to improve how the students function in small groups. What is the best strategy for improving the
WINSTONCH [101]

Answer:

I don't think you are asking a math question, but in my opinion you should choose a leader that you think would be responsible in a group, then tell them what the group needs to do.

Step-by-step explanation:

4 0
3 years ago
I need help with c please.
STALIN [3.7K]

Answer:

||w|| = 6

Step-by-step explanation:

||w|| = modulus or magnitude of vector w

Since formula to get the modulus of any vector = \sqrt{\text{(x-component)}^2+\text{(y-component)}^2}

Vector w = <-6, 0>

x-component of vector w = (-6)

y-component of vector w = 0

Therefore, ||w|| = \sqrt{(-6)^2)+(0)^2}

                         = 6

3 0
3 years ago
Please help me wit dis no one would help and i really need help
lukranit [14]

(4) 64 tiles,   (5) \mathrm{N}=\mathrm{s}^{2},   (6)  N^{2}=S^{4}

Step-by-step explanation:

Given data: Side length of square tile = 1 ft

(4)  Side length of square pool = 8 ft

\text {Number of tiles} =\frac{\text { Area of square pool }}{\text { area of square tile }}

=\frac{8 \times 8}{1 \times 1}

= 64

Hence 64 tiles needed for the border.

(5) Suppose number of square tiles be N and side of square tub is s ft.

So, N = \frac{s \times s}{1 \times 1}

\Rightarrow N=s^{2}

So, the expression is \mathrm{N}=\mathrm{s}^{2}.

(6) The equivalent expression is \mathrm{N}^{2}=\mathrm{S}^{4}.

5 0
3 years ago
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