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s344n2d4d5 [400]
3 years ago
14

ALGEBRAIC EXPRESSION, FACTORIZATION AND IDENTITIES... pls help me in this...​

Mathematics
1 answer:
zysi [14]3 years ago
8 0

Answer:

I solved question number 8.

a) p= 9

b) p = 17

I solved number 9 but I'm not sure about it

i) 119

ii) 14165

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Please helpp!!!!
fomenos

Answer:

Dom(gof)=Dom(f)={1,3,4}.

(gof)(1)=g{f(1)}=g(2)=3.

(gof)(3)=g{f(3)}=g(5)=1.

(gof)(4)=g{f(4)}=g(1)=3.

∴gof={(1,3),(3,1),(4,3)}.

7 0
2 years ago
Determine the range of the graph
alexandr402 [8]
Answer: Range: [0, 5]
8 0
3 years ago
What are some helpful tricks to know when solving non homogenous second order differential equation
DiKsa [7]
Science is  the answer
4 0
3 years ago
What represents the inverse of the function f(x)=4x
Yuki888 [10]

Answer:

\frac{x}{4}

Step-by-step explanation:

The given function is

f(x) = 4x

As we know

f(x) = y, which implies x= f^{-1} (y)

Therefore,

y = 4x

x = \frac{y}{4}

as x= f^{-1} (y)

So,

f^{-1} (y) = \frac{y}{4}

replacing y by x in the above expression

f^{-1} (x) = \frac{x}{4}

So our inverse function is  \frac{x}{4}

4 0
3 years ago
A homeowner is installing a swimming pool. You have been asked to install a circuit to operate a 600-watt underwater light and a
insens350 [35]

Answer:

The answer is given below

Step-by-step explanation:

The loads in the circuit are continuous duty, so the circuit can only loaded to 80% of its current rating.

Calculate 80% current rating of the circuit breaker.

Icb = [ 80/100] (20 A)

     =  (O.8) (20 A )

     = 16 A

Now we assume that the value of supply voltage is 120 V

Now we will calculate current flows via under water light which will be taken as  Load 1

<em>Ii = </em>500 W/120 V

  = 4.167 A

Current via the circulating pump which will be taken as Load 2 IS 8.5 A

<em>I = </em>load1 + load 2

<em>    = 4.167 A + 8.5 A</em>

<em>   = 12.67 A</em>

The total current in the circuit obtained which is less than the circuit breaker rating current. Therefore, the power can be operated for both the loads.

3 0
3 years ago
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