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Serga [27]
3 years ago
7

13. Find a cubic function with the given zeros. 7, -3, 2

Mathematics
1 answer:
emmainna [20.7K]3 years ago
5 0

\bf \begin{cases} x=7\implies &x-7=0\\ x=-3\implies &x+3=0\\ x=2\implies &x-2=0 \end{cases}~\hspace{7em}(x-7)(x+3)(x-2)=\stackrel{y}{0} \\\\\\ (x^2-4x-21)(x-2)=y \\\\\\ x^3-4x^2-21x-2x^2+8x+42=y\implies x^3-6x^2-13x+42=y

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When multiplying with decimals...
laiz [17]

Answer:

A

Step-by-step explanation:

Decimal place can be counted before multiplying. The multiply number without decimals, when you get the answer add the decimal place back.

8 0
3 years ago
What’s the correct answer
vitfil [10]

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6 0
3 years ago
Prove that: [1 + 1/tan²theta] [1 + 1/cot² thata] = 1/(sin²theta - sin⁴theta]
Stels [109]

Step-by-step explanation:

<h3><u>Given :-</u></h3>

[1+(1/Tan²θ)] + [ 1+(1/Cot²θ)]

<h3><u>Required To Prove :-</u></h3>

[1+(1/Tan²θ)]+[1+(1/Cot²θ)] = 1/(Sin²θ-Sin⁴θ)

<h3><u>Proof :-</u></h3>

On taking LHS

[1+(1/Tan²θ)] + [ 1+(1/Cot²θ)]

We know that

Tan θ = 1/ Cot θ

and

Cot θ = 1/Tan θ

=> (1+Cot²θ)(1+Tan²θ)

=> (Cosec² θ) (Sec²θ)

Since Cosec²θ - Cot²θ = 1 and

Sec²θ - Tan²θ = 1

=> (1/Sin² θ)(1/Cos² θ)

Since , Cosec θ = 1/Sinθ

and Sec θ = 1/Cosθ

=> 1/(Sin²θ Cos²θ)

We know that Sin²θ+Cos²θ = 1

=> 1/[(Sin²θ)(1-Sin²θ)]

=> 1/(Sin²θ-Sin²θ Sin²θ)

=> 1/(Sin²θ - Sin⁴θ)

=> RHS

=> LHS = RHS

<u>Hence, Proved.</u>

<h3><u>Answer:-</u></h3>

[1+(1/Tan²θ)]+[1+(1/Cot²θ)] = 1/(Sin²θ-Sin⁴θ)

<h3><u>Used formulae:-</u></h3>

→ Tan θ = 1/ Cot θ

→ Cot θ = 1/Tan θ

→ Cosec θ = 1/Sinθ

→ Sec θ = 1/Cosθ

<h3><u>Used Identities :-</u></h3>

→ Cosec²θ - Cot²θ = 1

→ Sec²θ - Tan²θ = 1

→ Sin²θ+Cos²θ = 1

Hope this helps!!

7 0
3 years ago
Find the remainder when f(x) = x3 − 14x2 51x − 22 is divided by x − 7.
Contact [7]
The remainder is -8.
5 0
3 years ago
Read 2 more answers
Solve for y: 1/2(4y+6)=4-2(y-6)/3​
gizmo_the_mogwai [7]

Answer:

0.09

Step-by-step explanation:

hope it helps thank you

8 0
3 years ago
Read 2 more answers
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