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sergeinik [125]
3 years ago
13

Find three consecutive odd integers such that the product of the second and the third integers is twenty-six more than three tim

es the first integer.
Mathematics
2 answers:
Ilia_Sergeevich [38]3 years ago
6 0

Step-by-step explanation:

first number=x

second number=x+2

third number=x+4

(x+2)(x+4)=3x+26

x(x+4)+2(x+4)=3x+26

x²+4x+2x+8=3x+26

x²+6x+8=3x+26

x²+6x-3x=26-8

x²+3x=18

x²+3x-18=0

you can use that to find the numbers

Step2247 [10]3 years ago
4 0

Answer:

6, 8 and 10

Step-by-step explanation:

Let the 3 consecutive odd numbers be x, x+2, x+4

product of the second and the third integer

(x+2)(x+4)

twenty-six more than three times the first integer is expressed as;

3x + 26

Equate

(x+2)(x+4) = 3x + 26

x²+4x+2x+8 = 3x + 26

x²+6x +8= 3x+26

x²+3x +8-26 = 0

x²+3x-18 = 0

Factorize

x²-6x+3x-18 = 0

x(x-6)+3(x-6) = 0

x+3 = 0 and x-6 = 0

x = -3 and 6

The first 3 number is are 6, 8 and 10

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finlep [7]

Answer:

50?

Step-by-step explanation:

I said 50 because there were 60 na butterflies so there was probably 50.

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3 years ago
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A manufacturer produces bolts of a fabric with a fixed width. A quantity q of this fabric (measured in yards) that is sold is a
BigorU [14]

Answer:

R'(20) = 2000

Step-by-step explanation:

We are given the following in the question:

Quantity, q

Selling price in dollars per yard, p

q=f(p)

Total revenue earned =

R(p)=pf(p)

f(20)=13000

This means that 13000 yards of fabric is sold when the selling price is 20 dollars per yard.

f′(20)=−550

This means that increasing the selling price by 1 dollar per yards there is a decrease in fabric sales by 550.

We have to find R'(20)

Differentiating the above expression, we have,

R'(p) = \displaystyle\frac{d(R(p))}{dp} = \frac{d(pf(p))}{dp} = f(p) + pf'(p)

Putting the values, we get,

R'(p) = f(p) + pf'(p)\\\\R'(20) = f(20) + 20(f'(20))\\\\R'(20) = 13000 + 20(-550) = 2000

7 0
4 years ago
What are the solutions of the equation x6 + 6x3 + 5 = 0? Use factoring to solve. mc017-1.jpg and x = 1 mc017-2.jpg and x = –1 mc
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We are given equation :x^6+6x^3+5=0

Use\:the\:rational\:root\:theorem

\mathrm{Therefore,\:we\:need\:to\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:5}{1}

-\frac{1}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+1

\mathrm{Compute\:}\frac{x^6+6x^3+5}{x+1}\mathrm{\:to\:get\:the\:rest\:of\:the\:eqution:\quad }x^5-x^4+x^3+5x^2-5x+5

Therefore, final factored form it

x^6\:+\:6x^3\:+\:5=\left(x+1\right)\left(x^5-x^4+x^3+5x^2-5x+5\right)

We can't factor it more.

Therefore,

x+1=0.

x=-1.

Therefore, the real solution of the equation would be -1.


5 0
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One-third (x minus 2) = one-fifth (x + 4) + 2?
Diano4ka-milaya [45]

Answer:

x = 26

Step-by-step explanation:

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