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3 years ago

Given f(x)= 4x^3 -x-7 , what is the value of x in the interval [1, 1.5] for which the function takes the value 1?

Mathematics

1 answer:

azamat3 years ago

8 0

Answer:

1.326

Step-by-step explanation:

4x³-x-7 = 1

4x³-x-8 = 0

x ≅ 1.326, -0.663+1.034i, -0.663-1.034i

-0.663+1.034i and -0.663-1.034i are extraneous solutions.

x ≅ 1.326

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Step-by-step explanation:

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V = 4/3 pi r^3

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Please please please help and solve this with steps, much help needed thank you :) 20 points for this!

leonid [27]

Answer:

$y=\sqrt{\frac{x^4-6x^2+12x+c_1}{2}},\:y=-\sqrt{\frac{x^4-6x^2+12x+c_1}{2}}$

(Please vote me Brainliest if this helped!)

Step-by-step explanation:

$\frac{dy}{dx}y=x^3+2x-5x+3$

$\mathrm{First\:order\:separable\:Ordinary\:Differential\:Equation}$

$\mathrm{A\:first\:order\:separable\:ODE\:has\:the\:form\:of}\:N\left(y\right)\cdot y'=M\left(x\right)$

1. $\mathrm{Substitute\quad }\frac{dy}{dx}\mathrm{\:with\:}y'\:$

$y'\:y=x^3+2x-5x+3$

2. $\mathrm{Rewrite\:in\:the\:form\:of\:a\:first\:order\:separable\:ODE}$

$yy'\:=x^3-3x+3$

$N\left(y\right)\cdot y'\:=M\left(x\right)$

$N\left(y\right)=y,\:\quad M\left(x\right)=x^3-3x+3$

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5. $\mathrm{Simplify}$

$y=\sqrt{\frac{x^4-6x^2+12x+c_1}{2}},\:y=-\sqrt{\frac{x^4-6x^2+12x+c_1}{2}}$

7 0

3 years ago

Read 2 more answers

Mr. Smith had 34 of a peach pie left over. He split the leftover pie among his 3 co-workers.

viktelen [127]

Answer:

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Step-by-step explanation:

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3 years ago

Solve each system using substitution: 1. 4x - 5y= -15 y= -3x + 22

Bumek [7]

4x - 5y = -15
y = -3x + 22

Plug in your y= equation into the y variable in the other equation

4x - 5 (y) = -15
^^
y = (-3x + 22) = -15

4x - 5 (-3x + 22) = -15

Distribute

4x + 15x - 110 = -15

Combine like terms and add 110 over to -15

19x = 95

Then, divide the whole equation by 19

x = 5

Then, plug in your x into your y= equation

y = -3 (5) + 22
y = -15 + 22
y = 7

x = 5
y = 7
(5, 7)

3 0

3 years ago