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4 years ago

Use the x-intercept method to find all real solutions of the equation x^3-10x^2+17x+28=0

Mathematics

2 answers:

hammer [34]4 years ago

7 0

Answer:

x = -1, x = 4 and x = 7

Step-by-step explanation:

x-intercept method consists in plotting the equation, in this case x^3-10x^2+17x+28, and see where the graph intersects the x-axis. These are the points where for a given value of x, y = 0, and they are called the solutions of the equation. In this case these points are: x = -1, x = 4 and x = 7 (see figure attached).

Alekssandra [29.7K]4 years ago

6 0

A graphing calculator shows the x-intercepts of the expression on the left to be -1, 4, 7.

The real solutions to the cubic equation are x ∈ {-1, 4, 7}.

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Step-by-step explanation:

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2 years ago

Find the multiplicative inverse of -2 in mod 7<br>

Elis [28]

Answer:

So something mod 7 is -2
Lets consider what mod does as a commands: If you have 7 and then mod it by 2 ie 7%2 you get 1
Essentially mod is seeing how many times a # goes into the other number and finding the remainder.
In this case we want the remainder of 2: -8%3 =-2
note there is also a mathematical procedural way to solve this.
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3 years ago

Solve y'' + 10y' + 25y = 0, y(0) = -2, y'(0) = 11 y(t) = Preview

svetlana [45]

Answer: The required solution is

$y=(-2+t)e^{-5t}.$

Step-by-step explanation: We are given to solve the following differential equation :

$y^{\prime\prime}+10y^\prime+25y=0,~~~~~~~y(0)=-2,~~y^\prime(0)=11~~~~~~~~~~~~~~~~~~~~~~~~(i)$

Let us consider that

$y=e^{mt}$ be an auxiliary solution of equation (i).

Then, we have

$y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.$

Substituting these values in equation (i), we get

$m^2e^{mt}+10me^{mt}+25e^{mt}=0\\\\\Rightarrow (m^2+10y+25)e^{mt}=0\\\\\Rightarrow m^2+10m+25=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow m^2+2\times m\times5+5^2=0\\\\\Rightarrow (m+5)^2=0\\\\\Rightarrow m=-5,-5.$

So, the general solution of the given equation is

$y(t)=(A+Bt)e^{-5t}.$

Differentiating with respect to t, we get

$y^\prime(t)=-5e^{-5t}(A+Bt)+Be^{-5t}.$

According to the given conditions, we have

$y(0)=-2\\\\\Rightarrow A=-2$

and

$y^\prime(0)=11\\\\\Rightarrow -5(A+B\times0)+B=11\\\\\Rightarrow -5A+B=11\\\\\Rightarrow (-5)\times(-2)+B=11\\\\\Rightarrow 10+B=11\\\\\Rightarrow B=11-10\\\\\Rightarrow B=1.$