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2 years ago

The equation x2 – 1x – 90 = 0 has solutions {a, b}. What is a + b

2 answers:

8 0

X^2 - x - 90 =0
This equation is in standard form ax^2+ bx + c
The sum of the solutions is -b/a (a and b are the coefficients from the equation above, not the solutions to the equation)
The answer is 1/1 = 1

REY [17]2 years ago

3 0

Answer:

Step-by-step explanation:

Alright, let get started.

The given equation is : $x^2-1x-90=0$

This is a quadratic equation.

The standard form of a quadratic equation is : $AX^2+BX+C=0$

Comparing the given equation with standard one, the values of A ,B and C are :

A is 1

B is -1

C is -90

Since a and b are the roots of the equation, so the formula for sum of roots a+b will be = $\frac{-B}{A}$

a+b= $-\frac{-1}{1} =1$

hence a+b will be 1 : Answer

Hope it will help :)

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3 years ago

Describe two methods for converting a mixed number to a decimal

uysha [10]

R example: 6 1/2 = 13/2 = 6.5

For the first example, six and a half is equal to thirteen halves, which is then equal to six point five. To do this, the rule to turn a mix number into a fraction is by multiplying the 2 with the 6 and then add the answer to 1, which gives 13/2 (Remember to always give the same denominator). Finally, thirteenth halves is equal to six point five (because when you divide 13 by 2, you get 6 and one left over. To continue dividing, add a 0 , and so 10 goes into 2 is five. so the decimal is 6.5

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3 years ago

PLZ HELP!!! Use limits to evaluate the integral.

Marrrta [24]

Split up the interval [0, 2] into <em>n</em> equally spaced subintervals:

$\left[0,\dfrac2n\right],\left[\dfrac2n,\dfrac4n\right],\left[\dfrac4n,\dfrac6n\right],\ldots,\left[\dfrac{2(n-1)}n,2\right]$

Let's use the right endpoints as our sampling points; they are given by the arithmetic sequence,

$r_i=\dfrac{2i}n$

where $1\le i\le n$ . Each interval has length $\Delta x_i=\frac{2-0}n=\frac2n$ .

At these sampling points, the function takes on values of

$f(r_i)=7{r_i}^3=7\left(\dfrac{2i}n\right)^3=\dfrac{56i^3}{n^3}$

We approximate the integral with the Riemann sum:

$\displaystyle\sum_{i=1}^nf(r_i)\Delta x_i=\frac{112}n\sum_{i=1}^ni^3$

Recall that

$\displaystyle\sum_{i=1}^ni^3=\frac{n^2(n+1)^2}4$

so that the sum reduces to

$\displaystyle\sum_{i=1}^nf(r_i)\Delta x_i=\frac{28n^2(n+1)^2}{n^4}$

Take the limit as <em>n</em> approaches infinity, and the Riemann sum converges to the value of the integral:

$\displaystyle\int_0^27x^3\,\mathrm dx=\lim_{n\to\infty}\frac{28n^2(n+1)^2}{n^4}=\boxed{28}$

Just to check:

$\displaystyle\int_0^27x^3\,\mathrm dx=\frac{7x^4}4\bigg|_0^2=\frac{7\cdot2^4}4=28$

4 0

2 years ago

Please help! Thanks in advance.

alukav5142 [94]

In this problem, we need to plug in the given x values for $f(x)=0$ and find a and b.

When we plug in 1, we get: $2 \times {1}^{4} - 5 \times {1}^{3} - 14 \times {1}^{2} + a \times 1 + b = 0$

Simplify:
$2 - 5 - 14 + a + b = 0$
$- 17 + a + b = 0$
$a + b = 17$

We got our first statement about the values of the variables. If we find one more we can find those 2 variables.

We have another given root: 4.

Plug it in:
$2 \times {4}^{4} - 5 \times {4}^{3} - 14 \times {4}^{2} + a \times 4 + b = 0$
$512 - 320 - 224 + 4a + b = 0$
$- 32 + 4a + b = 0$
$4a + b = 32$

Now we have our second one. We can combine them:
$a + b = 17 \\ 4a + b = 32$

I use elimination method which is easier here.

Multiply the top equation by -1: $- a - b = - 17 \\ 4a + b = 32$

Add them up:
$3a = 15$

Simplify:
$a = 5$

Now we have a, we can plug in one of those equations to find b:
$5 + b = 17$
$b = 17 - 5$
$b = 12$

So, the answers are $a=5$ and $b=12$ .

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3 years ago