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3 years ago

Solve for x.

Mathematics

1 answer:

Masteriza [31]3 years ago

7 0

Answer:

x = 2.5 or -0.8

Step-by-step explanation:

Here, we are to use completing the square method to solve for the values of x

Firstly, we divide through by 4

= x^2 -7/4x -2 = 0

Now, we move the two term to the right hand side

x^2 -7/4x = 2

Now, we divide the coefficient of x by 2 and square it;

That would be;

(-7/4 * 1/2)^2 = 49/64

we now add this value to both sides of the equation

x^2 -7/4x + 49/64 = 2 + 49/64

The right hand side can be rewritten as;

(x-7/8)^2 = 177/64

taking the square root of both sides

x-7/8 = √(177/64)

x = 7/8 ± √(177/64)

x = 7/8 + √(177/74) or 7/8 - √(177/64)

= x = 2.53802 or -0.788017

Which to the nearest tenth is

x = 2.5 or -0.8

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Writey=1/8 x+7 in standard form using integers.

sineoko [7]

Answer:

x - 8y = - 56

Step-by-step explanation:

The equation of a line in standard form is

Ax + By = C ( A is a positive integer and B, C are integers )

Given

y = $\frac{1}{8}$ x + 7

Multiply through by 8 to clear the fraction

8y = x + 56 ( subtract 8y from both sides )

0 = x - 8y + 56 ( subtract 56 from both sides )

- 56 = x - 8y, that is

x - 8y = - 56 ← in standard form

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3 years ago

How is solving inequalities with multiplication and division similar to solving equations with multiplication and division?

bija089 [108]

Because they’re both the opposites of eachothers quantities

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3 years ago

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Line segment AB has vertices A(-3, 4) and B(1, -2). A dilation centered at the orgin is applied to AB. The image has vertices A(

bixtya [17]

Answer:

option 2

Step-by-step explanation:

consider the coordinates A (- 3, 4 ) and A' (- 1, $\frac{4}{3}$ )

since the dilatation is centred at the origin, then corresponding coordinates are multiples/ divisors of each other, then image to original gives scale factor.

scale factor = $\frac{A'}{A}$ = $\frac{-1}{-3}$ = $\frac{1}{3}$ and $\frac{\frac{4}{3} }{4}$ = $\frac{1}{3}$

similarly B (1, - 2 ) and B' ( $\frac{1}{3}$ , - $\frac{2}{3}$ )

$\frac{\frac{1}{3} }{1}$ = $\frac{1}{3}$ and $\frac{-\frac{2}{3} }{-2}$ = $\frac{1}{3}$

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2 years ago

Which relations are functions?

guapka [62]

Answer:

The 1st and 2nd graph are functions while the 3rd and 4th aren't.

Step-by-step explanation:

The first and second graph doesn't have more than one output for each input. The third and fourth graph isn't a function since there's more than one output for the same input given. If you do the vertical line test, then you'll know which one is a function and which one is not a function. Hope this helps :)

3 0

3 years ago

Use the fundamental theorem of calculus to find the area of the region between the graph of the function x^5 + 8x^4 + 2x^2 + 5x

BaLLatris [955]

Answer:

The area of the region is 25,351 $units^2$ .

Step-by-step explanation:

The Fundamental Theorem of Calculus: if $f$ is a continuous function on $[a,b]$ , then

$\int_{a}^{b} f(x)dx = F(b) - F(a) = F(x) | {_a^b}$

where $F$ is an antiderivative of $f$ .

A function $F$ is an antiderivative of the function $f$ if

$F^{'}(x)=f(x)$

The theorem relates differential and integral calculus, and tells us how we can find the area under a curve using antidifferentiation.

To find the area of the region between the graph of the function $x^5 + 8x^4 + 2x^2 + 5x + 15$ and the x-axis on the interval [-6, 6] you must:

Apply the Fundamental Theorem of Calculus

$\int _{-6}^6(x^5+8x^4+2x^2+5x+15)dx$

$\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx\\\\\int _{-6}^6x^5dx+\int _{-6}^68x^4dx+\int _{-6}^62x^2dx+\int _{-6}^65xdx+\int _{-6}^615dx$

$\int _{-6}^6x^5dx=0\\\\\int _{-6}^68x^4dx=\frac{124416}{5}\\\\\int _{-6}^62x^2dx=288\\\\\int _{-6}^65xdx=0\\\\\int _{-6}^615dx=180\\\\0+\frac{124416}{5}+288+0+18\\\\\frac{126756}{5}\approx 25351.2$

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3 years ago