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3 years ago

Write the quadric equation in standard form and then choose the value of "b" (2x-1) (x+5)=0

Mathematics

1 answer:

ASHA 777 [7]3 years ago

6 0

Answer:

2x² + 9x - 5 = 0

b = 9

Step-by-step explanation:

(2x - 1) (x + 5) = 0

Expand

2x² + 10x - x - 5 = 0

Simplify

2x² + 9x - 5 = 0

b = 9

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Find f. (Use C for the constant of the first antiderivative and D for the constant of the second antiderivative. f"(x) = 2x + 9e

ad-work [718]

Answer:

$f(x)=\frac{x^3}{3}+9e^x+Cx+D$

Step-by-step explanation:

The function F(x) is an antiderivative of the function f(x) on an interval I if

F′(x) = f(x) for all x in I.

The function F(x) + C is the General Antiderivative of the function f(x) on an interval I if F′(x) = f(x) for all x in I and C is an arbitrary constant.

The Indefinite Integral of f(x) is the General Antiderivative of f(x).

$\int {f(x)} \, dx =F(x)+C$

To find the first antiderivative you must integrate the function $f''(x) = 2x + 9e^x$

$f'(x)=\int { 2x + 9e^x} \, dx \\\\\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx\\\\\int \:2xdx+\int \:9e^xdx = x^2+9e^x\\\\\mathrm{Add\:a\:constant\:to\:the\:solution}\\\\x^2+9e^x+C$

To find the second antiderivative you must integrate the function $f'(x) =x^2+9e^x+C$

$f(x)=\int {x^2+9e^x+C} \, dx\\\\\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx\\\\\int \:x^2dx+\int \:9e^xdx+\int \:Cdx= \frac{x^3}{3}+9e^x+Cx\\\\\mathrm{Add\:a\:constant\:to\:the\:solution}\\\\\frac{x^3}{3}+9e^x+Cx+D$

Therefore,

$f(x)=\frac{x^3}{3}+9e^x+Cx+D$

8 0

2 years ago

Help pls no links I'll give briliantest if you give me the right answer

timama [110]

Answer:

4 unit, 3 units

Step-by-step explanation:

It can be observed that there are 4 units on each side of the square and the triangle has a height of 3 units using the picture

3 0

3 years ago

Can anyone tell me the standard form of this parabola equation?

irina [24]

Answer:

$x=4y^2-24y+30$

$(x-4.5)^2+(y-2.5)^2=23$

Step-by-step explanation:

Standard form of a sideways parabola: $x=ay^2+by+c$

Given equation:

$4y^2-x-24y+30=0$

Add x to both sides:

$4y^2-24y+30=x$

$\implies x=4y^2-24y+30$

--------------------------------------------------------------------------------------------

Standard form of circle equation

$(x-a)^2+(y-b)^2=r^2$

(where (a,b) is the center and r is the radius)

Given equation:

$2x^2+2y^2-18x-10y+7=0$

Group like terms:

$2x^2-18x+2y^2-10y+7=0$

Divide by 2:

$x^2-9x+y^2-5y+3.5=0$

Factor by completing the square for each variable:

$(x-4.5)^2-20.25+(y-2.5)^2-6.25+3.5=0$

Rearrange into standard form:

$(x-4.5)^2+(y-2.5)^2=23$

Therefore, the circle has a center at (4.5, 2.5) and a radius of √23

7 0

2 years ago

An outlier is a data value that is numerically distant from ________ of the other data points in a set of data

laila [671]

In the statement regarding outlier we have to fill "most" to complete the statement.

Given An outlier is a data value that is numerically distant from_________ of the other data points in a set of data.

We have to fill the blank. We can fill " most" in the blank.

If a number in data is more than 1.5 times the length of the box away from either of the upper and lower quartile of the distribution then that number is said to be an outlier.

There are two types of outliers: upper outliers, lower.

If any values in the data set are greater than the upper boundary then those values are the upper outliers.

If any values in the data set are less than the lower boundary then those values are called lower outliers.

Lower quartile = $Q_{1}-1.5*IQR$