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3 years ago

Write the equation of the line in slope-intercept form of the statement

Mathematics

1 answer:

Law Incorporation [45]3 years ago

6 0

Answer:

Y= -2x +40

Step-by-step explanation:

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The sum of a rational number and an even integer is rational. A) Always True B) Sometimes True C) Usually True D) Never True

SOVA2 [1]

I think that the sum will always be a rational number
let's prove that

any rational number can be represented as a/b where a and b are integers and b≠0

and an integer is the counting numbers plus their negatives and 0
so like -4,-3,-2,-1,0,1,2,3,4....

so, 2 rational numbers can be represented as

a/b and c/d (where a,b,c,d are all integers and b≠0 and d≠0)

their sum is
a/b+c/d=
ad/bd+bc/bd=
(ad+bc)/bd

1. the numerator and denominator will be integers
2. that the denominator does not equal 0

alright
1.
we started with that they are all integers
ab+bc=?
if we multiply any 2 integers, we get an integer
like 3*4=12 or -3*4=-12 or -3*-4=12, etc.
even 0*4=0, that's an integer
the sum of any 2 integers is an integer
like 4+3=7, 3+(-4)=-1, 3+0=3, etc.
so we have established that the numerator is an integer

now the denominator
that is just a product of 2 integers so it is an integer

2. we originally defined that b≠0 and d≠0 so we're good

therefore, the sum of any 2 rational numbers will always be a rational number is the correct answer.

6 0

3 years ago

What is the area of this composite figure? A. 135 cm2 B. 207 cm2 C. 249 cm2 D. 279 cm2

mariarad [96]

Answer:

249 cm^2

Step-by-step explanation:

This problem becomes easier if we subdivide the figure, find the areas of the resulting figures and then sum them up.

Draw a vertical line straight down from the edge marked "4 cm" towards the edge marked "18 cm." The resulting rectangle on the left is 15.5 cm long and (18 - 7.5) cm wide, or 15.5 by 10.5 cm. Its area is 162.75 cm^2.

Next, find the area of the rectangle on the right of the line we drew. Its width is 7.5 cm and its height (15.5 - 4) cm, resulting in an area of 86.25 cm^2.

Last, add together these two subareas: combine 86.25 cm^2 and 162.75 cm^2. The total area of the composite figure is then 249 cm^2 (answer).

7 0

3 years ago

Quadrilateral ABCD undergoes a series of transformations to become A’’B’’C’’D’’. If the first transformation is a reflection ove

ale4655 [162]

C' (-6,3)
A 90 degree rotation counterclockwise around point A'

3 0

3 years ago

Read 2 more answers

Evaluate the integral, show all steps please!

Aloiza [94]

Answer:

$\displaystyle \int \dfrac{1}{(9-x^2)^{\frac{3}{2}}}\:\:\text{d}x=\dfrac{x}{9\sqrt{9-x^2}} +\text{C}$

Step-by-step explanation:

Fundamental Theorem of Calculus

$\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))$

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

Given indefinite integral:

$\displaystyle \int \dfrac{1}{(9-x^2)^{\frac{3}{2}}}\:\:\text{d}x$

Rewrite 9 as 3² and rewrite the 3/2 exponent as square root to the power of 3:

$\implies \displaystyle \int \dfrac{1}{\left(\sqrt{3^2-x^2}\right)^3}\:\:\text{d}x$

Integration by substitution

$\boxed{\textsf{For }\sqrt{a^2-x^2} \textsf{ use the substitution }x=a \sin \theta}$

$\textsf{Let }x=3 \sin \theta$

$\begin{aligned}\implies \sqrt{3^2-x^2} & =\sqrt{3^2-(3 \sin \theta)^2}\\ & = \sqrt{9-9 \sin^2 \theta}\\ & = \sqrt{9(1-\sin^2 \theta)}\\ & = \sqrt{9 \cos^2 \theta}\\ & = 3 \cos \theta\end{aligned}$

Find the derivative of x and rewrite it so that dx is on its own:

$\implies \dfrac{\text{d}x}{\text{d}\theta}=3 \cos \theta$

$\implies \text{d}x=3 \cos \theta\:\:\text{d}\theta$

Substitute everything into the original integral:

$\begin{aligned}\displaystyle \int \dfrac{1}{(9-x^2)^{\frac{3}{2}}}\:\:\text{d}x & = \int \dfrac{1}{\left(\sqrt{3^2-x^2}\right)^3}\:\:\text{d}x\\\\& = \int \dfrac{1}{\left(3 \cos \theta\right)^3}\:\:3 \cos \theta\:\:\text{d}\theta \\\\ & = \int \dfrac{1}{\left(3 \cos \theta\right)^2}\:\:\text{d}\theta \\\\ & = \int \dfrac{1}{9 \cos^2 \theta} \:\: \text{d}\theta\end{aligned}$

Take out the constant:

$\implies \displaystyle \dfrac{1}{9} \int \dfrac{1}{\cos^2 \theta}\:\:\text{d}\theta$

$\textsf{Use the trigonometric identity}: \quad\sec^2 \theta=\dfrac{1}{\cos^2 \theta}$

$\implies \displaystyle \dfrac{1}{9} \int \sec^2 \theta\:\:\text{d}\theta$

$\boxed{\begin{minipage}{5 cm}\underline{Integrating $\sec^2 kx$}\\\\$\displaystyle \int \sec^2 kx\:\text{d}x=\dfrac{1}{k} \tan kx\:\:(+\text{C})$\end{minipage}}$