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2 years ago

Find the coordinate plane that represents the solution of this system y > -x - 1 y + 4 ≤ x

Mathematics

1 answer:

gulaghasi [49]2 years ago

8 0

Answer:

The graph will open to the right and will be mostly in quadrant 1

Step-by-step explanation:

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Q # 7 please help to solve

Neko [114]

The fourth choice - 5/104.

In one cube, there are 3 even numbers - 2,4,6. For two cubes, the total probability is 36.
3*3 / 36 = 9/36

For letters, there are five vowels over 26 letters = 5 / 26

Multiply the two probabilities to get the probability of rolling two even numbers and selecting one of the vowels.
9/36 * 5/26 = 5/104

6 0

3 years ago

(d) what is the line integral of f⃗ f→ around the clockwise-oriented triangle with corners at the origin, pp, and qq? Hint: sket

tekilochka [14]

Idk, so good luck!;)

5 0

3 years ago

A 23.7-foot tall flagpole casts a 42-foot shadow. What is the angle that the sun hits the flagpole? Express your answer in DMS

svp [43]

Answer:

60.6°

Step-by-step explanation:

Using trigonometry :

Opposite = Height of shadow = 42 feets

Adjacent = height of flagpole = 23.7 feets

Tan θ = opposite / Adjacent

Tan θ = height of shadow / height of flagpole

Tan θ = 42 / 23.7

Tan θ = 1.7721518

θ = tan^-1(1.7721518)

θ = 60.56

θ = 60.6°

4 0

2 years ago

Write out the form of the partial fraction decomposition of the function. Do not determine the numerical values of the coefficie

Dvinal [7]

For part (a), you have

$\dfrac x{x^2+x-6}=\dfrac x{(x+3)(x-2)}=\dfrac a{x+3}+\dfrac b{x-2}$
$x=a(x-2)+b(x+3)$

If $x=2$ , then $2=b(2-3)\implies b=-2$ .

If $x=-3$ , then $-3=a(-3-2)\implies a=\dfrac35$ .

So,

$\dfrac x{x^2+x-6}=\dfrac 3{5(x+3)}-\dfrac 2{x-2}$

For part (b), since the degrees of the numerator and denominator are the same, you first need to find the quotient and remainder upon division.

$\dfrac{x^2}{x^2+x+2}=\dfrac{x^2+x+2-x-2}{x^2+x+2}=1-\dfrac{x+2}{x^2+x+2}$

In the remainder term, the denominator $x^2+x+2$ can't be factorized into linear components with real coefficients, since the discriminant is negative $(1-4\times1\times2=-7)$ . However, you can still factorized over the complex numbers, so a partial fraction decomposition in terms of complexes does exist.

$x^2+x+2=0\implies x=-\dfrac12\pm\dfrac{\sqrt7}2i$
$\implies x^2+x+2=\left(x-\left(-\dfrac12+\dfrac{\sqrt7}2i\right)\right)\left(x-\left(-\dfrac12-\dfrac{\sqrt7}2i\right)\right)$
$\implies x^2+x+2=\left(x+\dfrac12-\dfrac{\sqrt7}2i\right)\left(x+\dfrac12+\dfrac{\sqrt7}2i\right)$

Then you have

$\dfrac{x+2}{x^2+x+2}=\dfrac a{x+\dfrac12-\dfrac{\sqrt7}2i}+\dfrac b{x+\dfrac12+\dfrac{\sqrt7}2i}$
$x+2=a\left(x+\dfrac12+\dfrac{\sqrt7}2i\right)+b\left(x+\dfrac12-\dfrac{\sqrt7}2i\right)$

When $x=-\dfrac12-\dfrac{\sqrt7}2i$ , you have

$-\dfrac12-\dfrac{\sqrt7}2i+2=b\left(-\dfrac12-\dfrac{\sqrt7}2i+\dfrac12-\dfrac{\sqrt7}2i\right)$
$\dfrac32-\dfrac{\sqrt7}2i=-\sqrt7ib$
$b=\dfrac12+\dfrac3{2\sqrt7}i=\dfrac1{14}(7+3\sqrt7i)$

When $x=-\dfrac12+\dfrac{\sqrt7}2i$ , you have

$-\dfrac12+\dfrac{\sqrt7}2i+2=a\left(-\dfrac12+\dfrac{\sqrt7}2i+\dfrac12+\dfrac{\sqrt7}2i\right)$
$\dfrac32+\dfrac{\sqrt7}2i=\sqrt7ia$
$a=\dfrac12-\dfrac3{2\sqrt7}i=\dfrac1{14}(7-3\sqrt7i)$

So, you could write

$\dfrac{x^2}{x^2+x+2}=1-\dfrac{x+2}{x^2+x+2}=1-\dfrac {7-3\sqrt7i}{14\left(x+\dfrac12-\dfrac{\sqrt7}2i\right)}-\dfrac {7+3\sqrt7i}{14\left(x+\dfrac12+\dfrac{\sqrt7}2i\right)}$

but that may or may not be considered acceptable by that webpage.

5 0

3 years ago

Read 2 more answers

3(x -y) + 4 if X = 3 and y = 2

Alchen [17]

Answer:

Step-by-step explanation:

3(3-2)+4

3(1)+4

3+4

7 0

2 years ago