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3 years ago

Which equation below would be parallel to the line y-2x=8?

Mathematics

1 answer:

Neporo4naja [7]3 years ago

3 0

This line is y=2x+8
so you change the intersect and keep the slope
y=2x+ __ anything you want

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Consider the initial value problem:

Inessa [10]

Answer:

$\left \{ {{u'=v} \atop {v'=5v-6u-5\sin(2t)}} \right. \\u(0)=-5;u'(0)=2$

Step-by-step explanation:

The given initial value problem is;

$y''-5y'+6y=-5\sin(2t)---(1)\\y(0)=-5,y'(0)=2$

Let

$u(t)=y(t)----(2)\\v(t)=y'(t)----(3)$

Differentiating both sides of equation (1) with respect to $t$ , we obtain:

$u'(t)=y'(t)---(4)\\ \\\implies u'(t)=v(t)---(5)$

Differentiating both sides of equation (2) with respect to $t$ gives:

$v'(t)=y''(t)----(6)$

From equation (1),

$y''=5y'-6y-5\sin(2t)\\y''=5v(t)-6u(t)-5\sin(2t)\\\implies v'(t)=5v(t)-6u(t)-5\sin(2t)----(7)$

Putting t=0 into equation (2) yields

$u(0)=y(0)\\\implies u(0)=-5$

Also putting t=0 into equation (3)

$u'(0)=y'(0)\\u'(0)=2$

The system of first order equations is:

$\left \{ {{u'=v} \atop {v'=5v-6u-5\sin(2t)}} \right. \\u(0)=-5;u'(0)=2$

3 0

3 years ago

Evaluate 2313+1302 in base

Lelechka [254]

2313 in base10 - 183

6 0

3 years ago

Read 2 more answers

What is the length of the hypotenuse of the triangle?

atroni [7]

Step-by-step explanation:

Hey there!

Here;

ABC is a Right angled triangle.

Taking reference angle as angleA.

Now;

Hypotenuse (h) = AB

Perpendicular (BC)= 24 ft.

Base(AC)= 10 ft.

Now, Using Pythagoras relation;

$h = \sqrt{ {p}^{2} + {b}^{2} }$

Put all values.

$h = \sqrt{ {24}^{2} + {10}^{2} }$

Simplify it.

$h = \sqrt{576 + 100}$

$h = \sqrt{676}$

Therefore, h= 26ft.

ANS: OPTION A.

Hope it helps....

3 0

3 years ago

Read 2 more answers

What is the point-slope form equation of the graph?

sammy [17]

Answer:

y-4= 1/2(x+2)

Step-by-step explanation:

6 0

3 years ago

Which inequality represents all values of x for which the product below is defined

zavuch27 [327]

Answer:

(-∞, -2) ∪ (0, ∞)

Step-by-step explanation:

Multiply together the (7x) and the (x+2), obtaining 7x^2 + 14x. This is the correct argument (input) to Answer A.

Since the domain of the square root function includes "all real numbers equal to or greater than 0," we set 7x^2 + 14x = to 0. This factors to the original 7x*(x+2), set equal to 0. The roots are 0 and -2.

These two values create three intervals: (-∞, -2), (-2, 0) and (0, ∞).

Choosing test numbers, one from each interval: {-5, -1, 1}.

7x^2 + 14x is + at x = -5, - at x = -1, and + at x = 1.

Thus, the domain of this function is (-∞, -2) ∪ (0, ∞). In other words, the product shown is defined on (-∞, -2) ∪ (0, ∞).

5 0

3 years ago