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3 years ago

What is the equation of the line that passes through the points (-4,3) and (2,-6)

Mathematics

1 answer:

Nonamiya [84]3 years ago

4 0

Answer: y = -1.5x - 3

Step-by-step explanation:

We'll assume this is a straight line, with the form y=mx+b. m is the slope (the rate that y changes with a change in x) and b is the y-intercept (the value of y when x=0).

Once we can determine both m and b, we'll have an equation that answers this question. First, m.

m, the slope, is known as the Rise/Run. We can use the two given points to calculate this value. Take the two points, (-4,3) and (2,-6), and calculate the change in x (the Run), by subtrating the initial point (I'll choose (-4,3)) from the final point ((2,-6)). For x that is: 2-(-4) = 6. For y: -6-3 = -9. When x moved +6 units, y went down by 9, or -9. The slope is the Rise (-9) over the Run (6), or (-9/6). The slope of this line is - 1.5.

We now have y = -1.5x + b. To find b, use wither of the two points. I'll use (2,-6) because I like even numbers, but fell free to use the other. Both will give the same answer for b. [Try it].

Using (2,-6), we get:

y = -1.5x + b

-6 = -1.5*(2) + b

-6 = -3 + b

b = -3

The equation is y = -1.5x - 3

=====

For fun, try the other point, (-4,3) to find the value of b:

y = -1.5x + b

3 = -1.5*(-4) + b

3 = 6 + b

b = -3 The Same!

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1) Use Newton's method with the specified initial approximation x1 to find x3, the third approximation to the root of the given

neonofarm [45]

Answer:

Check below, please

Step-by-step explanation:

Hello!

1) In the Newton Method, we'll stop our approximations till the value gets repeated. Like this

$x_{1}=2\\x_{2}=2-\frac{f(2)}{f'(2)}=2.5\\x_{3}=2.5-\frac{f(2.5)}{f'(2.5)}\approx 2.4166\\x_{4}=2.4166-\frac{f(2.4166)}{f'(2.4166)}\approx 2.41421\\x_{5}=2.41421-\frac{f(2.41421)}{f'(2.41421)}\approx \mathbf{2.41421}$

2) Looking at the graph, let's pick -1.2 and 3.2 as our approximations since it is a quadratic function. Passing through theses points -1.2 and 3.2 there are tangent lines that can be traced, which are the starting point to get to the roots.

We can rewrite it as: $x^2-2x-4=0$

$x_{1}=-1.1\\x_{2}=-1.1-\frac{f(-1.1)}{f'(-1.1)}=-1.24047\\x_{3}=-1.24047-\frac{f(1.24047)}{f'(1.24047)}\approx -1.23607\\x_{4}=-1.23607-\frac{f(-1.23607)}{f'(-1.23607)}\approx -1.23606\\x_{5}=-1.23606-\frac{f(-1.23606)}{f'(-1.23606)}\approx \mathbf{-1.23606}$

As for

$x_{1}=3.2\\x_{2}=3.2-\frac{f(3.2)}{f'(3.2)}=3.23636\\x_{3}=3.23636-\frac{f(3.23636)}{f'(3.23636)}\approx 3.23606\\x_{4}=3.23606-\frac{f(3.23606)}{f'(3.23606)}\approx \mathbf{3.23606}\\$

3) Rewriting and calculating its derivative. Remember to do it, in radians.

$5\cos(x)-x-1=0 \:and f'(x)=-5\sin(x)-1$

$x_{1}=1\\x_{2}=1-\frac{f(1)}{f'(1)}=1.13471\\x_{3}=1.13471-\frac{f(1.13471)}{f'(1.13471)}\approx 1.13060\\x_{4}=1.13060-\frac{f(1.13060)}{f'(1.13060)}\approx 1.13059\\x_{5}= 1.13059-\frac{f( 1.13059)}{f'( 1.13059)}\approx \mathbf{ 1.13059}$

For the second root, let's try -1.5

$x_{1}=-1.5\\x_{2}=-1.5-\frac{f(-1.5)}{f'(-1.5)}=-1.71409\\x_{3}=-1.71409-\frac{f(-1.71409)}{f'(-1.71409)}\approx -1.71410\\x_{4}=-1.71410-\frac{f(-1.71410)}{f'(-1.71410)}\approx \mathbf{-1.71410}\\$

For x=-3.9, last root.

$x_{1}=-3.9\\x_{2}=-3.9-\frac{f(-3.9)}{f'(-3.9)}=-4.06438\\x_{3}=-4.06438-\frac{f(-4.06438)}{f'(-4.06438)}\approx -4.05507\\x_{4}=-4.05507-\frac{f(-4.05507)}{f'(-4.05507)}\approx \mathbf{-4.05507}\\$

5) In this case, let's make a little adjustment on the Newton formula to find critical numbers. Remember their relation with 1st and 2nd derivatives.

$x_{n+1}=x_{n}-\frac{f'(n)}{f''(n)}$

$f(x)=x^6-x^4+3x^3-2x$

$\mathbf{f'(x)=6x^5-4x^3+9x^2-2}$

$\mathbf{f''(x)=30x^4-12x^2+18x}$

For -1.2

$x_{1}=-1.2\\x_{2}=-1.2-\frac{f'(-1.2)}{f''(-1.2)}=-1.32611\\x_{3}=-1.32611-\frac{f'(-1.32611)}{f''(-1.32611)}\approx -1.29575\\x_{4}=-1.29575-\frac{f'(-1.29575)}{f''(-4.05507)}\approx -1.29325\\x_{5}= -1.29325-\frac{f'( -1.29325)}{f''( -1.29325)}\approx -1.29322\\x_{6}= -1.29322-\frac{f'( -1.29322)}{f''( -1.29322)}\approx \mathbf{-1.29322}\\$

For x=0.4

$x_{1}=0.4\\x_{2}=0.4\frac{f'(0.4)}{f''(0.4)}=0.52476\\x_{3}=0.52476-\frac{f'(0.52476)}{f''(0.52476)}\approx 0.50823\\x_{4}=0.50823-\frac{f'(0.50823)}{f''(0.50823)}\approx 0.50785\\x_{5}= 0.50785-\frac{f'(0.50785)}{f''(0.50785)}\approx \mathbf{0.50785}\\$

and for x=-0.4

$x_{1}=-0.4\\x_{2}=-0.4\frac{f'(-0.4)}{f''(-0.4)}=-0.44375\\x_{3}=-0.44375-\frac{f'(-0.44375)}{f''(-0.44375)}\approx -0.44173\\x_{4}=-0.44173-\frac{f'(-0.44173)}{f''(-0.44173)}\approx \mathbf{-0.44173}\\$

These roots (in bold) are the critical numbers

3 0

3 years ago

What is the product of (3y^-4)(2y^-4)?

borishaifa [10]

(x^a)(x^b)=x^(a+b)

(ab)(cd)=(a)(b)(c)(d)

x^-m=1/(x^m)

(3y^-4)(2y^-4)=
(3)(y^-4)(2)(y^-4)=
(6)(y^-8)=
6/(y^8)

4 0

3 years ago

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The number can be divided by the least common multiple of 3 and 5

kondaur [170]

9514 1404 393

Answer:

255

Step-by-step explanation:

If the sum of digits is 12, the number is divisible by 3. If the number ends in 0 or 5, it is divisible by 5. So, we're looking for ...

2x0 . . . where x is a digit and 2+x+0 = 12 . . . . . . not possible

2x5 . . . where x is a digit and 2+x+5 = 12 . . . . true for x=5

The number is 255.

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3 years ago

Mr. Anders was three times as old as Kate 5 years ago. Their total age now is 42 years. How old is Kate now?

katrin [286]

We can write this as two equations. Call Mr. Anders' age A and Kate's age K:
$A-5=3(K-5)$
$A+K=42$

Then solve for A in the second equation:
$A=42-K$

Substitute this into the first equation:
$(42-K)-5=3(K-5)$
$-K+37=3K-15$
$52=4K$
$K=13$

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3 years ago

16 is a factor of 24.<br> O A. True<br> O B. False

Alexxx [7]

The answer is b/false I hope this help

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3 years ago

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