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3 years ago

Are two lines with the same slope and different y-intercepts always, sometimes, or never perpendicular?

Mathematics

2 answers:

ryzh [129]3 years ago

4 0

They are never perpendicular. They will be parallel and will never meet!

lorasvet [3.4K]3 years ago

3 0

Answer:

Never perpendicular

Step-by-step explanation:

Two lines with the same slope and different y-intercepts on the Cartesian Plane, behave like that graph below. Since they do not form a 90º angle and do not have any common point, then they are neither perpendicular nor concurrent. So these are parallel lines.

Notice that parallel lines share the value for the slope. In this case, m=3

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What property is illustrated by the statement? 0 + x = x A. Identity property of addition. B.commutative property of addition. C

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A. identity property of addition

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4 years ago

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What are the roots of the polynomial equation x3 - 6x = 3x2 - 8? Use a graphing calculator and a system of equations.

makkiz [27]

Answer:

d : 4, 1, -2.

Step-by-step explanation:

There's no need for a calculator in my opinion because we can use the rational root theorem which states that the equations of this form:

$\sum_{i=0}^{n} a_{i}x^i = 0 = a_{0} + a_{1}x + a_2x^2 + ... + a_nx^n = 0\\ a_i \in \mathbb{Z}$

Have rational roots, than the roots are of the form: $\frac{k \cdot a_o}{p \cdot a_n}, \ where \ k, p \in \mathbb{Z}.$

Rewriting the equation we have:

$x^3 - 3x^2 -6x + 8 = 0\\By \ our\ previous \ claim \ x \in D_8 \ where \ D_n \ the \ set \ of \ divisors \ of \ n.\\D_8 = \{\pm 1, \pm 2, \pm 4, \pm 8\} \\We \ plug \ in \ some \ number from \ D_8.\\Letting \ x = -2;\\(-2)^3 -3(-2)^2 -6(-2) + 8 = 0 \ \ \ Thus \ x+2 \ is \ a \ factor.\\We \ can \ now \ simplify \ the \ eq. \ using \ Polynomial \ Long \ Division.$

$x^3 -3x^2 -6x + 8 = (x+2)(Q(x))\\To \ find \ Q \ we \ divide \ the \ original \ equation \ by \ x + 2; which \ yields:\\Q(x) = x^2 - 5x + 4.\\So \ we \ can \ use \ the \ quadratic \ formula \ to \ find \ the \ roots:\\x = 4,x = 1. \\Thus \ x \in \{-2, 1, 4\}. \ \ These \ are \ all \ the \ roots.$

5 0

4 years ago

Please help me I’m really stuck

dalvyx [7]

Answer:

/ 3=/ 5 { alternate anglea}

/ 5+/6 =180° {straight angle}

/ 6=180-26

/ 6=154°

hope it helps

<h3>stay safe healthy and happy.</h3>

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3 years ago

In 2014, the enrollment at an after-school childcare program was approximately 2,500 students. In 2016, the enrollment had incre

Ne4ueva [31]

Answer:

Step-by-step explanation:

In 2 years it grows by 1,500 so in 1 year it should be 750 because 1500 divided by 2 = 750

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3 years ago

Explain why it would not make sense to approximate 256 using the binomial theorem for the binomial (20 + 5)6.

Olenka [21]

The formula for the Binomial Theorem with a power 6 is as:

$(x+y)^6=x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+y^6$

Thus, if we plug in 20 for x and 5 for y, our first term itself will be $20^6=64000000$ which is much greater than 256 and thus it will not make any sense to use $(20+5)^6$ to approximate 256 using the binomial theorem.

Also, it will not make any sense to use $(20+5)6$ as that has no power and we know that Binomial Theorem makes use of Power. Anyway, $(20+5)6=150\neq 256$ .

Our best bet here would be to use the equation with power 8:

$(x+y)^8=x^8+8x^7y+28x^6y^2+56x^5y^3+70x^4y^4+56x^3y^8+28x^2y^6+7xy^7+y^8$

and have $x=1$ and $y=1$ which will give us

$(1+1)^8=1^8+8(1)^7(1)+28(1)^6(1)^2+56(1)^5(1)^3+70(1)^4(1)^4+56(1)^3(1)^8+28(1)^2(1)^6+7(1)(1)^7+(1)^8 =256$

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4 years ago

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