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3 years ago

the lines below are perpendicular. if the slope of the green line is 3/2 , what is the slope of the red line

Mathematics

1 answer:

storchak [24]3 years ago

7 0

Answer:

-2/3

Step-by-step explanation:

The slope of a line perpendicular to one with a slope of 3/2 will have a slope that is the negative reciprocal of that:

red slope = -1/(3/2) = -2/3

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What number is 12% of 315

goldenfox [79]

Answer:

12% of 315 is 37.8 :) Tag me if I'm wrong

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What is the length side of a triangle that has vertices at (-5, -1), (-5, 5), and (3, -1)?

Rasek [7]

The side lengths of triangle are 6 units, 8 units and 10 units.

<u>SOLUTION: </u>

Given that, we have to find what is the length side of a triangle that has vertices at (-5, -1), (-5, 5), and (3, -1)

We know that, distance between two points $P\left(x_{1}, y_{1}\right) \text { and } Q\left(x_{2}, y_{2}\right)$ is given by

$P Q=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

Now,

$\begin{array}{l}{\text { Distance between }(-5,-1) \text { and }(-5,5)=\sqrt{(-5-(-5))^{2}+(5-(-1))^{2}}} \\\\ {\qquad \begin{array}{l}{=\sqrt{(-5-(-5))^{2}+(5-(-1))^{2}}} \\\\ {=\sqrt{(0)^{2}+(5+1)^{2}}=\sqrt{(6)^{2}}=6} \\\\ {=\sqrt{(-5)^{2}+(5+1)^{2}+(5-(-1))^{2}}} \\\\ {=\sqrt{(-8)^{2}+(5+1)^{2}}=\sqrt{64+36}=\sqrt{100}=10} \\\\ {=\sqrt{(3-1)^{2}+(-1-(-1))^{2}}} \\\\ {=\sqrt{(5+3)^{2}+(0)^{2}}=\sqrt{(8)^{2}}=8}\end{array}}\end{array}$

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3 years ago

Let p and q be statements. Identify the truth table for q → p.

Serjik [45]

Answer:

T F T T.

Step-by-step explanation:

Let p and q be two statement

When the value of p is true and the value of q is true then the value of $q\implies p$ is true.

When the value of p is true and the value of q is false then the value of $q\implies p$ is true.

When the value p is false and the value of q is true then the value of $q\implies p$ is false.

When the value of p is false and the value of q is false then the value of $q\implies p$ is true.

Hence, the value of $q\implies p$ is false when q is true and p is false.

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If, P(x,y) is a point on the unit circle determined by real number δ, then tanδ = what

Lisa [10]

$\bf sin(\theta)=\cfrac{opposite}{hypotenuse}
=\cfrac{y}{r}
=\cfrac{1}{csc(\theta)}
\\ \quad \\\\
% cosine
cos(\theta)=\cfrac{adjacent}{hypotenuse}
=\cfrac{x}{r}
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\\ \quad \\\\
% tangent
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$\bf cot(\theta)=\cfrac{adjacent}{opposite}
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csc(\theta)=\cfrac{hypotenuse}{opposite}
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% secant
sec(\theta)=\cfrac{hypotenuse}{adjacent}
=\cfrac{r}{x}
=\cfrac{1}{cos(\theta)}\\\\
-------------------------------\\\\
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makvit [3.9K]

Answer:

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Step-by-step explanation:

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