Please help I have no clue what I'm doing!!! Thanks :)

Mathematics

1 answer:

Kamila [148]3 years ago

4 0

A^2 + b^2 = c^2 That's the Pythagorean Theorem
Sin B = b/c
m<A + m<B = 90° because there are 180° in a triangle

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I am trying to figure out what x is for this quadratic equation, could you help? The exact question on the assignment is

aleksandrvk [35]

There seems to be something missing. Either a missing diagram or more context to the problem, or perhaps more instructions. I'm not exactly sure.

Anyways, if you simply need to solve the equation for x, then...
3x+7 = 5x-9
3x+7-3x = 5x-9-3x ... subtract 3x from both sides
7 = 2x-9
7+9 = 2x-9+9 ... add 9 to both sides
16 = 2x
2x = 16
2x/2 = 16/2 ... divide both sides by 2
x = 8

I'm assuming after this point, you'll use x somehow to determine the length of segment AD. Unfortunately there isn't enough info provided to do that, but I'll let you take over from here.

7 0

4 years ago

Find the angle between u = (-2,-5) and v = (5,2)

OleMash [197]

The angle between u = (-2,-5) and v = (5,2) is 134 degrees approximately.

<u>Solution:</u>

Given, two vectors are u = (-2, -5) and v = (5, 2)

We have to find the angle between two vectors.

We know that,

$a. b=\|a\| .\|b\| \times \cos \theta$

where $\theta$ is angle between vectors a and b

$\text { Now vectors are }(-2 \vec{\imath}-5 \vec{\jmath}) \text { and }(5 \vec{\imath}+2 \vec{\jmath})$

$(-2 \vec{\imath}-5 \vec{\jmath}) \cdot(5 \vec{\imath}+2 \vec{\jmath})=\sqrt{(-2)^{2}+(-5)^{2}} \times \sqrt{5^{2}+2^{2}} \times \cos \theta$

$\text { since }\|a\|=\sqrt{x^{2}+y^{2}} \text { for } a=x \vec{\imath}+y \vec{\jmath}$

$\begin{array}{l}{-10-10=\sqrt{29} \times \sqrt{29} \times \cos \theta} \\\\ {-20=29 \times \cos \theta} \\\\ {\cos \theta=\frac{-20}{29}} \\\\ {\theta=\cos ^{-1} \frac{-20}{29}} \\\\ {\theta=133.60}\end{array}$