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3 years ago

I need help with with problem please thanks!

Mathematics

1 answer:

liraira [26]3 years ago

5 0

40 is the correct answer. Hope this helps :)

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BRAINLIEST!!!

Sonbull [250]

Answer:

$\sqrt{37}$

Step-by-step explanation:

The distance d between a point (m , n ) and a line in the form

Ax + By + C = 0 , is calculated as

d = $\frac{|Am+Bn+C|}{\sqrt{A^2+B^2} }$

Here (m, n ) = (6, 2 ) and

6x - y = - 3 ( add 3 to both sides )

6x - y + 3 = 0 → A = 6, B = - 1, C = 3

d = $\frac{|6(6)+-1(2)+3|}{\sqrt{6^2+(-1)^2} }$

= $\frac{|36-2+3|}{\sqrt{36+1} }$

= $\frac{|37|}{\sqrt{37} }$

= $\frac{37}{\sqrt{37} }$ × $\frac{\sqrt{37} }{\sqrt{37} }$

= $\frac{37\sqrt{37} }{37}$ ← cancel 37 on numerator/ denominator

= $\sqrt{37}$

7 0

2 years ago

1 -sin^2x÷csc^1x=sinx^2<br>

Stella [2.4K]

Answer: 37

Step-by-step explanation:

6 0

2 years ago

I need help with this problem

Arada [10]

Answer:

65.56°

Step-by-step explanation:

We know that if we take dot product of two vectors then it is equal to the product of magnitudes of the vectors and cosine of the angle between them

That is let p and q be any two vectors and A be the angle between them

So, p·q=|p|*|q|*cosA

⇒ $cosA=\frac{u.v}{|u||v|}$

Given u=-8i-3j and v=-8i+8j

$|u|=\sqrt{(-8)^{2}+ (-3)^{2}} =8.544$

$|v|=\sqrt{(-8)^{2}+ (8)^{2}} =11.3137$

let A be angle before u and v

therefore, $cosA=\frac{u.v}{|u||v|}=\frac{(-8)*(-8)+(-3)*(8)}{8.544*11.3137} =\frac{40}{96.664}$

⇒ $A=arccos(\frac{40}{96.664} )=arccos(0.4138 )=65.56$

Therefore angle between u and v is 65.56°

5 0

2 years ago

Find two consecutive numbers whose product is 210.

Rus_ich [418]

The two numbers are 14 and 15.

4 0

2 years ago

Read 2 more answers

Evaluating a function graphically? How do I solve this

Artyom0805 [142]

Answer:

f(1) = -6

Explanation:

When you look at the points with x = 1, you will see a point that is open (o) and closed (•) The point that is closed (the one you are looking for) is an actual point of the function. The open point is where the function is discontinuous.

6 0

2 years ago