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kaheart [24]
3 years ago
5

(-1,2) (5,-3) What is the slope of this equation

Mathematics
2 answers:
Mazyrski [523]3 years ago
5 0
I thinks it’s b I hope this helps
VashaNatasha [74]3 years ago
4 0

Slope is the picture.

Point-slope form is y-2=-5/6*(x+1)

The y=mx+b is y=-5/6x+7/6

i hope its not confusing lol

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What is the solution to the equation? 2 x squared plus 4 equals 20 Choose all answers that are correct.
Stolb23 [73]
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8 0
3 years ago
SIMPLIFY and SHOW WORK THANKS 3∙[ 9 – 2∙ (7 – 8)]
Zepler [3.9K]

Answer:

33

Step-by-step explanation:

3∙[ 9 – 2∙ (7 – 8)]

PEMDAS,

Parentheses, start from the inside out

3∙[ 9 – 2∙ (-1)]

3∙[ 9 +2]

3* 11

33

8 0
3 years ago
Let f(x) = 5x +7 and g(x) = x-1 . Find f(g(x))
Bogdan [553]

Answer:

f(g(x))=5×(x-1)+7=5x-2

6 0
3 years ago
According to ​Lambert's law​, the intensity of light from a single source on a flat surface at point P is given by Upper L equal
malfutka [58]

Answer:

(a) L = k*(1 - sin^{2}(\theta))        

(b) L reaches its maximum value when θ = 0 because cos²(0) = 1

Step-by-step explanation:

Lambert's Law is given by:

L = k*cos^{2}(\theta)   (1)

(a) We can rewrite the above equation in terms of sine function using the following trigonometric identity:

cos^{2}(\theta) + sin^{2}(\theta) = 1

cos^{2}(\theta) = 1 - sin^{2}(\theta)  (2)

By entering equation (2) into equation (1) we have the equation in terms of the sine function:

L = k*(1 - sin^{2}(\theta))        

(b) When θ = 0, we have:

L = k*cos^{2}(\theta) = k*cos^{2}(0) = k  

We know that cos(θ) is a trigonometric function, between 1 and -1 and reaches its maximun values at nπ, when n = 0,1,2,3...

Hence, L reaches its maximum value when θ = 0 because cos²(0) = 1.

I hope it helps you!

5 0
3 years ago
Read 2 more answers
No files please........
FinnZ [79.3K]

√7(-8√10+√14)

-8√70+√98

-8√70+√49.2

-8√70+7√2

Just comment if you want to ask

3 0
3 years ago
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