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yKpoI14uk [10]
2 years ago
12

Okkkkkkk jjjjjjjjhhhhh

Mathematics
1 answer:
JulsSmile [24]2 years ago
5 0
Okkkkkkkkkkkkkkkkkkkkkkkkkkk
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What is 2+2+3+4+4+4+4​
antiseptic1488 [7]

Answer:

23

Step-by-step explanation:

8 0
3 years ago
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A polynomial divided by a polynomial is a polynomial. always, sometimes, or never true?
Artemon [7]
I think it is always true. I am not sure but I hope I'm right
3 0
3 years ago
Barack walks 2 kilometers south and then x kilometers east. ​ ​He ends 5 kilometers away from his starting position. ​ ​ Find x
amid [387]

Answer:

4.6 km

Step-by-step explanation:

-This is a Pythagorean Theorem problem where the distances walked are effectively the base and height of the imaginary right triangle thereof.

#Apply Pythagorean theorem to solve:

b^2+h^2=H^2\\\\2^2+x^2=5^2\\\\x^2=5^2-2^2=21\\\\x=\sqrt{21}\\\\x=4.58\  km

\approx 4.6\ km

Hence, the distance x is 4.6 km

7 0
3 years ago
Which of the following is a point-slope equation of a line that passes through
Mademuasel [1]

Answer:

The Answer is: y = -2/3x + 10/3

Step-by-step explanation:

Given points (-1, 4) and (8, -2) first find the slope:

m = (y - y1)/(x - x1)

m = (4 - (-2)) / (-1 - 8)

m = 4 + 2 / -9

m = -6/9 = -2/3

Use the point slope form and point (8, -2) and slove for y:

y - y1 = m(x - x1)

y - (-2) = -2/3(x - 8)

y + 2 = -2/3x + 16/3

y = -2/3x + 16/3 - 2

y = -2/3x + 16/3 - 6/3

y = -2/3x + 10/3

Proof using the point (-1, 4):

f(-1) = -2/3(-1) + 10/3

= 2/3 + 10/3

= 12/3 = 4, giving point (-1, 4)

Hope this helps! Have an Awesome Day!! :-)

4 0
3 years ago
Find the value of the six trig functions given the triangle below.
Y_Kistochka [10]

\text{Here,}\\\\\text{Opposite side,}~ O = 2\\\\\text{Hypotenuse,}~ H   = \sqrt 5 \\\\\text{Adjacent side,}~ A = \sqrt{5-4} = \sqrt 1  =1  ~~ ;[\text{By using Pythagorean theorem}]\\\\\\\sin \theta = \dfrac{O}H = \dfrac 2{\sqrt 5}\\\\\\\cos \theta = \dfrac{A}{H} = \dfrac{1}{\sqrt 5} \\ \\\\ \tan \theta = \dfrac{O}{A} = \dfrac 21 = 2

csc \theta = \dfrac{1}{\sin \theta} = \dfrac{\sqrt 5}2\\\\\\\sec \theta = \dfrac{1}{\cos \theta} = \sqrt 5  \\\\\\\cot \theta = \dfrac 1{\tan \theta} = \dfrac 12

5 0
3 years ago
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