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suter [353]
3 years ago
6

CHECK ANSWERS EASY POINTS

Mathematics
2 answers:
olga55 [171]3 years ago
4 0

Answer:

Yes, you are correct.

Step-by-step explanation:

Nice penmanship by the way.

coldgirl [10]3 years ago
3 0
Yeah you right lol i thought u needed help
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Where is Point B on the number line
pishuonlain [190]
I think it’s -3 1/3 because if you count your lines after -3 to -2 you get 3 and it’s on the first line so 1/3! Hope this makes sense :)
3 0
2 years ago
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The diagram shows squares 1 2 and 3
nikklg [1K]

(Area of 1) + (Area of 2) = (Area of 3) this statement about the squares must be true.

What is Right angle traingle?

A right-angled triangle is a particular kind of triangle in which one of the angles is 90 degrees. The combined angles of the other two are 90 degrees. The triangle's base and perpendicular sides both include the right angle. The longest of the three sides, known as the hypotenuse, is the third side.

we know that;

By applying the Pythagorean Theorem;

c^2 = a^2 + b^2                                                                 .............................(1)

Area of the square says A1;

A1 = a^2 unit^2                                                           ...............................(2)

Area of the square says A2;

A2 = b^2 unit^2                                                         .................................(3)

Area of the square says A1;

A3 = c^2 unit^2                                                         ..................................(4)

Replace the A1, A2, and A3 values in the equation 1 ;

A3 = A1 + A2

therefore;

(Area of 1) + (Area of 2) = (Area of 3).

Learn more about right angled triangle click here:

brainly.com/question/22790996

#SPJ4

8 0
10 months ago
What is the slope of a line that is parallel to the line y = x + 2?
Sauron [17]

Answer:

m = - 1

Step-by-step explanation:

6 0
2 years ago
Read 2 more answers
The slope of the line tangent to the curve y^2 + (xy+1)^3 = 0 at (2, -1) is ...?
Lina20 [59]
\frac{d}{dx}(y^2 + (xy+1)^3 = 0)
\\
\\\frac{d}{dx}y^2 + \frac{d}{dx}(xy+1)^3 = \frac{d}{dx}0
\\
\\\frac{dy}{dx}2y + \frac{d}{dx}(xy+1)\ *3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(\frac{d}{dx}(xy)+\frac{d}{dx}1\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(\frac{d}{dx}xy+x\frac{d}{dx}y+\frac{dx}{dx}\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(\frac{dx}{dx}y+x\frac{dy}{dx}+\frac{dx}{dx}0\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(y+x\frac{dy}{dx}\right)\ * 3(xy+1)^2 = 0
\frac{dy}{dx}2y + \left(y+x\frac{dy}{dx}\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(3y+3x\frac{dy}{dx}\right)  (xy+1)^2 = 0
\\
\\2y\ y' + \left(3y+3x\ y'\right)  (xy+1)^2 = 0
\\
\\2y\ y' + \left(3y+3x\ y'\right)  ((xy)^2+2xy+1) = 0
\\
\\2y\ y' + \left(3y+3x\ y'\right)  ((xy)^2+2xy+1) = 0
\\
\\2y\ y' + 3x^2y^3 +6xy^2+3y+(3x y' +6x^2y y'+3x^2y y') = 0
\\
\\3x^2y^3 +6xy^2+3y+(3x y' +6x^2y y'+3x^2y y'+2yy' ) = 0
\\
\\y'(3x +6x^2y+3x^2y+2y ) = -3x^2y^3 -6xy^2-3y


y' = \frac{-3x^2y^3 -6xy^2-3y}{(3x +6x^2y+3x^2y+2y )};\ x=2,y=-1
\\
\\y' = \frac{-3(2)^2(-1)^3 -6(2)(-1)^2-3(-1)}{(3(2) +6(2)^2(-1)+3(2)^2(-1)+2(-1) )}
\\
\\y' = \frac{-3(4)(-1) -6(2)(1)+3}{(6 +6(4)(-1)+3(4)(-1)-2 )}
\\
\\y' = \frac{12 -12+3}{(6 -24-12-2 )}
\\
\\y' = \frac{3}{( -32 )}

7 0
3 years ago
I just wanna make sure my answer is right can anyone confirmmm
valina [46]
You would get it wrong cause you haven’t answered anything
3 0
2 years ago
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