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3 years ago

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Papessa [141]3 years ago

4 0

Answer: can you explain what you mean because its just numbers

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SBC is an isosceles triangle with main vertex S.

stira [4]

Answer + Step-by-step explanation:

1) D be the symmetric of B with respect to C then CD = BC

A the symmetric of C with respect to B then AB = BC

We obtain :

CD = BC

AB = BC

Then AB = CD

2) m∠SBA = 180 - SBC = 180 - SCB = m∠SCD

3) we have :

BA = CD

BS = CS

m∠SBA = m∠SCD

Then

the triangles SBA and SCD are congruent

4)

the triangles SBA and SCD are congruent Then SA = SD

Therefore SAD is an isosceles triangle.

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2 years ago

What is the area of this figure?<br><br> _______ square kilometers

Nimfa-mama [501]

Answer:

90 square kilometers

Step-by-step explanation:

1. split the figure in small definite sections

2. find the areas of the smaller sections.

3. find the sum of the areas.

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2 years ago

Read 2 more answers

Which relation is a function ?

Leokris [45]

Answer:

Bottom left

Step-by-step explanation:

If you can draw a vertical line through a graph and touch only one point, the relation is a function.

3 0

3 years ago

Can someone help me understand how to find a slope?

bezimeni [28]

Answer:

-7/6

Step-by-step explanation:

I have plotted 8,10 and 2,17 on a grid and connected them with a line to help visualize the slope of the line we are trying to find.

The answer can also be written as a decimal as shown below, but it is neater to use the fraction.

8 0

3 years ago

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