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mario62 [17]
3 years ago
5

Please help me ASAP

Mathematics
1 answer:
scZoUnD [109]3 years ago
5 0
Slope is 1/2 remember rise over run to find slope
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Can someone please help me and explain? please
mariarad [96]
Composite figure is made up of 4 triangles and a square.

Area of Square = Length x Length 
Area of Square = 12 x 12
Area of Square= 144 cm²

Area of triangle = 1/2 x base x height
Area of triangle  = 1/2 x 12 x 8 
Area of triangle  = 48 cm²

Area of 4 Triangles = 48 x 4
Area of 4 Triangles = 192 cm²

Total Area = 144 + 192 = 336 cm² 

\Longrightarrow \boxed { \boxed { Answer : 336 cm^{2} }}


4 0
3 years ago
After sharing with the class, Dana has 4/12 of her cupcakes left. What is the fractional part of the cupcakes did she share with
kotykmax [81]


She shared 2/3 of her cupcakes


7 0
3 years ago
Read 2 more answers
Va rog sa ma ajutați: 420/180x54/180
MAVERICK [17]
First, you want to divide 420 by 180. Then, divide 54 by 180. Lastly, you want to multiply the answers of the two together.
                    - 
420/180= 2.3 

54/180= 0.3

2.3 * 0.3 = 0.69

*You just take it step by step. Hope I helped. :)
4 0
3 years ago
Read 2 more answers
Help me please and thank you ​
goldenfox [79]
I believe the answer is 78.54 , D
4 0
3 years ago
Read 2 more answers
For the function P(x) = x3 − 9x, at the point (2, −10), find the following. (a) the slope of the tangent to the curve (b) the in
Shalnov [3]

Answer:

3, in both a), b)

Step-by-step explanation:

a) The slope of the line tangent to the curve that passes through the point (2,-10) is equal to the derivative of p at x=2.

Using differentiation rules (power rule and sum rule), the derivative of p(x) for any x is p'(x)=3x^2-9. In particular, the value we are looking for is p'(2)=3(2^2)-9=12-9=3.

If you would like to compute the equation of the tangent line, we can use the point-slope equation to get y=3(x-2)-10=3x-16

b) The instantaneus rate of change is also equal to the derivative of P at the point x=2, that is, P'(2). This is equal to p'(2)=3.

4 0
3 years ago
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