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Xelga [282]
3 years ago
9

7m 10 m 7 m 9 m 34 m What the is the area

Mathematics
1 answer:
notka56 [123]3 years ago
5 0

Answer:

10m

hope this helped you alot

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(pls also provide an explanation for how i do this)
Marizza181 [45]

Answer:

slope = \frac{3}{4}

Step-by-step explanation:

Calculate the slope m using the slope formula

m = \frac{y_{2}-y_{1}  }{x_{2}-x_{1}  }

with (x₁, y₁ ) = (5, 3 ) and (x₂, y₂ ) = (9, 6 )

m = \frac{6-3}{9-5} = \frac{3}{4}

6 0
3 years ago
How is the graph of y = (-3x)^2 a transformation of the graph of y = x^2?
Lynna [10]
The graph of y = (-3x)^2 is much narrower than it's original graph, but still keeps all of the other properties of it's parent parable y = x^2. The new graph's with is that of the original parabola's with divided by 3. 
3 0
3 years ago
Segments AB, EF, and CD intersect at point C, and angle ACD is a right angle. Find the value of g. Please help me
iogann1982 [59]

Answer:

37 degree

Step-by-step explanation:

AB = 180 it is a straight line

ACD = 90 degree

DE+ DCB are complimentary angles they add up to 90 they dont need to be next to each other.

DE = 53 deg

DCB = 90-53= 37 degree

g = 37 and F is its vertical angle as they share the same corner point and lay opposite each other and each have same angle 37 degree.

8 0
3 years ago
A multiplication table. In the row labeled 3, the numbers 9, 12, 15, and 18 are highlighted. Consider the highlighted products.
Basile [38]
3, 3x3=9, 3x4=12, 3x5=15, 3x6=18
3 0
3 years ago
Can someone please help me
agasfer [191]

Answer:

See below

Step-by-step explanation:

If |x| < |y|   and both are negative the x will be  to the right of y on the number line.  The point x is loser to the zero on the number line. We can illustrate this by supposing x = -3 and y = -6.

|x| = 3 and |y| = 6  and 3 < 6  but -3 on the number line is higher ( to the right) of -6 and closer to the zero.


|-8 1/3| = 8 1/3

|7  2/3|  = 7 2/3

The absolute values of 8 1/3 > than absolute value of 7 2/3.  The point 7 2/3 is closer to the zero on a number line.

The absolute value of F is greater than absolute value of E so E is closer to the zero line than F.


6 0
3 years ago
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