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artcher [175]
2 years ago
11

Find the area of the figure. A-56cm^2 B-88cm^2 C-112cm^2 D-384cm^2

Mathematics
1 answer:
julia-pushkina [17]2 years ago
4 0
  • firstly find the area of the square which is sidexside=8x8=64cm^2
  • then find the area of the triangle which is 1/2xbasexhight=1/2x8x6=24cm^2
  • THEN add both areas (64+24)=88cm^2

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If the endpoints of the diameter of a circle are (−10, −8) and (−6, −2), what is the standard form equation of the circle? A) (x
kolbaska11 [484]
Selections B and D both appear to be appropriate.
  B) (x+8)² + (y+5)² = 13
  D) (x+8)² + (y+5)² = 13

_____
The center is at the midpoint of the diameter, ((-10-6)/2, (-8-2)/2) = (-8, -5). For center (h, k) and radius r, the equation is
  (x -h)² +(y -k)² = r²
  (x -(-8))² +(y -(-5))² = (√13)²
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4 0
3 years ago
Read 2 more answers
How do you solve 39 equal to 1 3/10b
elena-s [515]
If 39 = (1 3/10)b, we could simplify this by writing 39 - (13/10)b.

Solve this for b by mult. both terms by (10/13):

(10*39)
----------- = (10/13)(13/10)b.  Then b = 30 (answer)
      13
5 0
2 years ago
A ramp slopes from a warehouse door down to the street. The height of the ramp is 5 feet, and the ramp ends 10 feet from the bas
Tomtit [17]
If you graph this, your first point will be at (0,5), as the ramp is at the warehouse door but 5 feet up. The second point is (10,0) as it is touching the ground but it's 10 feet away from the warehouse.
To find slope, you do (y2-y1)/(x2-x1).
When substituting in the variables, you get (0-5)/(10-0), which is -5/10, which is simplified to -1/2. Of course, that is when the warehouse is Quadrant II. If you look at it from another point of view, the slope will be positive so your answer is A) 1/2.
If this was unclear feel free to comment :)
3 0
2 years ago
Read 2 more answers
Please help math
Naddika [18.5K]
   
\displaystyle\\
15(2a - 2) = 5(a^2 -1) \\  \\ 
30a - 30 = 5a^2-5\\\\
-5a^2 + 30a -30 + 5 =0\\\\
-5a^2 + 30a -25 =0 ~~~~~\Big| \times (-1) \\\\
5a^2 - 30a +25 = 0~~~~~\Big| : 5 \\\\
a^2 - 6a +5 = 0 \\  \\ 
a_{12}= \frac{-b\pm \sqrt{b^2-4ac}}{2a}= \frac{6\pm \sqrt{36-20}}{2}= \frac{6\pm \sqrt{16}}{2}=\frac{6\pm 4}{2}= \boxed{3\pm 2} \\  \\ 
a_1 = 3+ 2 = \boxed{5}\\\\
a_2 = 3-2 = \boxed{1}



8 0
3 years ago
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Step 1: (6×10^3) ÷ (2×10^2)
Nookie1986 [14]

Answer:

Error is in between the second and third step

the exponents must be subtracted

Step-by-step explanation:

8 0
2 years ago
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