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lutik1710 [3]
3 years ago
15

What are the solution points for the system graph?

Mathematics
1 answer:
kykrilka [37]3 years ago
3 0

Answer:

if i right it (0,2)

Step-by-step explanation:

i hope u get it right

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15, Yes: a^2 +b^2 = c^2
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The square root of 225 is 15, therefore x=15.
Answer to question is yes.
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Solve for x by finding the missing side of the triangle. Round your answer to the nearest tenth ​
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tankabanditka [31]
1)
Area of largest circle - 2 * Area of one smaller circle = Area of the shaded region

AE = diameter of large circle = 48cm
radius of larger circle = diameter / 2 = 48cm / 2 = 24cm

4 circles fit across the diameter of the circle, so the diameter of the larger circle = 4 * diameter of the smaller circle
diameter of larger circle = 48cm = 4 * diameter of the smaller circle
diameter of the smaller circle = 48cm / 4 = 12cm
radius of smaller circle = diameter / 2 = 12cm / 2 = 6cm

Area of a circle = pi * r^2

Now plug the circle area equation into the first equation:
A_{shaded}=A_{l} - 2*A_{s}\\\\A_{shaded}=[\pi (r_{l})^{2}]-2*[\pi (r_{s})^{2}]\\\\A_{shaded}=[\pi (48cm)^{2}]-2*[\pi (6cm)^{2}]\\\\A_{shaded}=2304\pi-72\pi\\\\Area\ of\ shaded\ region\ is\ 2232\pi.


2)
Area of the shaded region = 2/7 * Area of the smaller circle
Area of the unshaded region = Area of larger circle + Area of smaller circle - Area of shaded region * 2
A_{unshaded}=[\pi (r_{1})^{2}]+[\pi (r_{2})^{2}]-2*[\pi (r_{2})^{2}]*\frac{2}{7}\\\\A_{unshaded}=[\pi (10cm)^{2}]+[\pi (7cm)^{2}] -\frac{4}{7}[\pi (7cm)^{2}]\\\\A_{unshaded}=100\pi\ cm^{2}+49\pi\ cm^{2}-\frac{4*49\pi\ cm^{2}}{7}\\\\A_{unshaded}=149\pi\ cm^{2}-(4*7*\pi\ cm^{2})\\\\A_{unshaded}=149\pi\ cm^{2}-28\pi\ cm^{2}\\\\\\A_{unshaded}=121\pi\ cm^{2}
3 0
4 years ago
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