I=prt,,262.5=3000*3.5 r,,r=262.5/3000*3.5= use the calculator
Answer:
7. ∠CBD = 100°
8. ∠CBD = ∠BCE = 100°; ∠CED = ∠BDE = 80°
Step-by-step explanation:
7. We presume the angles at A are congruent, so that each is 180°/9 = 20°.
Then the congruent base angles of isosceles triangle ABC will be ...
∠B = ∠C = (180° -20°)/2 = 80°
The angle of interest, ∠CBD is the supplement of ∠ABC, so is ...
∠CBD = 180° -80°
∠CBD = 100°
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8. In the isosceles trapezoid, base angles are congruent, and angles on the same end are supplementary:
∠CBD = ∠BCE = 100°
∠CED = ∠BDE = 80°
No it can not be formed because the sum of the two smaller side lengths is not greater than the longest length
Answer:
x=-8
Step-by-step explanation:
![\sqrt{4x+41}=x+5](https://tex.z-dn.net/?f=%5Csqrt%7B4x%2B41%7D%3Dx%2B5)
on squaring both sides
![4x+41=(x+5)^2\\\\4x+41=x^2+10x+25\\\\x^2+6x-16=0\\\\x^2+8x-2x-16=0\\\\x(x+8)-2(x+8)=0\\\\(x-2)(x+8)=0](https://tex.z-dn.net/?f=4x%2B41%3D%28x%2B5%29%5E2%5C%5C%5C%5C4x%2B41%3Dx%5E2%2B10x%2B25%5C%5C%5C%5Cx%5E2%2B6x-16%3D0%5C%5C%5C%5Cx%5E2%2B8x-2x-16%3D0%5C%5C%5C%5Cx%28x%2B8%29-2%28x%2B8%29%3D0%5C%5C%5C%5C%28x-2%29%28x%2B8%29%3D0)
either x-2=0 or x+8=0
either x=2 or x=-8
Putting x=2 in original equation, we get
![\sqrt{4\times 2+41}=2+5](https://tex.z-dn.net/?f=%5Csqrt%7B4%5Ctimes%202%2B41%7D%3D2%2B5)
7=7 hence, it is not an extraneous solution
Putting x=-8 in original equation
![\sqrt{4\times (-8)+41}=-8+5](https://tex.z-dn.net/?f=%5Csqrt%7B4%5Ctimes%20%28-8%29%2B41%7D%3D-8%2B5)
i.e. 3=-3
Hence, x=-8 is an extraneous solution (since it doesn't satisfy the original equation)