Answer:
12 centimetres and a 15 degree angle
Step-by-step explanation:
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13. Four observations:
1. m<SWT = 90° (given)
2. QW = TW (given)
3. m<RWS = m<UWV (vertical angle pair)
4. m<RWU = 180° (straight angle)
Answer:
We know that the equation of the circle in standard form is equal to <em>(x-h)² + (y-k)² = r²</em> where (h,k) is the center of the circle and r is the radius of the circle.
We have x² + y² + 8x + 22y + 37 = 0, let's get to the standard form :
1 - We first group terms with the same variable :
(x²+8x) + (y²+22y) + 37 = 0
2 - We then move the constant to the opposite side of the equation (don't forget to change the sign !)
(x²+8x) + (y²+22y) = - 37
3 - Do you recall the quadratic identities ? (a+b)² = a² + 2ab + b². Now that's what we are trying to find. We call this process <u><em>"completing the square"</em></u>.
x²+8x = (x²+8x + 4²) - 4² = (x+4)² - 4²
y²+22y = (y²+22y+11²)-11² = (y+11)²-11²
4 - We plug the new values inside our equation :
(x+4)² - 4² + (y+22)² - 11² = -37
(x+4)² + (y+22)² = -37+4²+11²
(x+4)²+(y+22)² = 100
5 - We re-write in standard form :
(x-(-4)²)² + (y - (-22))² = 10²
And now it is easy to identify h and k, h = -4 and k = - 22 and the radius r equal 10. You can now complete the sentence :)
Answer:
<u>Given:</u>
- DC ║ AB
- CM = MB as M is midpoint of BC
i) <u>Since DN and BC are transversals, we have:</u>
- ∠DCM ≅ ∠NBM and
- ∠CDM ≅ ∠BNM as alternate interior angles
<u>As two angles and one side is congruent, the triangles are also congruent:</u>
- ΔDCM ≅ ΔNBM (according to AAC postulate)
So their areas are same.
ii)
<u>The quadrilateral has area of:</u>
- A(ADCB) = A(ADMB) + A(DCM)
<u>And the triangle has area of:</u>
- A(ADN) = A(ADMB) + A(NBM)
Since the areas of triangles DCM and NBM are same, the quadrilateral ADCB has same area as triangle ADN.
Answer: 63
Step-by-step explanation:
14.5x1.5=21.75
15x2.75=41.25
21.75+41.25=63