The definition of similar triangles says that 2 triangles are similar if they have the same shape but different size. There are two criteria to check for this:
1) If all angles in one triangle are equal to the angles in another one, then the 2 are equal.
2) If the sides have the same proportions, then the 2 triangles are similar.
1) We have that all the angles of the 2 triangles have an equal angle in the other triangle. In specific, Q is matched to B, P to A and R to C. Hence, since corresponding angles are congruent, the two triangles are similar.
2) Here we are given information about the sides of the triangles, so we will check the second criterion. We form the ratio of the largest sides of each trangle and the shortest sides. 30/5=6. For the shortest sides, 18/3=6. Finally for the middle sides, 24/4=6. Hence, we have that the triangles are similar since the ratios are equal. (it doesn't matter whether we take the bigger or the smaller side as a numerator, as long as we are consistent).
Answer:
A: b^3. B. X^3
Step-by-step explanation:
When dividing subtract the exponents is the property.
Answer: a) y = f(x - 6)
b) y = f(x) - 2
<u>Step-by-step explanation:</u>
For transformations we use the following formula: y = a f(x - h) + k
- a = vertical stretch
- h = horizontal shift (positive = right, negative = left)
- k = vertical stretch (positive = up, negative = down)
a) f(x) has a vertex at (-1, 1)
M has a vertex at (5, 1)
The vertex shifted 6 units to the right → h = +6
Input h = +6 into the equation and disregard "a" and "k" since those didn't change. ⇒ y = f(x - 6)
b) f(x) has a vertex at (-1, 1)
N has a vertex at (-1, -1)
The vertex shifted down 2 units → k = -2
Input k = -2 into the equation and disregard "a" and "h" since those didn't change. ⇒ y = f(x) - 2
Answer:
-11
8 = 2x + 30
8 - 30 = -22
-22 ÷ 2 = -11
x = -11
Also if you place 2×(-11) + 30 you get 8
Step-by-step explanation:
<h2>
<em><u>You can solve this using the binomial probability formula.</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows:</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: </u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) </u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157when x=3 (4 3)(1/6)^3(5/6)^4-3 = 0.0154</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157when x=3 (4 3)(1/6)^3(5/6)^4-3 = 0.0154when x=4 (4 4)(1/6)^4(5/6)^4-4 = 0.0008</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157when x=3 (4 3)(1/6)^3(5/6)^4-3 = 0.0154when x=4 (4 4)(1/6)^4(5/6)^4-4 = 0.0008Add them up, and you should get 0.1319 or 13.2% (rounded to the nearest tenth)</u></em></h2>
