Please provide your work for your answer, the answer is not much important, the work that led to the answer is what matters the
most. Provide answer too tho. Please help I dont have much time left. Will give brainliest to most thoughtful and correct answer. Thanks and have a great day everyone!
2 answers:
Answer:
It gives you 28 different 10-digit numbers
Step-by-step explanation:
The leftmost position digit can not be 0 --- hence, it is 1, since there is no other possibilities.
The rightmost position (the "ones" position) must be 0 to provide an even number.
Hence, you have 8 remaining positions to fill with six ones and two zeros by an arbitrary way.
You can place two digits of zero among 8 positions by =
different ways = 4*7 = 28 different ways,
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Answer:
3a) The value of x = 56
3b) The measure of ∠ H T M = 90°
3c) The radius of the circle = 53
Step-by-step explanation:
3a) ∵ A F is a tangent to the circle O at point F
∵ Secant AH intersects circle O at point T
∴ (A F)² = (A T)(A H)
∴ 7( x + 7) = (21)² ⇒ ÷ 7
∴ x + 7 = 63
∴ x = 63 - 7 = 56
3b) ∵ HM is a diameter
∴ The measure of the arc HM = 180° ⇒ semi-circle
∵ ∠ H T M is inscribed angle subtended by the arc HM
∴ m ∠ H T M = half the measure of arc HM
∴ m ∠ H T M = 180° ÷ 2 = 90°
3c) ∵ Δ H T M is a right angle triangle at T
∴ (H M)² = (M T)² + (H T)² ⇒ Pythagorean theorem
∴ (H M)² = (90)² + (56)²
∴ (H M)² = 11236
∴ HM = = 106
∴ OM = 106 ÷ 2 = 53
∵ OM is the radius of the circle O
∴ The radius = 53