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aliina [53]
3 years ago
11

Factorise a squared - b squared

Mathematics
1 answer:
marusya05 [52]3 years ago
5 0
(a-b)(a+b) because when you reverse the sum you do axa and ax-b xx
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Derrick made a goal for 2 points during the basketball game. The path of the ball's height in feet can be modeled by y=-x^2-2x+2
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2 years ago
What is the area of the trapezoid with height 13 units? Enter your answer in the box. units2 A rectangle with a length of 15 and
jolli1 [7]

Answer:

Step-by-step explanation:

Area of the trapezoid = area of rectangle + 2 * area of triangle

Area of a triangle = 1/2 bh

= 1/2 * 13 * 7

= 45.5 units²

Area of a rectangle = L × B

= 15 × 13

= 195 units²

Area of the trapezoid = [195 + 2 * (45.5)]

= 195 + 91

= 286 units²

3 0
3 years ago
Read 2 more answers
A recipe calls for 125 cups of sugar. If jasmine wants to make 13 of the recipe, how many cups of sugar will she need?
iragen [17]

Answer: 1625

Step-by-step explanation: So if you want to make 13 of one recipe it said that in 1 recipe there's 125 cups of sugar so you need to multiply 123 by 13 so do this 125 x 13 = 1625 and then there, you have your answer 1,625 cups of sugar for 13 things of the recipe.

7 0
3 years ago
2783 and 7283. The value of 2 in _ is _ times the value of two in _. The 2 is the underlined digit.
NARA [144]

The value of 2 in 2783 (2000) is 10 times the value of two in 7283 (200).

Hope this helped!

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3 0
3 years ago
Problem 8-4 A computer time-sharing system receives teleport inquiries at an average rate of .1 per millisecond. Find the probab
Sphinxa [80]

Answer:  a) 0.9980, b) 0.0013, c) 0.0020, d) 0.00000026, e) 0.0318

Step-by-step explanation:

Problem 8-4 A computer time-sharing system receives teleport inquiries at an average rate of .1 per millisecond. Find the probabilities that the number of inquiries in a particular 50-millisecond stretch will be:

Since we have given that

\lambda=0.1\ per\ millisecond=5\ per\ 50\ millisecond=5

Using the poisson process, we get that

(a) less than or equal to 12

probability=  P(X\leq 12)=\sum _{k=0}^{12}\dfrac{e^{-5}(-5)^k}{k!}=0.9980

(b) equal to 13

probability= P(X=13)=\dfrac{e^{-5}(-5)^{13}}{13!}=0.0013

(c) greater than 12

probability= P(X>12)=\sum _{k=13}^{50}\dfrac{e^{-5}.(-5)^k}{k!}=0.0020

(d) equal to 20

probability= P(X=20)=\dfrac{e^{-5}(-5)^{20}}{20!}=0.00000026

(e) between 10 and 15, inclusively

probability=P(10\leq X\leq 15)=\sum _{k=10}^{15}\dfrac{e^{-5}(-5)^k}{k!}=0.0318

Hence, a) 0.9980, b) 0.0013, c) 0.0020, d) 0.00000026, e) 0.0318

6 0
3 years ago
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