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nirvana33 [79]
3 years ago
10

Let f: 4 → and c(t): → 4. suppose ∇f(1, 1, π, e6) = (0, 1, 2, −4), c(π) = (1, 1, π, e6), and c'(π) = (13, 12, 0, 1). find d(f ∘

Mathematics
1 answer:
Tems11 [23]3 years ago
6 0
3tt hope it helps just starting out
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The mean temperature for the first 7 days in January was 6 °C.
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Hope this help!!!

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2 years ago
Can someone help me please!
Alex777 [14]

Answer:

4x^2+17x+15

Step-by-step explanation:

(4x-3)(x+5)

= 4x(x+5) -3(x+5)

= 4x^2+ 20x -3x + 15

= 4x^2 + 17x + 15

7 0
3 years ago
Find the slope of every line that is parallel to the graph of the equation x+7y=6
raketka [301]
x+7y=6

7y=-x+6

(7y)/7=(-x+6)/7

y=(-1/7)x+6/7


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2 years ago
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Amanda [17]

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yes

Step-by-step explanation:

3 0
1 year ago
A snack-size bag of M&Ms candies is opened. Inside, there are 12 red candies, 12 blue, 7 green, 13 brown, 3 orange, and 10 y
Anestetic [448]

Answer:

\frac{1}{4180}

Step-by-step explanation:

GIVEN: A snack-size bag of M&Ms candies is opened. Inside, there are 12 red candies, 12 blue, 7 green, 13 brown, 3 orange, and 10 yellow. Three candies are pulled from the bag in succession, without replacement.

TO FIND: What is the probability that the first two candies drawn are orange and the third is green.

SOLUTION:

Total  candies in the bag =57

Probability that first ball is orange, P(A)=\frac{\text{total orange candies in bag}}{\text{total candies in bag}}

                                                       =\frac{3}{57}=\frac{1}{19}

Probability that second ball is orange, P(B)=\frac{\text{total orange candies in bag}}{\text{total candies in bag}}

                                                        =\frac{2}{56}=\frac{1}{28}

Probability that third ball is green, P(C)=\frac{\text{total green candies in bag}}{\text{total candies in bag}}

                                                                 =\frac{7}{55}

Now, probability that first two balls are orange and third is green is

=P(A)\times P(B)\times P(C)

=\frac{1}{19}\times\frac{1}{28}\times\frac{7}{55}

=\frac{1}{19}\times\frac{1}{4}\times\frac{1}{55}

=\frac{1}{4180}

Hence,  probability that first two balls are orange and third is green is \frac{1}{4180}

3 0
3 years ago
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