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qaws [65]
3 years ago
14

Which is larger, 7 feet or 84 inches?

Mathematics
1 answer:
Anna35 [415]3 years ago
6 0

Answer:

They are equal

Step-by-step explanation:

12 inches is equal to 1 foot. 7 feet is equal to 12x7 which is equal to 84. Therefore, they are the same.

If this answer has helped you, please mark this as brainliest

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Give a direct proof for each of the following statements about the integers.
kkurt [141]

Answer:

See explanation below

Step-by-step explanation:

If x is even, then we know that x = 2k for a k ≥1

If y is odd, then we know that y = 2j -1 for a j≥1

Therefore, x + y can be written similarly as:

x+y

2k +(2j-1)\\2(k+j)-1

And clearly, 2(k + j) -1 is odd.

Therefore, the sum x + y is an odd number.

6 0
3 years ago
I honestly need help with these
Brilliant_brown [7]

9. The curve passes through the point (-1, -3), which means

-3 = a(-1) + \dfrac b{-1} \implies a + b = 3

Compute the derivative.

y = ax + \dfrac bx \implies \dfrac{dy}{dx} = a - \dfrac b{x^2}

At the given point, the gradient is -7 so that

-7 = a - \dfrac b{(-1)^2} \implies a-b = -7

Eliminating b, we find

(a+b) + (a-b) = 3+(-7) \implies 2a = -4 \implies \boxed{a=-2}

Solve for b.

a+b=3 \implies b=3-a \implies \boxed{b = 5}

10. Compute the derivative.

y = \dfrac{x^3}3 - \dfrac{5x^2}2 + 6x - 1 \implies \dfrac{dy}{dx} = x^2 - 5x + 6

Solve for x when the gradient is 2.

x^2 - 5x + 6 = 2

x^2 - 5x + 4 = 0

(x - 1) (x - 4) = 0

\implies x=1 \text{ or } x=4

Evaluate y at each of these.

\boxed{x=1} \implies y = \dfrac{1^3}3 - \dfrac{5\cdot1^2}2 + 6\cdot1 - 1 = \boxed{y = \dfrac{17}6}

\boxed{x = 4} \implies y = \dfrac{4^3}3 - \dfrac{5\cdot4^2}2 + 6\cdot4 - 1 \implies \boxed{y = \dfrac{13}3}

11. a. Solve for x where both curves meet.

\dfrac{x^3}3 - 2x^2 - 8x + 5 = x + 5

\dfrac{x^3}3 - 2x^2 - 9x = 0

\dfrac x3 (x^2 - 6x - 27) = 0

\dfrac x3 (x - 9) (x + 3) = 0

\implies x = 0 \text{ or }x = 9 \text{ or } x = -3

Evaluate y at each of these.

A:~~~~ \boxed{x=0} \implies y=0+5 \implies \boxed{y=5}

B:~~~~ \boxed{x=9} \implies y=9+5 \implies \boxed{y=14}

C:~~~~ \boxed{x=-3} \implies y=-3+5 \implies \boxed{y=2}

11. b. Compute the derivative for the curve.

y = \dfrac{x^3}3 - 2x^2 - 8x + 5 \implies \dfrac{dy}{dx} = x^2 - 4x - 8

Evaluate the derivative at the x-coordinates of A, B, and C.

A: ~~~~ x=0 \implies \dfrac{dy}{dx} = 0^2-4\cdot0-8 \implies \boxed{\dfrac{dy}{dx} = -8}

B:~~~~ x=9 \implies \dfrac{dy}{dx} = 9^2-4\cdot9-8 \implies \boxed{\dfrac{dy}{dx} = 37}

C:~~~~ x=-3 \implies \dfrac{dy}{dx} = (-3)^2-4\cdot(-3)-8 \implies \boxed{\dfrac{dy}{dx} = 13}

12. a. Compute the derivative.

y = 4x^3 + 3x^2 - 6x - 1 \implies \boxed{\dfrac{dy}{dx} = 12x^2 + 6x - 6}

12. b. By completing the square, we have

12x^2 + 6x - 6 = 12 \left(x^2 + \dfrac x2\right) - 6 \\\\ ~~~~~~~~ = 12 \left(x^2 + \dfrac x2 + \dfrac1{4^2}\right) - 6 - \dfrac{12}{4^2} \\\\ ~~~~~~~~ = 12 \left(x + \dfrac14\right)^2 - \dfrac{27}4

so that

\dfrac{dy}{dx} = 12 \left(x + \dfrac14\right)^2 - \dfrac{27}4 \ge 0 \\\\ ~~~~ \implies 12 \left(x + \dfrac14\right)^2 \ge \dfrac{27}4 \\\\ ~~~~ \implies \left(x + \dfrac14\right)^2 \ge \dfrac{27}{48} = \dfrac9{16} \\\\ ~~~~ \implies \left|x + \dfrac14\right| \ge \sqrt{\dfrac9{16}} = \dfrac34 \\\\ ~~~~ \implies x+\dfrac14 \ge \dfrac34 \text{ or } -\left(x+\dfrac14\right) \ge \dfrac34 \\\\ ~~~~ \implies \boxed{x \ge \dfrac12 \text{ or } x \le -1}

13. a. Compute the derivative.

y = x^3 + x^2 - 16x - 16 \implies \boxed{\dfrac{dy}{dx} = 3x^2 - 2x - 16}

13. b. Complete the square.

3x^2 - 2x - 16 = 3 \left(x^2 - \dfrac{2x}3\right) - 16 \\\\ ~~~~~~~~ = 3 \left(x^2 - \dfrac{2x}3 + \dfrac1{3^2}\right) - 16 - \dfrac13 \\\\ ~~~~~~~~ = 3 \left(x - \dfrac13\right)^2 - \dfrac{49}3

Then

\dfrac{dy}{dx} = 3 \left(x - \dfrac13\right)^2 - \dfrac{49}3 \le 0 \\\\ ~~~~ \implies 3 \left(x - \dfrac13\right)^2 \le \dfrac{49}3 \\\\ ~~~~ \implies \left(x - \dfrac13\right)^2 \le \dfrac{49}9 \\\\ ~~~~ \implies \left|x - \dfrac13\right| \le \sqrt{\dfrac{49}9} = \dfrac73 \\\\ ~~~~ \implies x - \dfrac13 \le \dfrac73 \text{ or } -\left(x-\dfrac13\right) \le \dfrac73 \\\\ ~~~~ \implies \boxed{x \le 2 \text{ or } x \ge \dfrac83}

5 0
2 years ago
John believes that the amount of sleep he gets the night before a test and his score on that test are inversely related. On his
eimsori [14]

Answer:

sometimes it be like that

Step-by-step explanation:

y

e

s

8 0
3 years ago
2. Alayna draws 18 more stars than Pearl. Alayna draws © Assessment 37 stars. How many stars does Pearl draw? 37718 = 55 Can you
Evgen [1.6K]

Answer:

hvcr;vg;ie

Step-by-step explanation:

8 0
3 years ago
Nancy and Karen went on a trip to the mountain. Nancy started her trip 30 minutes later than Karen. They started from the same p
Katyanochek1 [597]
Distance = (rate)(time)
In this problem, note that both girls traveled the same distance.

Karens distance = r * 1.5 hrs          we dont know her rate
Nancy distance = 50 * 1 hr              we know she started 30 min later
                                                         but arrived at the same time

since the distances are equal
r*1.5 = 50 * 1          all you have to do is solve for r

1.5r = 50
r = 50/1.5
r = 33 1/3 miles per hour
3 0
3 years ago
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