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3 years ago

Need help with 4 questions

Mathematics

1 answer:

Galina-37 [17]3 years ago

3 0

Answer:

Step-by-step explanation:

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A motorboat can maintain a constant speed of 44 miles per hour relative to the water. The boat makes a trip upstream to a certai

serg [7]

Answer:

Step-by-step explanation:

Distance upstream = (34-c)(39/60) miles

--------------------------

Distance downstream = (34+c)(29/60) miles

-----------------------------------------------

Equation:

distance = rate*time

distance up = distance down

(34-c)(39/60) = (34+c)(29/60)

Multiply both sides by 60 to get:

39(34-c) = 29(34+c)

39*34 - 39c = 29*34 + 29c

68c = 340

c = 5 mph speed of the current

4 0

3 years ago

h(n)=-13nh(n)=−13nh, left parenthesis, n, right parenthesis, equals, minus, 13, n Complete the recursive formula of h(n)h(n)h, l

mr Goodwill [35]

Answer:

h(1)=-13

$h(n)=h(n-1)-13, n\geq 2$

Step-by-step explanation:

We are given that

$h(n)=-13n$

We have to find the recursive formula of h(n).

Substitute n=1

$h(1)=-13$

n=2

$h(2)=-13-13=-26=h(1)-13$

n=3

$h(3)=-13(3)=-39=-26-13=h(2)-13$

$h(4)=-13(4)=-52=-39-13=h(3)-13$

$h(n)=h(n-1)-13$

Therefore, the recursive formula is given by

h(1)=-13

$h(n)=h(n-1)-13, n\geq 2$

5 0

4 years ago

What is the measure of angle A in the triangle below? Right triangle A B C is shown. Side B C has a length of 9 and hypotenuse A

klasskru [66]

Answer:

The answer is 30°

Step-by-step explanation:

I finished the test on edgen and passed

8 0

3 years ago

Suppose r(t)=costi+sintj+3tk represents the position of a particle on a helix, where z is the height of the particle above the g

Ilia_Sergeevich [38]

a. The $\vec k$ component tells you the particle's height:

$3t=16\implies t=\dfrac{16}3$

b. The particle's velocity is obtained by differentiating its position function:

$\vec v(t)=\dfrac{\mathrm d\vec r(t)}{\mathrm dt}=-\sin t\,\vec\imath+\cos t\,\vec\jmath+3\,\vec k$

so that its velocity at time $t=\frac{16}3$ is

$\vec v\left(\dfrac{16}3\right)=-\sin\dfrac{16}3\,\vec\imath+\cos\dfrac{16}3\,\vec\jmath+3\,\vec k$

c. The tangent to $\vec r(t)$ at $t=\frac{16}3$ is

$\vec T(t)=\vec r\left(\dfrac{16}3\right)+\vec v\left(\dfrac{16}3\right)t$

4 0

3 years ago

-2(9 + 3n) - 9 (1 - 10n)*

GuDViN [60]

Answer:

-27+84n

Step-by-step explanation:

4 0

3 years ago

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