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3 years ago

What is the 14th term of the sequence?

Mathematics

1 answer:

fomenos3 years ago

3 0

Answer:

D. 16,384

Step-by-step explanation:

The given sequence (including "6") is neither arithmetic nor geometric. The 14th term cannot be predicted. (It can be anything you like.)

Based on the answer choices, we assume the intended sequence is the geometric sequence ...

2, 4, 8, 16, 32, ...

which has formula ...

an = 2^n

Then the 14th term is ...

a14 = 2^14 = 16,384

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Ms. Graves gave her class 12 minutes to read. Carrie read 5 1/2 pages in that time. At what rate, in pages per hour, did Carrie

max2010maxim [7]

To make this easier, you want to do 5 1/2 into a decimal. 5.5, next divide to find her pace per minute. 12/ 5.5 =2.1818
Multiply that by 60, and your answer is 130

4 0

3 years ago

Julie keeps rabbits and hamsters as pets she has four rabbits and an odd number of hamsters she has more hamsters than rabbits t

son4ous [18]

Answer:

H is greater than 4, less than 6, and is odd. H, therefore, must be 5.

Step-by-step explanation:

Let's write down what we know:

Julie keeps Rabbits ("R") and Hamsters ("H") as pets.

R=4

She has more H than R, so:

H>4

and H is odd.

The total of R and H is less than 10, so

R+H<10

we can substitute in for R:

4+H<10

and now we can solve for H:

H<10−4=6

So we know that H is greater than 4, less than 6, and is odd. H, therefore, must be 5.

3 0

3 years ago

Find the directional derivative of the function at the given point in the direction of the vector v. f(x, y, z) = xe^y + ye^z +

LUCKY_DIMON [66]

Answer:

$D_{\vec{u}}f(0,0,0)=\frac{6}{\sqrt{54}}$

Step-by-step explanation:

We need to find the directional derivative of the function at the given point in the direction of the vector v.

$f(x, y, z)=xe^{y} + ye^{z} + ze^{x}$ ,point (0, 0, 0) and $v=$

By Theorem: If f is a differentiable function of x , y and z , then f has a directional derivative for any unit vector $\overrightarrow{v} =$ and

$D_{\overrightarrow{u}}f(x,y,z)=f_{x}(x,y,z)u_1+f_{y}(x,y,z)u_2+f_{z}(x,y,z)u_3$

where $\overrightarrow{u}=\frac{\overrightarrow{v}}{||v||}$

since, $v=$

then $\overrightarrow{u}=\frac{\overrightarrow{v}}{||v||}$

$\overrightarrow{u}=< \frac{6}{\sqrt{6^{2}+3^{2}+(-3)^{2}}},\frac{3}{\sqrt{6^{2}+3^{2}+(-3)^{2}}},\frac{-3}{\sqrt{6^{2}+3^{2}+(-3)^{2}}} >$

$\overrightarrow{u}=< \frac{6}{\sqrt{54}},\frac{3}{\sqrt{54}},\frac{-3}{\sqrt{54}} >$

The partial derivatives are

$f_{x}(x,y,z)=e^{y}+ze^{x}$

$f_{y}(x,y,z)=xe^{y}+e^{z}$

$f_{z}(x,y,z)=ye^{z}+e^{x}$

Then the directional derivative is

$D_{\vec{u}}f(x,y,z)=(e^{y}+ze^{x})(\frac{6}{\sqrt{54}})+(xe^{y}+e^{z})(\frac{3}{\sqrt{54}})+(ye^{z}+e^{x})(\frac{-3}{\sqrt{54}})$

so, directional derivative at point (0,0,0)

$D_{\vec{u}}f(0,0,0)=(e^{0}+0e^{0})(\frac{6}{\sqrt{54}})+(0e^{0}+e^{0})(\frac{3}{\sqrt{54}})+(0e^{0}+e^{0})(\frac{-3}{\sqrt{54}})$

$D_{\vec{u}}f(0,0,0)=\frac{6}{\sqrt{54}}+\frac{3}{\sqrt{54}}+\frac{-3}{\sqrt{54}}$

$D_{\vec{u}}f(0,0,0)=\frac{6+3-3}{\sqrt{54}}$

$D_{\vec{u}}f(0,0,0)=\frac{6}{\sqrt{54}}$

3 0

4 years ago

Inez and Joel work at a store that sells

faltersainse [42]

Answer:

2 hrs and 32 mins

Step-by-step explanation:

Step 1. Because Inez worked more we are going to start by subtracting the the minutes. 23-51= -28.

Step 2. Now we move on to the hours. 7-4= 3. Know we know that the hour time is going to be three. But wait!

Step 3. Because we can't have negative numbers, we need to subtract some more. 60-28= 32. Now I know that my minutes will be 32.

Step 4. But because we used a negative number we have to subtract another hour from our time.

Answer: 2 hrs and 32 mins

Hope this helps!

7 0

3 years ago

Use Green's Theorem to evaluate the line integral along the given positively oriented curve.

Archy [21]

Answer: ∫c ((3y + 7e^(√x) dx + (8x + 5 cos (y²)) dy) = ∫∫ₐ 5dA = 5/3

Step-by-step explanation:

Given that;

∫c ((3y + 7e^(√x) dx + (8x + 5 cos (y²)) dy)

Green's Theorem is given as;

∫c (P(x,y)dx + Q(x,y)dy) = ∫∫ₐ { (-β/βy) P(x,y) + (β/βy) Q(x,y) } dA

Now our P(x,y) = 3y + 7e^(√x) and our Q(x,y) = 8x + 5 cos (y²)

Since we know this, therefore; we substitute

∫c ((3y + 7e^(√x) dx + (8x + 5 cos (y²)) dy) = ∫∫ₐ { (-β/βy) (3y + 7e^(√x)) + (β/βy) (8x + 5 cos (y²)) } dA

∫c ((3y + 7e^(√x) dx + (8x + 5 cos (y²)) dy) = ∫∫ₐ ( 8-3) dA

∫c ((3y + 7e^(√x) dx + (8x + 5 cos (y²)) dy) = ∫∫ₐ 5dA

from the question, our region is defined by a lower bound: y = x² and an upper bound of y = √x

going from x = 0 to x = 1

Now calculating ∫∫ₐ 5dA by means of the description of the region, we say;

∫∫ₐ 5dA = 5¹∫₀ ₓ²∫^(√x) dydx

∫∫ₐ 5dA = 5¹∫₀ (y)∧(y-√x) ∨(y-x²) dx

∫∫ₐ 5dA = 5¹∫₀ (√x-x²) dx

∫∫ₐ 5dA = 5 [ ((x^(3/2))/(3/2)) - x³/3]¹₀ NOW since ∫[f(x)]ⁿ dx = ([f(x)]ⁿ⁺¹ / n+1) + C

then

∫∫ₐ 5dA = 5 [ ((1^(3/2))/(3/2)) - 1³ / 3) - ((0^(3/2))/(3/2)) - 0³ / 3) ]

∫∫ₐ 5dA = 5 [ ((1^(3/2))/(3/2)) - 1³ / 3)

∫∫ₐ 5dA = 5/3

Therefore ∫c ((3y + 7e^(√x) dx + (8x + 5 cos (y²)) dy) = ∫∫ₐ 5dA = 5/3

5 0

4 years ago