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nydimaria [60]
3 years ago
10

Write an expression for the 15th partial sum of the series 5/3 + 4/3 + 1 using summation notation.

Mathematics
1 answer:
mafiozo [28]3 years ago
3 0

Answer:

Answer choice C

Step-by-step explanation:

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Alberto spent $115 on shirts T-shirts cost $10 and long sleeve shirts cost $15 if he bought a total of nine then how many of eac
Step2247 [10]

Answer:

4 T-shirts and 5 long sleeves shirts.

Step-by-step explanation:

T-shirt cost =$10

Long sleeve shirt cost = $15

Total number of shirts = 9

Total number of T-shirts = 4

Cost of 4 T-shirts = 4 × 10 =$ 40

Total number of long sleeve shirt = 5

Cost of 5 long sleeve shirt = 5 × 15 =$ 75

Verification:

75+40 = $115 proved!

8 0
3 years ago
(1) The ratios 18:3 and 6:1 are equivalent. Why?
aliya0001 [1]

Answer:

Because if you take 18:3 and divide both sides by 3, you will get 6:1

Step-by-step explanation:

Because 18:3 divided by 3 on both sides equals 6:1, also 6:1 times 3 on both sides equals 18:3.

6 0
3 years ago
How to draw a picture that represents 24 divided by 6
olasank [31]
Draw a cake or a pie cut into 24 pieces. Then draw 6 figures around the cake. color code each group of 4 pieces of cake and coordinate it with figure.

Ex: If 4 pieces of the cake were red, color one of the figures red so each figure gets 4 pieces of the cake.
8 0
3 years ago
Read 2 more answers
Jake is employed by don’s auto sales and earns 4% commission on each car that he sells. if jake sold a car for $6,700, how much
Kazeer [188]
The answer to this question is 268 your welcome 

6 0
3 years ago
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Let S be the surface defined by x 2 + 2y 3 + 3z 4 = 6. Let T be the surface defined parametrically by r(u, v) = (1+ln u, 2e v+u−
aleksandrvk [35]

The tangent to C through (1, 1, 1) must be perpendicular to the normal vectors to the surfaces S and T at that point.

Let f(x,y,z)=x^2+2y^3+3z^4. Then S is the level curve f(x,y,z)=6. Recall that the gradient vector is perpendicular to level curves; we have

\nabla f(x,y,z)=(2x,6y,12z^2)

so that the gradient of f at (1, 1, 1) is

\nabla f(1,1,1)=(2,6,12)

For the surface T, we have

\begin{cases}1+\ln u=1\\2e^v+u-2=1\\uv+1=1\end{cases}\implies u=1,v=0

so that \vec r(1,0)=(1,1,1). We can obtain a vector normal to T by taking the cross product of the partial derivatives of \vec r(u,v), and evaluating that product for u=1,v=0:

\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}=\left(u-2ve^v,-1,\dfrac{2e^v}u\right)

\left(\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}\right)(1,0)=(1,-1,2)

Now take the cross product of the two normal vectors to S and T:

(2,6,12)\times(1,-1,2)=(24,8,-8)

The direction of vector (24, 8, -8) is the direction of the tangent line to C at (1, 1, 1). We can capture all points on the line containing this vector by scaling it by t\in\Bbb R. Then adding (1, 1, 1) shifts this line to the point of tangency on C. So the tangent line has equation

\vec\ell(t)=(1,1,1)+t(24,8,-8)=(1+24t,1+8t,1-8t)

7 0
3 years ago
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