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julsineya [31]
3 years ago
11

Part a : solve - vp + 40 < 65 for v .

Mathematics
1 answer:
miskamm [114]3 years ago
6 0

Answer:

\large \text{Part a:}\\\\for\ d-\dfrac{25}{d}}

\large\text{Part b:}\\\\\boxed{r=\dfrac{7}{3}w-5}

Step-by-step explanation:

\text{Part a}\\\\-vp+400,\ \text{then}\ \boxed{v>-\dfrac{25}{d}}\\\\\text{if}\ d

\text{Part b}\\\\7w-3r=15\qquad\text{subtract 7w from both sides}\\\\-3r=-7w+15\qquad\text{divide both sides by (-3)}\\\\\boxed{r=\dfrac{7}{3}w-5}

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How do u solve 2x3/5+1/5
Sergeeva-Olga [200]

Answer:

1 2/5

Step-by-step explanation:

The answer to the above question is 1 2/5

7 0
3 years ago
Please answer correctly !!!! Will mark brainliest !!!!!!!!!!!!!!
love history [14]

Answer:

A. 2.2

Step-by-step explanation:

The average rate of change between 2 x values is the total change of the y values (output values) divided by the change in the x values.Slope is special situation of that.

point 1 (x₁,y₁) → ( 2,7.5)

point 2 (x₂,y₂) → ( 18,42.5)

slope = 42.5 -7.5/18-2 =35 / 16 = 2.1875 ≈ 2.2

6 0
3 years ago
Read 2 more answers
Linda sells her bracket for $6 And her necklace for $10 what is the expression or equation
rosijanka [135]

The equation is 6x+10y

Step-by-step explanation:

The bracket is denoted by the variable x.

The necklace is denoted by the variable y.

From given, Linda sells her bracket for $6.

Thus, the expression for bracket is 6x.

Also, Linda sells her necklace for $10.

Thus, the expression for necklace is 10y.

The equation for Linda sells her bracket and necklace is 6x+10y

4 0
3 years ago
Write the equation of the line that is parallel to y=4x-3 and passes through the point (5,12)
katovenus [111]

Answer

y=4x-8

Step-by-step explanation:

m = 4

y- 12 = 4(x-5)

simplify and keep it in point-slop form

y=4x-8

6 0
3 years ago
1-What is the sum of the series? ​∑j=152j​ Enter your answer in the box.
tangare [24]

Answer:

Please see the Step-by-step explanation for the answers

Step-by-step explanation:

1)

∑\left \ {{5} \atop {j=1}} \right. 2j

The sum of series from j=1 to j=5 is:

∑ = 2(1) + 2(2) + 2(3) + 2(4) + 2(5)

  =  2 + 4 + 6 + 8 + 10

∑ = 30

2)

This question is not given clearly so i assume the following series that will give you an idea how to solve this:

∑\left \ {{4} \atop {k=1}} \right. 2k²

The sum of series from k=1 to j=4 is:

∑ = 2(1)² + 2(2)² + 2(3)² + 2(4)²

  = 2(1) + 2(4) + 2(9) + 2(16)

  =  2 + 8 + 18 + 32

∑ = 60

∑\left \ {{4} \atop {k=1}} \right. (2k)²

∑ = (2*1)² + (2*2)² + (2*3)² + (2*4)²

  = (2)² + (4)² + (6)² + (8)²

  = 4 + 16 + 36 + 64

∑ = 120

∑\left \ {{4} \atop {k=1}} \right. (2k)²- 4

∑ = (2*1)²-4 + (2*2)²-4 + (2*3)²-4 + (2*4)²-4

  = (2)²-4 + (4)²-4 + (6)²-4 + (8)²-4

  = (4-4) + (16-4) + (36-4) + (64-4)

  = 0 + 12 + 32 + 60

∑ = 104

∑\left \ {{4} \atop {k=1}} \right. 2k²- 4

∑ = 2(1)²-4 + 2(2)²-4 + 2(3)²-4 + 2(4)²-4

  = 2(1)-4 + 2(4)-4 + 2(9)-4 + 2(16)-4

  = (2-4) + (8-4) + (18-4) + (32-4)

  = -2 + 4 + 14 + 28

∑ = 44

3)

∑\left \ {{6} \atop {k=3}} \right. (2k-10)

∑ = (2×3−10) + (2×4−10) + (2×5−10) + (2×6−10)  

  = (6-10) + (8-10) + (10-10) + (12-10)

  = -4 + -2 + 0 + 2  

∑ = -4

4)

1+1/2+1/4+1/8+1/16+1/32+1/64

This is a geometric sequence where first term is 1 and the common ratio is 1/2 So

a = 1

This can be derived as

1/2/1 = 1/2 * 1 = 1/2

1/4/1/2 = 1/4 * 2/1 = 1/2

1/8/1/4 = 1/8 * 4/1  = 1/2

1/16/1/8 = 1/16 * 8/1  = 1/2

1/32/1/16 = 1/32 * 16/1  = 1/2

1/64/1/32 = 1/64 * 32/1  = 1/2

Hence the common ratio is r = 1/2

So n-th term is:

ar^{n-1} = 1(\frac{1}{2})^{n-1}

So the answer that represents the series in sigma notation is:

∑\left \ {{7} \atop {j=1}} \right. (\frac{1}{2})^{j-1}

5)

−3+(−1)+1+3+5

This is an arithmetic sequence where the first term is -3 and the common difference is 2. So  

a = 1

This can be derived as

-1 - (-3) = -1 + 3 = 2

1 - (-1) = 1 + 1 = 2

3 - 1 = 2

5 - 3 = 2

Hence the common difference d = 2

The nth term is:

a + (n - 1) d

= -3 + (n−1)2

= -3 + 2(n−1)

= -3 + 2n - 2

= 2n - 5

So the answer that represents the series in sigma notation is:

∑\left \ {{5} \atop {j=1}} \right. (2j−5)

6 0
3 years ago
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