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3 years ago

If y represents a number, write an expression that represents the sum of y and a number one greater than y

Mathematics

2 answers:

Liula [17]3 years ago

4 0

Answer:

2y+1=0

Step-by-step explanation:

a number one greater than y is y+1

so sum of y and y+1=2y+1

hope its helpful

Dimas [21]3 years ago

3 0

Answer:

2y+1=0

Step-by-step explanation:

a number one greater than y is y+1

so sum of y and y+1= 2y+1

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Juan was given a gift card for a coffee shop. Each morning, Juan uses the card to buy one cup of coffee. Each cup of coffee cost

Black_prince [1.1K]

Answer:

$A=30-2x$

Step-by-step explanation:

Cost of one cup of coffee=$2

The original amount of money on the gift card= $30

Juan's expenses in buying coffee, after $x$ morning are $\$2\times x$ .

The amount of money left, $A$ , on the card $=$ (Original amount) $-$ (Total expenses)

$A=30-2x$

Hence, the equation for, $A$ , the amount of money remaining on the card is

$A=30-2x.$

7 0

3 years ago

The Board of Directors of a small company consists of five people. Two of those directors are considered "strong leaders". If th

Pavel [41]

1: 40%
2: 30%
3: 20%

If we know one of the succeeded, then
1: 44.44%
2: 33.33%
3: 22.22%

If under the circumstance that they all fail to meet a strong leader, the board picks one of them at random,
1: 43.33%
2: 33.33%
3: 23.33%

The first person has a 2 in 5 chance of randomly getting a strong leader. If the first person doesn't find a strong leader, the second person has a 2/4 chance of getting a strong leader. If neither the first nor second person gets a strong leader, the third sales person has a 2/3 chance of getting a strong leader.

1: 2/5=40% Chance
2: (3/5)*(2/4)=30% Chance You have to take the probability the first rep failed to get a strong leader, then multiply by the probability the second rep gets a strong leader.
3: (3/5)*(2/4)*(2/3)=20% You have to take the probability both the other people failed, then multiply by the probability they succeeded.

7 0

3 years ago

how much simple interest is earned on an investment of $1250 if the money is invested for 5 years at an annual interest rate of

andrew-mc [135]

Approximately $307.73 is the total amount of interest earned for 5 years at an annual interest rate of 4.5%, with an initial investment of $1,250.

7 0

3 years ago

Which one would it be?

Leni [432]

C is the answer.
in the original function, when x=0, y=-1
in the new function, to make y=-1, x+3=0, x=-3. From the original 0 to -3 is a shift of 3 units to the left.

6 0

3 years ago

Evaluate the line integral by the two following methods. xy dx + x2 dy C is counterclockwise around the rectangle with vertices

Airida [17]

Answer:

25/2

Step-by-step explanation:

Recall that for a parametrized differentiable curve C = (x(t), y(t)) with the parameter t varying on some interval [a, b]

$\large \displaystyle\int_{C}[P(x,y)dx+Q(x,y)dy]=\displaystyle\int_{a}^{b}[P(x(t),y(t))x'(t)+Q(x(t),y(t))y'(t)]dt$

Where P, Q are scalar functions

We want to compute

$\large \displaystyle\int_{C}P(x,y)dx+Q(x,y)dy=\displaystyle\int_{C}xydx+x^2dy$

Where C is the rectangle with vertices (0, 0), (5, 0), (5, 1), (0, 1) going counterclockwise.

a) Directly

Let us break down C into 4 paths $\large C_1,C_2,C_3,C_4$ which represents the sides of the rectangle.

$\large C_1$ is the line segment from (0,0) to (5,0)

$\large C_2$ is the line segment from (5,0) to (5,1)

$\large C_3$ is the line segment from (5,1) to (0,1)

$\large C_4$ is the line segment from (0,1) to (0,0)

Then

$\large \displaystyle\int_{C}=\displaystyle\int_{C_1}+\displaystyle\int_{C_2}+\displaystyle\int_{C_3}+\displaystyle\int_{C_4}$

Given 2 points P, Q we can always parametrize the line segment from P to Q with

r(t) = tQ + (1-t)P for 0≤ t≤ 1

Let us compute the first integral. We parametrize $\large C_1$ as

r(t) = t(5,0)+(1-t)(0,0) = (5t, 0) for 0≤ t≤ 1 and

r'(t) = (5,0) so

$\large \displaystyle\int_{C_1}xydx+x^2dy=0$

Now the second integral. We parametrize $\large C_2$ as

r(t) = t(5,1)+(1-t)(5,0) = (5 , t) for 0≤ t≤ 1 and

r'(t) = (0,1) so

$\large \displaystyle\int_{C_2}xydx+x^2dy=\displaystyle\int_{0}^{1}25dt=25$

The third integral. We parametrize $\large C_3$ as

r(t) = t(0,1)+(1-t)(5,1) = (5-5t, 1) for 0≤ t≤ 1 and

r'(t) = (-5,0) so

$\large \displaystyle\int_{C_3}xydx+x^2dy=\displaystyle\int_{0}^{1}(5-5t)(-5)dt=-25\displaystyle\int_{0}^{1}dt+25\displaystyle\int_{0}^{1}tdt=\\\\=-25+25/2=-25/2$

The fourth integral. We parametrize $\large C_4$ as

r(t) = t(0,0)+(1-t)(0,1) = (0, 1-t) for 0≤ t≤ 1 and

r'(t) = (0,-1) so

$\large \displaystyle\int_{C_4}xydx+x^2dy=0$

$\large \displaystyle\int_{C}xydx+x^2dy=25-25/2=25/2$

Now, let us compute the value using Green's theorem.

According with this theorem

$\large \displaystyle\int_{C}Pdx+Qdy=\displaystyle\iint_{A}(\displaystyle\frac{\partial Q}{\partial x}-\displaystyle\frac{\partial P}{\partial y})dydx$

where A is the interior of the rectangle.

so A={(x,y) | 0≤ x≤ 5, 0≤ y≤ 1}

We have

$\large \displaystyle\frac{\partial Q}{\partial x}=2x\\\\\displaystyle\frac{\partial P}{\partial y}=x$

$\large \displaystyle\iint_{A}(\displaystyle\frac{\partial Q}{\partial x}-\displaystyle\frac{\partial P}{\partial y})dydx=\displaystyle\int_{0}^{5}\displaystyle\int_{0}^{1}xdydx=\displaystyle\int_{0}^{5}xdx\displaystyle\int_{0}^{1}dy=25/2$

3 0

3 years ago