All categories

3 years ago

Name a pair of non adjacent complementary angels.

Mathematics

2 answers:

Jobisdone [24]3 years ago

4 0

Answer:

Step-by-step explanation:

mash [69]3 years ago

4 0

Angle KJP
Angle KJL
Are the possible answers if you had shown the answer choices

You might be interested in

Carlos deposited $7,924 into a savings account 30 years ago. The account has an interest rate of 4.6% and the balance is current

exis [7]

Answer:

n = 1, this means the interest compounds ANNUALLY.

Step-by-step explanation:

Carlos deposited $7,924 into a savings account 30 years ago. The account has an interest rate of 4.6% and the balance is currently $30,541.83. How often does the interest compound?

Compound Interest Formula

: A = P(1 + r/n)^nt

A = Amount after time t

P = Principal (Initial Amount Invested)

r = Interest rate

n = Number of times the interest is compounded

t = time in years

A = $30,541.83

r = 4.6% = 0.046

t = 30

P = $7,924

Hence,

$30,541.83 = $7924(1 + 0.046/n)^30n

Divide both sides by 7924

$30,541.93/$7924 = (1 + 0.046/n)^30n

$30,541.93/$7924 = (n + 0.046/n)^30n

3.8543576477 = (n + 0.046/n)^30n

We take the logarithm of both sides

log 3.8543576477 = log (n + 0.046/n)^30n

Solving for n,

n = 1

Therefore, from the calculation above, since n = 1, this means the interest compounds ANNUALLY.

6 0

2 years ago

Drag the slider to 2 inches. How many centimeters are in 2 inches? There are centimeters in 2 inches.

choli [55]

Answer: 5

Step-by-step explanation: There are 2.5 cm per in, so its just 2.5*2

4 0

3 years ago

Read 2 more answers

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

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

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