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3 years ago

Please Help Me!!!!

1 answer:

5 0

2. WXZ=YXZ
3. Reflexive property of equality
(if they are angles)-
4. WXZ=YXZ ~~Reason- Congruent angles have equal number of degrees
(if they are triangles-more likely judging by the problem)-
4. Triangle WXZ=YXZ ~ Reason- SAS Congruence Postulate

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Which numbers are a distance of 3 units from 12 on the a number line?

Mamont248 [21]

It would be B, hope i could help out.

3 0

3 years ago

Read 2 more answers

Verify the identity. <br><br>cot. (x - (pi/2)) = -tan x<br><br>Please, help me!!!

Leya [2.2K]

Answer:

cot . (x- π/2)= - tanx

Step-by-step explanation:

cot . (x- π/2)= - tanx

cot .x - cot π/2= - tanx

cot π/2= 1/tanπ/2

But tan π/2= ∞

cot π/2 = 1/∞= 0

cotx - 0= - tanx

cot x = - tanx

cosx/sinx= - sinx/cosx Putting values of cot and tan

cosx/ sinx ( sinx/cos x) = -sinx/cos x (sinx/cos x )

Multiplying both sides with sinx/cos x

1= - sin²x/cos²x

cos²x= - sin²x as cosx = -sinx

5 0

3 years ago

The tyre of a car wheel has an outer diameter of 30 cm . How many times will the wheel rotate on a journey of 5 km

faltersainse [42]

Answer:

5305.1647 times

Step-by-step explanation:

The Circumference of a circle : πd

d = diameter of circle = 30 cm

Circumference of wheel = π * 30cm = 94.247779 cm

1km to cm

1km = 100000cm

5 km = 500000 cm

Hence, number of times wheel rotate on a 5km journey :

500000 cm / 94.247779 cm

= 5305.1647 times

3 0

3 years ago

4. Using the geometric sum formulas, evaluate each of the following sums and express your answer in Cartesian form.

nikitadnepr [17]

Answer:

$\sum_{n=0}^9cos(\frac{\pi n}{2})=1$

$\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=0$

$\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})=\frac{1}{2}$

Step-by-step explanation:

$\sum_{n=0}^9cos(\frac{\pi n}{2})=\frac{1}{2}(\sum_{n=0}^9 (e^{\frac{i\pi n}{2}}+ e^{\frac{i\pi n}{2}}))$

$=\frac{1}{2}(\frac{1-e^{\frac{10i\pi}{2}}}{1-e^{\frac{i\pi}{2}}}+\frac{1-e^{-\frac{10i\pi}{2}}}{1-e^{-\frac{i\pi}{2}}})$

$=\frac{1}{2}(\frac{1+1}{1-i}+\frac{1+1}{1+i})=1$

2nd

$\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=\frac{1-e^{\frac{i2\pi N}{N}}}{1-e^{\frac{i2\pi}{N}}}$

$=\frac{1-1}{1-e^{\frac{i2\pi}{N}}}=0$

3th

$\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})==\frac{1}{2}(\sum_{n=0}^\infty ((\frac{e^{\frac{i\pi n}{2}}}{2})^n+ (\frac{e^{-\frac{i\pi n}{2}}}{2})^n))$