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3 years ago

14

Determine the relationship between the two triangles and whether or not they can be proven to be congruent.

1 answer:

Jlenok [28]3 years ago

6 0

Answer:

only five points uneceptableeeeee

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The simplified form of an expression is 1/256 t28 which expression was simplified?

bonufazy [111]

Answer:

1/258 *(t^28)

= t^28 / 4^4

= t^28 4^-4

= (t^-7 * 4)^-4

= (4t^-7)^-4

Step-by-step explanation:

5 0

4 years ago

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A printer that originally cost $200 is on sale for 25% off.

umka21 [38]

I think the correct answer is $50

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3 years ago

PLEASE HELP!!!!! Chloe stops for lunch in a town that has no meal tax. She has $20 and tips 15%. Select all the prices of meals

lord [1]

Answer:

$16.25 $15.95 $15.75

Step-by-step explanation:

because a 15% tip is $3.00

4 0

3 years ago

Show that a = −1 + √3i and b = 2 satisfy 1/a+b=1/a + 1/b

Zarrin [17]

Answer:

LHS = $\frac{1 - \sqrt3i}{4}$ = RHS = $\frac{1 - \sqrt3i}{4}$

Step-by-step explanation:

Data provided in the question:

a = −1 + √3i and b = 2

to prove:

$\frac{1}{a+b}=\frac{1}{a} + \frac{1}{b}$

Considering the LHS

⇒ $\frac{1}{a+b}$

substituting the value of a and b, we get

⇒ $\frac{1}{−1 + \sqrt3i+2}$

or

⇒ $\frac{1}{1 + \sqrt3i}$

on multiplying and dividing by conjugate ( 1 - √3i )

we get

$\frac{1}{1 + \sqrt3i}\times\frac{1 - \sqrt3i}{1 - \sqrt3i}$

or

$\frac{1 - \sqrt3i}{(1^2 - (\sqrt3i)^2}$

or

$\frac{1 - \sqrt3i}{1 + 3}$ (as (√i)² = -1 )

or

$\frac{1 - \sqrt3i}{4}$

Now,

considering the RHS

$\frac{1}{a} + \frac{1}{b}$

substituting the value of a and b, we get

⇒ $\frac{1}{-1 + \sqrt3i} + \frac{1}{2}$

or

⇒ $\frac{2\times1 + ( -1 + \sqrt3i)\times1}{(-1 + \sqrt3i)\times2}$

or

⇒ $\frac{2 + ( -1 + \sqrt3i)}{(-1 + \sqrt3i)\times2}$

or

⇒ $\frac{1 + \sqrt3i}{(-1 + \sqrt3i)\times2}$

now,

on multiplying and dividing by conjugate ( -1 - √3i )

we get

$\frac{1 + \sqrt3i}{(−1 + \sqrt3i)\times2}\times\frac{-1 - \sqrt3i}{-1 - \sqrt3i}$

or

$\frac{1 + \sqrt3i}{(−1 + \sqrt3i)\times2}\times\frac{-1( 1 + \sqrt3i)}{-1 - \sqrt3i}$

or

$\frac{(1 + \sqrt3i}^2\times(-1){((-1)^2 - (\sqrt3i)^2)\times2}$

or

$\frac{(1^2 + (\sqrt3i)^2+2(1)(\sqrt3i)\times(-1)}{(1 + 3)\times2}$

or

$\frac{(1 - 3 + 2\sqrt3i)\times(-1)}{(4)\times2}$

or

$\frac{(-2 + 2\sqrt3i)\times(-1)}{(4)\times2}$

or

$\frac{-2( 1 - 2\sqrt3i)\times(-1)}{(4)\times2}$

or

$\frac{( 1 - 2\sqrt3i)}{(4)}$

Since, LHS = RHS

hence satisfied

3 0

3 years ago

Assume that θ is an acute angle in a right triangle satisfying the given conditions. Evaluate the 5 remaining trigonometric func

neonofarm [45]

Answer: See step by step. Replace the x with theta

Step-by-step explanation: Since sin is less than zero, and secant is positive. This means the trig values are in Quadrant 4.

We know that secФ=17/15. We need to find sin,cos,tan,csc, and cot.

We can find tangent by using the identity

$tan^{2} (x)+ 1=sec^{2} (x)$ where x is theta.

$tan^{2} (x)+1=\frac{289}{225}$

$tan^{2} (x)+\frac{225}{225} =\frac{289}{225}$

$tan^{2}(x)=\frac{64}{225}$

tan x= $\frac{8}{15}$ , Tangent in the fourth quadrant is negative so the answer is instead

$tan x= -\frac{8}{15}$

We can find cotangent by taking the recipocial of tan so

$cot=-\frac{15}{8}$

We can find cos by taking reciprocal of sec remeber that cosine is positive in 4th quadrant so the answer is

$cos =\frac{15}{17}$

We can find sin by doing quoteint identies.

$\frac{sin x}{cos x} =tan x$

$\frac{ sin x}{15/17} =-\frac{8}{15}$

$sin x= -\frac{8}{17}$

To find csc, take reciprocal of sine.

$csc=\frac{-17}{8}$

7 0

3 years ago