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Bess [88]
3 years ago
5

Help what’s this answer

Mathematics
1 answer:
Kryger [21]3 years ago
6 0
The answer is 9 and that on period
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ABCD is a parallelogram.
Dmitry_Shevchenko [17]

Answer:

Using SAS they are congruent.

Step-by-step explanation:

E is the point where the diagonals AC and BD meet.

Side: AB = CD

Angle: ∠ABE = ∠EDC

Side: BE = DE

Hence by SAS theorem the two triangles are congruent.

7 0
3 years ago
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Sin theta+costheta/sintheta -costheta+sintheta-costheta/sintheta+costheta=2sec2/tan2 theta -1
sleet_krkn [62]

\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}=\dfrac{2sec^2\theta}{tan^2\theta-1}

From Left side:

\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}\bigg(\dfrac{sin\theta+cos\theta}{sin\theta+cos\theta}\bigg)+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}\bigg(\dfrac{sin\theta-cos\theta}{sin\theta-cos\theta}\bigg)

\dfrac{sin^2\theta+2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}+\dfrac{sin^2\theta-2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}

NOTE: sin²θ + cos²θ = 1

\dfrac{1 + 2cos\theta sin\theta}{sin^2\theta-cos^2\theta}+\dfrac{1-2cos\theta sin\theta}{sin^2\theta-cos^2\theta}

\dfrac{1 + 2cos\theta sin\theta+1-2cos\theta sin\theta}{sin^2\theta-cos^2\theta}

\dfrac{2}{sin^2\theta-cos^2\theta}

\dfrac{2}{\bigg(sin^2\theta-cos^2\theta\bigg)\bigg(\dfrac{cos^2\theta}{cos^2\theta}\bigg)}

\dfrac{2sec^2\theta}{\dfrac{sin^2\theta}{cos^2\theta}-\dfrac{cos^2\theta}{cos^2\theta}}

\dfrac{2sec^2\theta}{tan^2\theta-1}

Left side = Right side <em>so proof is complete</em>

8 0
3 years ago
Read 2 more answers
MATH IS HARD. HELP<br> Change the following into a single fraction: 1− (a+b/a−b)
lianna [129]

Answer:

-2b/(a-b)

Step-by-step explanation:

1− (a+b/a−b)


We need to get a common denominator of a-b

1 = (a-b)/(a-b)

Replace 1 with (a-b)/(a-b)

(a-b)/(a-b) - (a+b)/(a-b)

Put over the common denominator

(a-b) - (a+b)

-----------------

(a-b)


Distribute the minus sign.

(a-b) - a-b)

-----------------

(a-b)

a-a-b-b

---------------

a-b


-2b

----------

a-b

6 0
3 years ago
Steven invests $20,000 in an account earning 3% interest, compounded annually for 10 years. Three years after Stevens's initial
Effectus [21]

The formula to find the amount is

here A is amount

P is the principal

'r' is the rate of interest

n is the number of years.

Case 1.

Stevan invests

P =$ 20,000

r = 3% = 0.03

n = 10 years

Hence the interest earned

= A - P = 26878.33 - 20000 = $6878.33

Case 2.

Evan invests

P = $10,000

r = 7% = 0.07

n = 7 years

Hence the interest earned

= A - P = 16057.81 - 10000 = 6057.81

Difference in the interest = 6878.33 - 6057.81 = $820.52

Rounded to the nearest dollar difference in interest = $821

6 0
3 years ago
Please help me with this <br><br> 5<br> 9<br> 10<br> All real numbers
Sati [7]

i think either 5 or 9

6 0
3 years ago
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