A number cube has faces numbered from 1 to 6. When the number cube is rolled, the theoretical probability of it

Please help me!!!!!!!!!!!!!

monitta

Answer: see proof below

Step-by-step explanation:

Use the following Half-Angle Identities: tan (A/2) = (sinA)/(1 + cosA)

cot (A/2) = (sinA)/(1 - cosA)

Use the Pythagorean Identity: cos²A + sin²B = 1

Use Unit Circle to evaluate: cos 45° = sin 45° = $\frac{\sqrt2}{2}$

Proof LHS → RHS

Given: $cot\ (22\frac{1}{2})^o-tan\ (22\frac{1}{2})^o$

Rewrite Fraction: $cot\ (\frac{45}{2})^o-tan\ (\frac{45}{2})^o$

Half-Angle Identity: $\dfrac{sin(45)^o}{1-cos(45)^o}-\dfrac{sin(45)^o}{1+cos(45)^o}$

Substitute: $\dfrac{\frac{\sqrt2}{2}}{1-\frac{\sqrt2}{2}}-\dfrac{\frac{\sqrt2}{2}}{1+\frac{\sqrt2}{2}}$

Simplify: $\dfrac{\frac{\sqrt2}{2}}{\frac{2-\sqrt2}{2}}-\dfrac{\frac{\sqrt2}{2}}{\frac{2+\sqrt2}{2}}$

$=\dfrac{\sqrt2}{2-\sqrt2}-\dfrac{\sqrt2}{2+\sqrt2}$

$=\dfrac{\sqrt2}{2-\sqrt2}\bigg(\dfrac{2+\sqrt2}{2+\sqrt2}\bigg)-\dfrac{\sqrt2}{2+\sqrt2}\bigg(\dfrac{2-\sqrt2}{2-\sqrt2}\bigg)$

$=\dfrac{2\sqrt2+2}{4-2}-\dfrac{2\sqrt2-2}{4-2}$

$=\dfrac{4}{2}$

= 2

LHS = RHS: 2 = 2 $\checkmark$

7 0

4 years ago

Please help i will brainly Δdef and Δxyz are similar isosceles triangles.

aivan3 [116]

Answer:

Part 1) The ratio of the sides is 1.5

Part 2) see the explanation

Step-by-step explanation:

Part 1) we know that

If two figures are similar, then the ratio of its corresponding sides is proportional and its corresponding angles are congruent

If Δdef and Δxyz are similar (given problem)

then

$\frac{XY}{DE}=\frac{ZY}{FE}$

substitute the values

$\frac{15}{10}=\frac{ZY}{6}$

solve for FY

$ZY=15(6)/10=9\ cm$

When two figures are similar, then the ratio of its corresponding sides is proportional and this ratio is called the scale factor

In this problem

The scale factor is equal to

$\frac{15}{10}=\frac{9}{6}=1.5$

Part 2) How you could verify that these two triangles are indeed similar.

Remember that

In similar figures the corresponding angles are congruent

so

If

m∠D≅m∠X

m∠W≅m∠Y

m∠F≅m∠Z

then

Triangles are similar

7 0

4 years ago

Always solve an algebraic expression by working from the left to right true or false

Gemiola [76]

The answer is FALSE hope this helps

7 0

3 years ago

Given: PQ ⊥ QR , PR=20, SR=11, QS=5 Find: The value of PS.

dlinn [17]

Answer:

The value of the side PS is 26 approx.

Step-by-step explanation:

In this question we have two right triangles. Triangle PQR and Triangle PQS.

Where S is some point on the line segment QR.

Given:

PR = 20

SR = 11

QS = 5

We know that QR = QS + SR

QR = 11 + 5

QR = 16

Now triangle PQR has one unknown side PQ which in its base.

Finding PQ:

Using Pythagoras theorem for the right angled triangle PQR.

PR² = PQ² + QR²

PQ = √(PR² - QR²)

PQ = √(20²+16²)

PQ = √656

PQ = 4√41

Now for right angled triangle PQS, PS is unknown which is actually the hypotenuse of the right angled triangle.

Finding PS:

Using Pythagoras theorem, we have:

PS² = PQ² + QS²

PS² = 656 + 25

PS² = 681

PS = 26.09

PS = 26

7 0

3 years ago

The ratios of the angles in ABC is 3:5:7. Find the

Kazeer [188]

So you have a diagram or picture of the triangle?

Anyways...

Answer: 36

Explanation:

3:5:7 is the ratio of the triangle's angles

The sum of all angles = 180 degrees.

3 + 5 + 7 = 15

Lets assume 'a' is the smallest angle with the ratio of 3

b is the second largest with a ratio of 5

And c is the largest angle with the ratio of 7

In other words:

a + b + c = 180

For 15 to get to 180 => 180 ÷ 15 = 12

So 15 × 12 = 180

So for 3 to get to a => 3 × 12 = 36

Thus the smallest angle is 36 degrees

8 0

3 years ago

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