All categories

3 years ago

I need help badly I don't really know how to finish

Mathematics

1 answer:

wlad13 [49]3 years ago

3 0

X=55, since straight lines are 180°. Y=70, since the triangle is isosceles, and the 2 bottom Angles are equal. Z=125, since the 2 opposite angles(x and y) add up to 125°

You might be interested in

Andrea purchased 2.8 pounds of bananas that cost $0.60 per pound

iren [92.7K]

Answer:

Total cost = $5.18.

Step-by-step explanation:

Given the following data;

A pound of banana = $0.60

Quantity of banana = 2.8 pounds

Bag of apples = $3.50

To find the amount spent on banana;

Cost of banana = Quantity of banana * cost of each pound of banana

Cost of banana = 2.8 * 0.6 = $1.68

Total amount spent on banana = $1.68

Therefore, the total amount spent by Andrea on both banana and apples;

Total cost = cost of apple + cost of banana

Total cost = $3.5 + $1.68

Total cost = $5.18.

7 0

3 years ago

Read 2 more answers

I can't figure this one out.

mezya [45]

Answer:

I'm pretty sure the answer is B. x+6

Step-by-step explanation:

a polynomial with two terms (x and 6) is called a binomial. bi means two, so your answer would have two terms! hope this helped :)

5 0

3 years ago

Please help me to prove this! I need is no.(c). So, please help me do it.

zloy xaker [14]

Answer: see proof below

Step-by-step explanation:

Given: A + B + C = 90° → A + B = 90° - C

→ C = 90° - (A + B)

Use the Double Angle Identity: cos 2A = 1 - 2 sin² A

→ sin² A = (1 - cos 2A)/2

Use Sum to Product Identity: cos A + cos B = 2 cos [(A + B)/2] · cos [(A - B)/2]

Use the Product to Sum Identity: cos (A - B) - cos (A + B) = 2 sin A · sin B

Use the Cofunction Identities: cos (90° - A) = sin A

sin (90° - A) = cos A

Proof LHS → RHS:

LHS: sin² A + sin² B + sin² C

$\text{Double Angle:}\qquad \dfrac{1-\cos 2A}{2}+\dfrac{1-\cos 2B}{2}+\sin^2 C\\\\\\.\qquad \qquad \qquad =\dfrac{1}{2}\bigg(2-\cos 2A-\cos 2B\bigg)+\sin^2 C\\\\\\.\qquad \qquad \qquad =1-\dfrac{1}{2}\bigg(\cos 2A+\cos 2B\bigg)+\sin^2 C$

$\text{Sum to Product:}\quad 1-\dfrac{1}{2}\bigg[2\cos \bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A-2B}{2}\bigg)\bigg]+\sin^2 C\\\\\\.\qquad \qquad \qquad =1-\cos (A+B)\cdot \cos (A-B)+\sin^2 C$

Given: 1 - cos (90° - C) · cos (A - B) + sin² C

Cofunction: 1 - sin C · cos (A - B) + sin² C

Factor: 1 - sin C [cos (A - B) + sin C]

Given: 1 - sin C[cos (A - B) - sin (90° - (A + B))]

Cofunction: 1 - sin C[cos (A - B) - cos (A + B)]

Sum to Product: 1 - sin C [2 sin A · sin B]

= 1 - 2 sin A · sin B · sin C

LHS = RHS: 1 - 2 sin A · sin B · sin C = 1 - 2 sin A · sin B · sin C $\checkmark$

6 0

3 years ago

Let X and Y be random variables with the following joint moment generating function: M_{X,Y}(s,t)=0.1+0.2e^s+0.3e^t+0.4e^{s+t}

MA_775_DIABLO [31]

We obtain the joint PMF directly from the joint MGF:

$M_{X,Y}(s,t)=0.1+0.2e^s+0.3e^t+0.4e^{s+t}$

$\implies\mathrm{Pr}[X=x,Y=y]=\begin{cases}0.1&\text{for }x=y=0\\0.2&\text{for }x=1,y=0\\0.3&\text{for }x=0,y=1\\0.4&\text{for }x=y=1\\0&\text{otherwise}\end{cases}$