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zloy xaker [14]
3 years ago
6

If Jada has 1980 apples and she takes out 1665 how many are left?

Mathematics
1 answer:
Alexandra [31]3 years ago
7 0
Answer: 315!
1980 subtracted by 1665
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Which describes a person designated to receive money from a life insurance
Rainbow [258]

Answer: C

Step-by-step explanation: Beneficiary

8 0
3 years ago
Read 2 more answers
{20, 68, 70, 70, 71, 72, 75, 76}
guapka [62]

Answer:

Mean: 65.25

Median: 70.5

Mode: 70

Step-by-step explanation:

Mean is the sum of the set of numbers divided by the number of numbers in the set.

20+68+70+70+71+72+75+76=522

522/8=65.25

The median is the number that is in the middle of the set when put in numerical order.

Since there is an even amount of numbers, the two middle numbers are 70 and 71, so you would take the mean of these two numbers to find the median.

70+71=141

141/2=70.5

The mode is the number that occurs most frequently in the set.

70 appears more times than any other number, therefore it is the mode.

5 0
4 years ago
Read 2 more answers
Please help me to prove this!<br>I need is no.(c). So, please help me do it.<br>​
zloy xaker [14]

Answer:  see proof below

<u>Step-by-step explanation:</u>

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

<u>Proof LHS → RHS:</u>

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
‼️‼️any help would be appreciated
never [62]
Because the triangle is equilateral we know that z=60, therefore:

60=\frac{1}{3}x-7
67=\frac{x}{3}
67 \times 3 = x
x=201 <span />
6 0
3 years ago
|(1)
Yanka [14]
So first you would divide 1950/6 to find the amount for one year of their age.
Then you would multiply that by the ages, which should get you 4 numbers, then you all those numbers together. Try 9750.
3 0
3 years ago
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