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3 years ago

How do you show your work for 800-374 on a number line

Mathematics

1 answer:

Sergio039 [100]3 years ago

7 0

Answer:

make the numberline 800 to 0 and do bunny hop to a different part of the number line to show subtraction.

Step-by-step explanation:

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Any help would be great

nadya68 [22]

Answer:

Step-by-step explanation:

38=10+13+c

c=38-10-13=15

Hope this helps!

8 0

3 years ago

Given the equation p2 = a3, what is the orbital period, in days, for the planet venus? (venus is located 0.72 au from the sun?)

Gelneren [198K]

Hello,
P=0.72^(3/2)=0.610940...years=223.1398...days

5 0

3 years ago

Read 2 more answers

5 and 1/7 * 8 HELP ME PLZ

il63 [147K]

Answer:

41 1/7

Step-by-step explanation:

5 1/7 x 8

Convert 5 1/7 to improper fraction

5 1/7 = 36/7

Multiply.

36 x 8

---- --

7 x 1

288/7

Simplify.

41 1/7

---

hope it helps

8 0

3 years ago

What is the arithmetic mean of the following numbers? 8 10 8 5 4 7 5 10 8

Bas_tet [7]

I re-orders as 4,5,5,7,8,8,8,10,10.

Mean 7.2222222222222
Median 8
Mode 8
Range 6
Minimum 4
Maximum 10
Count n 9
Sum 65
Quartiles Quartiles:
Q1 --> 5
Q2 --> 8
Q3 --> 9
Interquartile
Range IQR 4
Outliers none

4 0

3 years ago

Please prove this........

Crazy boy [7]

Answer: see proof below

Step-by-step explanation:

Given: A + B + C = π → C = π - (A + B)

→ sin C = sin(π - (A + B)) cos C = sin(π - (A + B))

→ sin C = sin (A + B) cos C = - cos(A + B)

Use the following Sum to Product Identity:

sin A + sin B = 2 cos[(A + B)/2] · sin [(A - B)/2]

cos A + cos B = 2 cos[(A + B)/2] · cos [(A - B)/2]

Use the following Double Angle Identity:

sin 2A = 2 sin A · cos A

Proof LHS → RHS

LHS: (sin 2A + sin 2B) + sin 2C

$\text{Sum to Product:}\qquad 2\sin\bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A - 2B}{2}\bigg)-\sin 2C$

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

$\text{Simplify:}\qquad \qquad 2\sin (A + B)\cdot \cos (A - B)-2\sin C\cdot \cos C$

$\text{Given:}\qquad \qquad \quad 2\sin C\cdot \cos (A - B)+2\sin C\cdot \cos (A+B)$

$\text{Factor:}\qquad \qquad \qquad 2\sin C\cdot [\cos (A-B)+\cos (A+B)]$

$\text{Sum to Product:}\qquad 2\sin C\cdot 2\cos A\cdot \cos B$

$\text{Simplify:}\qquad \qquad 4\cos A\cdot \cos B \cdot \sin C$

LHS = RHS: 4 cos A · cos B · sin C = 4 cos A · cos B · sin C $\checkmark$

7 0

3 years ago