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Lena [83]
3 years ago
15

Can somebody help me with this?

Mathematics
1 answer:
N76 [4]3 years ago
8 0
Rise is 35, run is 1
35 is the answer

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You have 59 to spend on lip balm and hand sanitizer. The equation 1.5x + 2.5y = 9 represents this situation, where x is tubes of
il63 [147K]

Answer:

Step-by-step explanation:

you have to keep the x and the y sepret and then you can divied one by nine then you get the answer so there you go

3 0
2 years ago
Convert 5x + 6y = -5 to slope-intercept form. Simplify your answer
ruslelena [56]
Slope intercept form is y=mx+b. In this case, I believe it's y= -5/6x -5/6.
7 0
2 years ago
15%of 250 <br> Pls give and step by step
pychu [463]

Answer:

Step-by-step explanation:

15% = 15/100 = .15

250 x .15 = 37.5

5 0
3 years ago
Read 2 more answers
Y=3x-2<br> Y=2x-5<br> Find X and Y using substitution
tiny-mole [99]

y = 3x -2~~~.....(i)\\\\y = 2x-5~~~.....(ii)\\\\\text{Substitute equation (i) in equation (ii):}\\\\3x-2 = 2x -5\\\\\implies 3x -2x = -5+2\\\\\implies x = -3\\\\\text{Substitute}~ x =-3,~ \text{in equation (i):}\\\\y = 3(-3) -2 = -9 -2 = -11\\\\\text{Hence,}~ (x,y) = (-3,-11)

4 0
2 years ago
Please prove this........​
Crazy boy [7]

Answer:  see proof below

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

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

<u>Proof LHS → RHS</u>

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
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