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weeeeeb [17]
3 years ago
14

How do I solve this math equation: 7=8-p

Mathematics
1 answer:
STALIN [3.7K]3 years ago
4 0

Answer:

p = 1

Step-by-step explanation:

7 = 8 - p

7 + p = 8

p = 8-7

p = 1

Answered by Gauthmath

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2 years ago
F(x)=2x+5 find f(-6)
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Answer:

Step-by-step explanation:

f(x) = 2x + 5

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3 years ago
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mario62 [17]

Answer: 64 miles

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5 0
3 years ago
Read 2 more answers
Let θ
konstantin123 [22]

Answer:

θ is decreasing at the rate of \frac{21}{16} units/sec

or \frac{d}{dt}(θ) = \frac{-21}{16}

Step-by-step explanation:

Given :

Length of side opposite to angle θ is y

Length of side adjacent to angle θ is x

θ is part of a right angle triangle

At this instant,

x =  8 , \frac{dx}{dt} = 7

( \frac{dx}{dt} denotes the rate of change of x with respect to time)

y = 8 , \frac{dy}{dt} = -14

( The negative sign denotes the decreasing rate of change )

Here because it is a right angle triangle,

tanθ = \frac{y}{x}-------------------------------------------------------------------1

At this instant,

tanθ = \frac{8}{8} = 1

Therefore θ = π/4

We differentiate equation (1) with respect to time in order to obtain the rate of change of θ or \frac{d}{dt}(θ)

\frac{d}{dt} (tanθ) = \frac{d}{dt} (y/x)

( Applying chain rule of differentiation for R.H.S as y*1/x)

sec^{2}θ\frac{d}{dt}(θ) = \frac{1}{x}\frac{dy}{dt} - \frac{y}{x*x}\frac{dx}{dt}-----------------------2

Substituting the values of x , y , \frac{dx}{dt} , \frac{dy}{dt} , θ at that instant in equation (2)

2\frac{d}{dt}(θ) = \frac{1}{8}*(-14)- \frac{8}{8*8}*7

\frac{d}{dt}(θ) = \frac{-21}{16}

Therefore θ is decreasing at the rate of \frac{21}{16} units/sec

or  \frac{d}{dt}(θ) = \frac{-21}{16}

3 0
3 years ago
Can someone help me with this one. I’m stuck
zmey [24]

you should use this one app called photomath it will give you  all the right answers for math... it works....

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