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Zigmanuir [339]
3 years ago
13

Please help me to figure out this math graph problem.

Mathematics
1 answer:
Monica [59]3 years ago
6 0

Answer:

Step-by-step explanation:

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The simplest way to teach middle school math scale drawings is to use real pictures to relate to each other to explain the concept. For example, scaling a red ball of 1" to a ball of 2" and so on. This will show how the ball increases by size by adding 1" each time.

4 0
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Which statements must be true it says check alll that apply ?
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The statements that must be true are c,e,f, but these are the statements that I think are correct
4 0
3 years ago
5×-3(4×-8)=-11 helppppppp
Damm [24]

5x - 3(4x-8) = -11

5x - 12x  + 24 = -11

-7x + 24 = -11

-7x = - 35

  x = 5

4 0
2 years ago
Find the differential of each function. (a) y = tan( 7t ) dy = Incorrect: Your answer is incorrect. (b) y = 4 − v2 4 + v2 dy =
Kisachek [45]

Answer:

(a) dy= 7sec^2(7t)dt\\\\(b) dy = \frac{-16v}{(4+v^2)^2} dv

Step-by-step explanation:

Given;

(a) y = tan (7t)

let u = 7t

y = tan(u)

du/dt = 7

dy/du = sec²(u)

\frac{dy}{dt} = \frac{dy}{du} *\frac{du}{dt}\\\\\frac{dy}{dt} = sec^2(u)* 7\\\\\frac{dy}{dt} =7sec^2(u)\\\\\frac{dy}{dt} = 7sec^2(7t)\\\\dy = 7sec^2(7t)dt

(b)  

y = \frac{4-v^2}{4+v^2}\\\\

let u = 4 - v²

du/dv = -2v

let v = 4 + v²

dv/du = 2v

y = \frac{4-v^2}{4+v^2}\\\\\frac{dy}{dv} = \frac{vdu-udv}{v^2} \\\\\frac{dy}{dv} =\frac{-2v(4+v^2)-2v(4-v^2)}{(4+v^2)^2}\\\\\frac{dy}{dv} =\frac{-8v-2v^3-8v+2v^3}{(4+v^2)^2}\\\\\frac{dy}{dv} =\frac{-16v}{(4+v^2)^2}\\\\dy = \frac{-16v}{(4+v^2)^2}dv

6 0
3 years ago
204% as a fraction? who every now put it in comments
ExtremeBDS [4]

Answer:

204/100 = 51/25= 2 1/25 that is the answer

6 0
2 years ago
Read 2 more answers
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