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3 years ago

What is the value of -3 over 4

Mathematics

2 answers:

vovangra [49]3 years ago

7 0

Answer:

-0.75

Step-by-step explanation:

NeTakaya3 years ago

3 0

Answer:

-0.75

Step-by-step explanation:

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For f(x) = 3x +1 and g(x) = x2 – 6, find (f- g)(x).

Fiesta28 [93]

Answer:

$\boxed{\sf (f-g)(x) = -{x}^{2} + 3x + 7}$

Given:

$\sf f(x) = 3x + 1 \\ \sf g(x) = {x}^{2} - 6$

To find:

$\sf (f - g)(x) = f(x) - g(x)$

Step-by-step explanation:

$\sf \implies(f - g)x = f(x) - g(x) \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = (3x + 1) - ( {x}^{2} - 6) \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = (3x + 1) + (- {x}^{2} + 6) \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 3x + 1 - {x}^{2} + 6 \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = - {x}^{2} + 3x + 1 + 6 \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = -{x}^{2} + 3x + 7$

8 0

3 years ago

Tell whether the statement is true or false.<br>{ {4, 16, 24, 7, 37} = (37, 16, 7,42,4}

nikklg [1K]

Answer:

false

Step-by-step explanation:

i dont understand kinda

but if you add it,

88<106

so false

8 0

3 years ago

AB has endpoints at (-4,5) and (4,-3) using the midpoint formula what is the midpoint segment

Mamont248 [21]

Answer:

(0,1)

Step-by-step explanation:

5 0

3 years ago

Outside temperature over a day can be modelled as a sinusoidal function. Suppose you know the high temperature for the day is 10

WINSTONCH [101]

Answer:

The function for the outside temperature is represented by $T(t) = 85\º + 15\º \cdot \sin \left[\frac{t-6\,h}{24\,h} \right]$ , where t is measured in hours.

Step-by-step explanation:

Since outside temperature can be modelled as a sinusoidal function, the period is of 24 hours and amplitude of temperature and average temperature are, respectively:

Amplitude

$A = \frac{100\º-70\º}{2}$