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2 years ago

27a+1/2b=4 solve for a

Mathematics

2 answers:

sladkih [1.3K]2 years ago

5 0

Answer:

a = 4/27 - b/54

Step-by-step explanation:

27a+1/2b=4

Subtract 1/2 b from each side

27a+1/2b - 1/2 b=4-1/2 b

27a = 4 - 1/2 b

Divide each side by 27

27a/27 = 4/27 - 1/2 b/27

a = 4/27 - b/54

GarryVolchara [31]2 years ago

5 0

Answer: a= - 1/54b+4/27

Step-by-step explanation:
primero multiplique ambos lados de la ecuación por 2
54a+b=8
mueve la variable al lado derecho y cambie su signo
54a=8-b
dividí ambos lados de la ecuación entre 54
a=4/27 - 1/54b
use la propiedad conmutativa para reorganizar los términos
a= 1/54b + 4/27 y listo

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Find an exponential function that passes through each pair of points.<br><br>(2,1.75) & (-2,28)

dusya [7]

Answer:

$f(x) = 7\, ((1/2)^{x})$ .

Step-by-step explanation:

An exponential function is typically in the form $f(x) = a\, (b^{x})$ , where $a$ and $b$ ( $b > 0$ ) are constants to be found.

In this question:

$f(2) = 1.75$ means that $a\, (b^{2}) = 1.75$ .

$f(-2) = 28$ means that $a\, (b^{-2}) = 28$ .

Divide one of the two equations by the other to eliminate $a$ and solve for $b$ .

The number of the right-hand side of the second equation is larger than that of the first equation. Hence, divide the second equation with the first:

$\displaystyle \frac{a\, (b^{-2})}{a\, (b^{2})} = \frac{28}{1.75}$ .

$\displaystyle b^{-4} = 16$ .

$b^{-1} = 2$ .

$\displaystyle b = \frac{1}{2}$ .

Substitute $b = (1/2)$ back into either equation (for example, the first equation) and solve for $a$ :

$a\, ((1/2)^{2}) = 1.75$ .

$a = 7$ .

Substitute $a = 7$ and $b = (1/2)$ into the other equation. That equation should also be satisfied.

Therefore, this function would be:

$f = 7\, ((1 / 2)^{x})$ .

8 0

2 years ago

Use the unit circle to find sec (-90)

andreyandreev [35.5K]

Using the unit circle, you can visualise the secant as follows:

Draw the vertical line passing trough (1,0)
Extend the radius until it meets this line at point P
The length of OP is the secant of your angle.

Since -90 means that the radius points downwards, it means that the radius is vertical as well, and thus it never meets the vertical line through (1,0).

Thus, the secant is not defined at -90.

In fact, if we switch back to the function notation, we can confirm this claim, since we have

$\sec(x)=\dfrac{1}{\cos(x)}\implies \sec(-90)=\dfrac{1}{\cos(-90)}=\dfrac{1}{0}$