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3 years ago

Which equation has a solution of -2?

Mathematics

1 answer:

kykrilka [37]3 years ago

8 0

Solution:

I'm not sure if there is law for it or not, but you can replace each x to -2 if both sides are equal then the equation is true.

$7x - 4 = 6x + 5\\7(-2) - 4 = 6(-2) + 5\\-14-4=-12+5\\-18\neq -7$

A. is incorrect.

$4x + 3 = 8x - 5\\4(-2) + 3 = 8(-2) - 5\\-8+3=-16-5\\-5\neq -21$

B. is incorrect.

$6x + 4 = 3x + 10\\6(-2) + 4 = 3(-2) + 10\\-12+4=-6+10\\-8\neq 4$

C. is incorrect.

$-5x - 3 = -7(x +1)\\-5(-2) - 3 = -7((-2) +1)\\10-3 =-7(-1)\\7=7$

D. is correct.

Answer:

D. $-5x-3=-7(x+1)$

<em>Hope this was helpful.</em>

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Ayudemen porfavor . Determine 4 ala potencia x + 4 ala potencia -x, si se sabe que 2ala potencia x + 2 ala potencia -x =3

Afina-wow [57]

Answer:

Es igual a 7.

Step-by-step explanation:

Ok, sabemos que:

$2^x + 2^{-x} = 3$

Queremos calcular:

$4^x + 4^{-x}$

Tambien sabemos que 4 = 2*2

Entonces:

$4^x + 4^{-x} = (2*2)^x + (2*2)^{-x}$

y sabemos que 2*2 = 2^2

Entonces:

$4^x + 4^{-x} = (2*2)^x + (2*2)^{-x} = (2^2)^x + (2^2)^{-x}$

y ahora podemos usar la relación:

$(a^n)^m = (a^m)^n = a^{m*n}$

entonces:

$(2^2)^x + (2^2)^{-x} = (2^x)^2 + (2^{-x})^2$

Ahora podemos completar cuadrados sumando y restando el termino:

$2*(2^x)*(2^{-x})$

Asi tendremos:

$(2^x)^2 + (2^{-x})^2 = (2^x)^2 + (2^{-x})^2 + 2*(2^x)*(2^{-x}) - 2*(2^x)*(2^{-x}) = (2^x + 2^{-x})^2 - 2*(2^x)*(2^{-x})$

Entonces de momento tenemos que:

$4^x + 4^{-x} = (2^x + 2^{-x})^2 - 2*(2^x)*(2^{-x})$

Sabemos que el termino que esta dentro del parentesis es igual a 3.

Y tambien podemos usar la propiedad:

$a^n*a^m = a^{n + m}$

en el termino de la derecha.

Asi tendremos:

$(2^x + 2^{-x})^2 - 2*(2^x)*(2^{-x}) = (3)^2 - 2*2^{x + (-x)} = 9 - 2 = 7$

7 0

3 years ago

What is the exact circumference of the circle?

cestrela7 [59]

Hello!

What is the exact circumference of the circle?

10π feet

20π feet

40π feet

60π feet

Solving: Using the data (Circle with center A and diameter of 20 feet labeled).

diameter = 2*radius

$20 = 2*r$

$2\:r = 20$

$r = \dfrac{20}{2}$

$\boxed{r = 10\:ft}$

Formula: (Circle length)

$\boxed{C = 2* \pi *r}$

Now, replace in the circumference length formula:

$C = 2\:\pi *r$

$C = 2\:\pi *10$

$\boxed{\boxed{C = 20\:\pi\:ft}}\end{array}}\qquad\quad\checkmark$

Answer:

B. 20π feet

$\underline{The\:exact\:circumference\:of\:the\:circle\:is\:\:20\:\pi\:ft}$

______________________________

I Hope this helps, greetings ... Dexteright02! =)

4 0

3 years ago

Which table of ordered pairs represents a proportional relationship?

gladu [14]

The table that represents a proportional relationship is:

x = -1, -3, -5

y = 1, 3, 5

<h3>Which table represents a proportional relation?</h3>

A proportional relationship is written as:

y = k*x

Where k is the constant of proportionality.

Notice that for equidistant increases in x, we should have equidistant increases on y. Also, proportional relations always have the point (0, 0)

Then the table that represents a proportional relationship is:

x = -1, -3, -5

y = 1, 3, 5

Where the proportional relation is:

y = (-1)*x

When x = -1

y = (-1)*(-1) = 1

When x = -3

y = (-1)*-3 = 3

When x = -5

y = (-1)*(-5) = 5

So the correct option is the second one.

If you want to learn more about proportional relationships:

brainly.com/question/12242745

#SPJ1

6 0

2 years ago

Find the whole. 20% of is 18.

Stolb23 [73]

Since 20% can be rewritten as 1/5, then we can multiply 18 by 5 to get 90, which is your answer. Hope this helps!

8 0

3 years ago

Solve the equation 2 sec² x = 3 - tan x for the domain 0° ≤ x ≤ 360°

vodka [1.7K]

$2 \sec^2 x = 3- \tan x \\\\\implies 2(1 +\tan^2 x) = 3- \tan x \\\\\implies 2 + 2 \tan^2 x +\tan x -3 =0\\\\\implies 2 \tan^2 x + \tan x -1 =0\\\\\implies 2u^2 + u -1 =0~~~ ;[\text{set} \tan x = u]\\\\\implies u = \dfrac{-1\pm \sqrt{1-4\cdot 2 \cdot (-1)}}{2(2)}\\\\\implies u =\dfrac{-1 \pm\sqrt{9}}{4}\\\\\implies u = \dfrac{-1 \pm 3}{4}\\\\\implies u = \dfrac{2}{4}=\dfrac 12~~ \text{or} ~~u =\dfrac{-4}{4} =-1\\\\$

$\implies \tan x = \dfrac 12 ~~ \text{or} ~~ \tan x =-1~~~ ;[\text{Substitute back u =tan x}]\\\\\text{Now,}~ \\\\\tan x = -1,\\\\\implies x = n\pi - \dfrac{\pi}4\\\\\\\text{For interval,}~ [0,2\pi) \\\\x = \dfrac{3\pi}4, \dfrac{7\pi}4\\\\\text{In degrees,}~ x = 135^{\circ}, x =315^{\circ}\\\\\tan x = \dfrac 12\\\\\implies x = n\pi + \tan^{-1} \left(\dfrac 12 \right)\\\\\text{For interval} ~[0,2\pi),\\\\$

$x=\tan^{-1} \left(\dfrac 12 \right), \left[\pi +\tan^{-1} \left( \dfrac 12 \right)\right]\\\\\text{in degrees,} ~~x=\tan^{-1} \left(\dfrac 12 \right), \left[180^{\circ} +\tan^{-1} \left( \dfrac 12 \right)\right]\\\\\\\\\text{Combine all solutions:}\\\\x=\dfrac{3\pi}4, \dfrac{7\pi}4, \tan^{-1} \left(\dfrac 12 \right), \left[\pi +\tan^{-1} \left( \dfrac 12 \right)\right]\\\\\\$

$\text{In degrees,}~ x = 135^{\circ},~315^{\circ},~\tan^{-1} \left(\dfrac 12 \right), \left[180^{\circ} +\tan^{-1} \left( \dfrac 12 \right)\right]$

4 0

2 years ago