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2 years ago

What is - 1/8 as a terminating decimal

Mathematics

1 answer:

wolverine [178]2 years ago

8 0

Answer:

1/8 as a terminating decimal is 0.125

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Solve for x use the completing the square method x^2+6x=5

Mrac [35]

Answer:

$x_{1} =-3 +\sqrt{14} \\\\x_{2} =-3 -\sqrt{14}$

Step-by-step explanation:

$x^{2}$ +6x-5=0

we divide the coefficient of the X by half :

in this case: 6/2 = 3 , then we do the following

The result obtained is raised to square power: 3^2=9

we sum and subtract by 9 to maintain the balance of the equation:

$x^{2}$ +6x+9-9-5=0

we have:

$(x+3)^{2}$ -9-5=0

$(x+3)^{2}$ = 14

lets apply square root on both sides of the equation:

$\sqrt{(x+3)^{2}}=\sqrt{14}$

we know:

$\sqrt{a^{2}} = abs(a)$

so we have:

abs(x+3)= $\sqrt{14}$

from where two solutions are obtained

$x_{1} + 3 =\sqrt{14} \\\\x_{2} + 3 =-\sqrt{14}$

finally we have:

$x_{1} =-3 +\sqrt{14} \\\\x_{2} =-3 -\sqrt{14}$

5 0

3 years ago

Read 2 more answers

Someone help me please

gogolik [260]

x/5 = 6/15
15x = 30
x = 2

answer
x = 2 ft

y/10 = 6/15
15y = 60
y = 4

answer
y = 4 ft

3 0

3 years ago

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What is the slope-intercept equation of this line?

PIT_PIT [208]

Answer:

Y=-4/2 + 6

Step-by-step explanation:

It is +6 because the line intercepts the Y axis at the point 6. It is a negative slope. It is a slope of -4/2 because from point (0,6) it goes down 4 and 2 right.

8 0

3 years ago

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write a linear equation that intersects y=x^2 at two points. Then write a second linear equation that intersects y=x^2 at just o

Liula [17]

We know that $y = x^2$ is a parabola, concave up, with vertex in the origin $(0,0)$

So, we may use three horizontal lines for our purpose: any horizontal line above the x axis will intersect the parabola twice. The x axis itself intersects the parabola once on the vertex, while any horizontal line below the x axis won't intercept the parabola.

Here's the examples:

The horizontal line $y = 4$ intercepts the parabola twice: the system $y = x^2,\ y = 4$ is solved by $x^2=4 \implies x = \pm 2$
The horizontal line $y=0$ intercepts the parabola only once: the system is $y=x^2,\ y=0$ which yields $x^2=0\implies x=0$ which is a repeated solution
The horizontal line $y=-5$ intercepts the parabola only once: the system is $y=x^2,\ y=-5$ which yields $x^2=-5$ which is impossible, because a squared number can't be negative.