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3 years ago

-8 equals x + 11 / -2

Mathematics

1 answer:

slava [35]3 years ago

5 0

If the equation is $-8=x+\frac{11}{-2}$ :

$-8=x+\frac{11}{-2}\\-8=x-\frac{11}{2}\\-8+\frac{11}{2}=x-\frac{11}{2}+\frac{11}{2}\\-\frac{16}{2}+\frac{11}{2}=x\\x=-\frac{5}{2}=-2\frac{1}{2}$

If the equation is $-8=\frac{x+11}{-2}$ :

$-8=\frac{x+11}{-2}\\-8(-2)=\frac{x+11}{-2}(-2)\\16=\frac{-2(x+11)}{-2}\\16=x+11\\16-11=x+11-11\\x=5$

Hopefully the math was self-explanatory. If it is not, please feel free to ask for a step-by-step, in depth explanation, and I will edit my answer to accommodate.

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3 years ago

Combine the like terms to create an equivalent expression. 7k-k+19

Harlamova29_29 [7]

Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

Find the roots of h(t) = (139kt)^2 − 69t + 80

Sonbull [250]

Answer:

The positive value of $k$ will result in exactly one real root is approximately 0.028.

Step-by-step explanation:

Let $h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80$ , roots are those values of $t$ so that $h(t) = 0$ . That is:

$19321\cdot k^{2}\cdot t^{2}-69\cdot t + 80=0$ (1)

Roots are determined analytically by the Quadratic Formula:

$t = \frac{69\pm \sqrt{4761-6182720\cdot k^{2} }}{38642}$

$t = \frac{69}{38642} \pm \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$

The smaller root is $t = \frac{69}{38642} - \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$ , and the larger root is $t = \frac{69}{38642} + \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$ .

$h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80$ has one real root when $\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} = 0$ . Then, we solve the discriminant for $k$ :