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3 years ago

7ab/a A does not equal 0 Simplify the expression

1 answer:

4 0

$\frac{7ab}{a}=7b*\frac{a}{a}=7b$

Field of the functions in this case are:a is not equal zero. <span>Equates to zero only the denominator in this casebecause there are other terms that could have an impact on the division by zero.</span>

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Heyyyy I really need help with dis. If u can please answer and explain

Arisa [49]

Answer:

This should help

Step-by-step explanation:

4 0

3 years ago

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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 $k$ :

$\frac{80\cdot k^{2}}{19321} = \frac{4761}{1493204164}$

$k \approx \pm 0.028$

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

7 0

2 years ago

If the trapezoid below is reflected across the x-axis, what are the coordinates of B'?

Grace [21]

The x-coordinate remains the same as the x-coordinate of point B.
The y-coordinate becomes the additive inverse of the y-coordinate of point B.

Answer: B. (3, -8)

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3 years ago

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Simplify<br> 5 x 5 squared

schepotkina [342]

Answer:
125

Explanation:
5^2 = 25
25 x 5 = 125

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3 years ago

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goldfiish [28.3K]

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