Pls show me how u did it

Mathematics

2 answers:

eduard2 years ago

7 0

Answer:

(2,1)

Step-by-step explanation:

This is because when you type the equations into desmos, this is the cordinate that they collide.

Hope this answers your question.

topjm [15]2 years ago

7 0

Answer:

(2,1)

Step-by-step explanation:

This is because when you type the equations into desmos, this is the cordinate that they collide.

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out of 35 students In a class 21 opted for basketball the remaining opted for volleyball what is the probability of choosing a s

jok3333 [9.3K]

Answer:

15 student

Step-by-step explanation:

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4 0

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 $k$ :