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r-ruslan [8.4K]

3 years ago

9

Help would be so appreciated thanks

1 answer:

Natasha_Volkova [10]3 years ago

3 0

Answer:

75°

:)

:<

;)

...................

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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

3 years ago

Natasha2012 [34]

137.2
30
$298.30
390
5970
3050
16
150
5500000

8 0

3 years ago

Given that F( x ) = x 2 + 2, evaluate F(1) + F(5).

stealth61 [152]

Hi there! :)

Answer:

$\huge\boxed{f(1) + f(5) = 30}$

Given:

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

Evaluate $f(1) + f(5)$

Solve for each:

$f(1) = (1)^{2} + 2 = 3$

$f(5) = (5)^{2} + 2 = 27$

$27 + 3 = 30$

3 0

3 years ago

Read 2 more answers

Evaluate 2^−1 + 2^−1.

Diano4ka-milaya [45]

Answer:

1

Step-by-step explanation:

2^−1 + 2^−1

2x2^-1

1

4 0

3 years ago

Hellohello, if you able to help me then please do. (:

madreJ [45]

Answer:

true duhh :))

Step-by-step explanation:

7 0

3 years ago