20% of a is 11. What is a?

Mathematics

1 answer:

a_sh-v [17]3 years ago

7 0

Answer:

a is 5.

b is 16.

c is 24.

D is 9.

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CAN YOU HELP ME WITH THIS

____ [38]

the sum of the expanded angle is 180⇒

$40+40+2x+30=180\\2x=70\\x=35$

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3 years ago

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Titus is asked to prove hexagon FEDCBA is congruent to hexagon

Lyrx [107]

Answer:

correct

Step-by-step explanation:

he is correct because the transformation is a translation and under the translation the image and preimage are congruent.

the measure of the sides are preserved, and the peasure of the angles are preserved so if all the corsponding sides and angles are congruent the hexagons are congruent too

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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 $k$ :