Geometry math question

Mathematics

1 answer:

enyata [817]3 years ago

5 0

The answer is C. Dilation

The other transformations do not change the size of the figure, but rather how it is portrayed.

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How to solve for linear equation 2/x+y/4=3/2

riadik2000 [5.3K]

This problem is Not Linear

6 0

3 years ago

Read 2 more answers

Caleb is putting tile down in his bathroom and needs to know the perimeter of the floor. Two sides rectangular floor are 5 1/3 f

ladessa [460]

Answer:

20 1/6 ft

Step-by-step explanation:

The perimeter is the sum of the lengths of the sides:

2×(5 1/3 ft) + 2×(4 3/4 ft) = 10 2/3 ft + 8 6/4 ft

= 10 2/3 ft + 9 2/4 ft . . . . change 6/4 ft to 1 1/2 ft

= 10 8/12 ft + 9 6/12 ft = 19 14/12 ft . . . . . use common denominator of 1/12 ft

= 20 2/12 ft . . . . . . change 14/12 ft to 1 2/12 ft; next reduce that fraction

= 20 1/6 ft . . . . perimeter of Caleb's bathroom

4 0

3 years ago

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 $k$ :