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3 years ago

Pls answer this don’t know

Mathematics

2 answers:

Keith_Richards [23]3 years ago

5 0

Answer:

-3, 12

This should be the correct answer

____ [38]3 years ago

3 0

Answer:

(-3, 12)

Step-by-step explanation:

Up five and right five is the same as saying plus 5. The X-axis goes side to side, the y-axis goes up and down.

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Plz help me this is already suppose to be done lol

nignag [31]

Answer:

1. 247 mg ÷ 10 = 24.7 cg

2. 247 mg ÷ 100 = 2.47 dg

247 mg ÷ 1000 = 0.247 g

Step-by-step explanation:

1. To convert from 'mg' to 'cg', you must divide by 10 because 'cg' consists of 10 mg.

2. To convert from 'mg' to 'dg', you must divide by 100 because 'dg' consists of 100 mg.

3. To convert from 'mg' to 'g', you must divide by 1000 because 'g' consists of 1000 mg.

8 0

4 years ago

√448<br><img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B488%7D%20" id="TexFormula1" title=" \sqrt{488} " alt=" \sqrt{488} " align

bazaltina [42]

The answer is 21.17 or 21.16601049.

8 0

4 years ago

HELP PLS IS DUE TODAY PLS!!

LenKa [72]

Sorry it’s so sloppy, hope it helps

6 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 $k$ :