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creativ13 [48]
3 years ago
15

Find the equation of the line. GIVING BRAINIEST TO WHOEVER HELPS ME AND GETS IT RIGHT!!!

Mathematics
2 answers:
Triss [41]3 years ago
4 0

Answer:

gradient  =  \frac{7 - 1}{0 - 2}  \\  =  - 3 \\ y = mx + c \\ at \: (1, \: 2) \\ 2 = ( - 3 \times 1) + c \\ c = 5 \\ substitute \: in \: y = mx + c \\ y =  - 3x + 5

Anestetic [448]3 years ago
3 0

Answer:

y= -3x + 7

Step-by-step explanation:

y intercept= 7

slope= -3

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An engineer is building a bridge that should be able to hold a maximum weight of 1 ton. He builds a model of the bridge that is
MrRissso [65]

Answer:

First blank- 32000 ounces

Second blank- 2000 pounds

Yes the bridge can hold 1 ton.

Step-by-step explanation:

The ratio of the scale of the model to the real bridge = 1:4

The test model shows the model can take 8000 ounces

The real bridge will therefore take 8000 x 4 = 32000 ounces

16 ounces = 1 pound

32000 ounces =  x pounds

==>  = 32000/16 = 2000 pounds

2000 pounds = 1 ton

therefore the bridge holds 1 ton

6 0
2 years ago
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iragen [17]
A:line and a prism.
B:line and rectangle
5 0
3 years ago
Someone pls show me
vlada-n [284]

Answer:

  x = 4

Step-by-step explanation:

An exterior angle of a triangle is equal to the sum of the remote interior angles.

  18x +5 = (46) +(-1 +8x)

  10x = 40 . . . . . . . . . . . . . subtract (5+8x) from both sides, simplify

  x = 4 . . . . . . divide by 10

__

<em>Additional comment</em>

The exterior angle is (18)(4) +5 = 77°. The marked unknown interior angle is -1+(8)(4) = 31°. The sum of the two remote interior angles is 46°+31° = 77°. The unmarked interior angle is 180°-77° = 103°.

7 0
3 years ago
A survey was administered to high school seniors in Anytown. According to the survey results, fewer than 0.5% of the students dr
Feliz [49]

Answer: high test-retest reliability

Step-by-step explanation: this is because the result of the survey was thesame with the previous result, despite the space of time between when the first survey was conducted and when the second survey was conducted. There was know observable difference in result and if conducted in the next 3 months again, it will give same result, this strongly indicate that the survey has high test-retest reliability.

8 0
3 years ago
adiocarbon dating of blackened grains from the site of ancient Jericho provides a date of 1315 BC ± 13 years for the fall of the
Zigmanuir [339]

Answer:

\left(\frac{m(t)}{m_{o}} \right)_{min} \approx 0.659 and \left(\frac{m(t)}{m_{o}} \right)_{max} \approx 0.661

Step-by-step explanation:

The equation of the isotope decay is:

\frac{m(t)}{m_{o}} = e^{-\frac{t}{\tau} }

14-Carbon has a half-life of 5568 years, the time constant of the isotope is:

\tau = \frac{5568\,years}{\ln 2}

\tau \approx 8032.926\,years

The decay time is:

t = 1315\,years + 2007\,years \pm 13\,years (There is no a year 0 in chronology).

t = 3335 \pm 13\,years

Lastly, the relative amount is estimated by direct substitution:

\frac{m(t)}{m_{o}} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{\mp\frac{13\,years}{8032.926\,years} }

\left(\frac{m(t)}{m_{o}} \right)_{min} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{-\frac{13\,years}{8032.926\,years} }

\left(\frac{m(t)}{m_{o}} \right)_{min} \approx 0.659

\left(\frac{m(t)}{m_{o}} \right)_{max} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{\frac{13\,years}{8032.926\,years} }

\left(\frac{m(t)}{m_{o}} \right)_{max} \approx 0.661

4 0
3 years ago
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