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Law Incorporation [45]
2 years ago
10

HELP ME PLEASEEE! Ty:)

Mathematics
1 answer:
Jlenok [28]2 years ago
5 0
I can help you if you mark me as braì LG
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I need help with this ASAP! (Click on the image I uploaded)
3241004551 [841]

Answer:

<em>d</em> = 24.63

Step-by-step explanation:

82.53 - (21.7 + 17.3 + 18.9) =

24.63

4 0
2 years ago
Which statement about the ordered pairs (–2, 9), (0, 9) and (3, 9) is not true?
Lelu [443]

Answer:

The slope of these coordinates is 0.

It is a horizontal line.

Step-by-step explanation:

8 0
2 years ago
A bottle contains 1.6L of water a second bottle constrains 390 mL of water what is the total number of mL in both bottles
patriot [66]

Answer:

1990 ml is the answer thank you

5 0
3 years ago
20 POINTS
Brut [27]

Answer:

Substitute (x+4)2−16 ( x + 4 ) 2 - 16 for x2+8x x 2 + 8 x in the equation x2+y2+8x−6y=24 x 2 + y 2 + 8 x - 6 y = 24 . (x+4)2−16+y2−6y=24 ( x + 4 ) 2 - 16 + y 2 ...

Step-by-step explanation:

5 0
3 years ago
Find the x- and y- intercepts of parabola y=5x^2-16x+10
____ [38]

Y-INTERCEPT

y = 5x^2 - 16x + 10

The y-intercept is where the equation/curve/parabola cosses the y-axis.

The y-axis is where x = 0. (The x-axis is where y = 0)

To find the y-intercept:

\text{y-axis} \rightarrow \text{x = 0} \rightarrow y = 5(0)^2 -16(0) + 10 = 10

The y-intercept must be at (0, 10)

X-INTERCEPT (ROOTS/SOLUTIONS)

y = 5x^2 - 16x + 10\\\text{make it equal 0}\\y = 0\\\therefore 5x^2 - 16x + 10 = 0

We need to use the quadratic formula

The quadratic formula helps us find what values of x make the equation = 0

Quadratic formula: x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

x=\frac{-(-16) + \sqrt{(-16)^2-4(5)(10)}}{2(5)}\\\\x = \frac{16 + \sqrt{256-200}}{10}\\x = \frac{16 + \sqrt{56}}{10}\\x = \frac{16 + 2\sqrt{14}}{10}\\x = \frac{8 + \sqrt{14}}{5}\\\\\\x=\frac{-(-16) - \sqrt{(-16)^2-4(5)(10)}}{2(5)}\\\text{doing the same thing...}\\\text{end up with...}\\x = \frac{8 - \sqrt{14}}{5}\\

The x-intercepts are at:

(\frac{8 + \sqrt{14}}{5}, 0)\\(\frac{8 - \sqrt{14}}{5}, 0)

5 0
2 years ago
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