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3 years ago

The sum of 12 and the product of 5 and a number.

Mathematics

1 answer:

natulia [17]3 years ago

7 0

Answer:

Step-by-step explanation:

We need to write this as an equation:

12 + 5x = 0

We then solve for x.

5x = -12

x = -2.4

<h2>___________________________________________________</h2><h3>I AM ALWAYS HAPPY TO HELP :)</h3>

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Jajajaja helppppppppp <3

Anna11 [10]

you can subtract from point 1 to point 2

(9,-2) - (8,3)

9-8= 1

-2-3= -5

therefore horizontal leg is 1

vertical leg is -5

you can also sketch and count the units :)

3 0

2 years ago

Gino bought 1/2 lb of candy, 3/4 lb of nuts, and 7/8 lb of granola to make a trail mix. How many pounds of trail mix did Gino

Yakvenalex [24]

1/2 + 3/4 + 7/8
4/4(1/2)+ 2/2(3/4) + 7/8
4/8 + 6/8 + 7/8

4+ 6 + 7
—————
8

17 1
——. Or 2—-
8 8

He prepared 2 1/8 lbs of trail mix

4 0

2 years ago

7. Use the quadratic formula to find the solution(s). x² + 2x - 8 = 0

morpeh [17]

Hey there!

Use the quadratic formula to find the solution(s). x² + 2x - 8 = 0

Answer :

x = -4 or x = 2 ✅

Explanation :

Quadratic formula : ax² + bx + c = 0 where a ≠ 0

The number of real-number solutions (roots) is determined by the discriminant (b² - 4ac) :

If b² - 4ac > 0 , There are 2 real-number solutions

If b² - 4ac = 0 , There is 1 real-number solution.

If b² - 4ac < 0 , There is no real-number solution.

The roots of the equation are determined by the following calculation:

$x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a}$

Here, we have :

a = 1
b = 2
c = -8

1) Calculate the discriminant :

b² - 4ac ⇔ 2² - 4(1)(-8) ⇔ 4 - (-32) ⇔ 36

b² - 4ac = 36 > 0 ; The equation admits two real-number solutions

2) Calculate the roots of the equation:

▪️ (1)

$x_1 = \frac{ - b - \sqrt{ {b}^{2} - 4ac} }{2a} \\ \\ x_1 = \frac{ - 2 - \sqrt{36} }{2(1) } \\ \\ x_1 = \frac{ - 2 - 6}{2} \\ \\ x_1 = \frac{ - 8}{2} \\ \\ \blue{\boxed{\red{x_1 = -4}}}$

▪️ (2)

$x_2 = \frac{ - b + \sqrt{ {b}^{2} - 4ac } }{2a} \\ \\ x_2 = \frac{ - 2 + \sqrt{36} }{2(1)} \\ \\ x_2 = \frac{ - 2 + 6}{2} \\ \\ x_2 = \frac{4}{2} \\ \\ \red{\boxed{\blue{x_2 = 2}}}$