All categories

3 years ago

How many solutions does the equation have?

Mathematics

2 answers:

timurjin [86]3 years ago

7 0

$4(2x+3)=2(3x-4)\\
8x+12=6x-8\\
2x=-20\\
x=-10 \Rightarrow A$

creativ13 [48]3 years ago

6 0

A. One Solution.
If you distribute the 4 and 2 you're left with
8x+12=6x-8
+8 to both sides
8x+20=6x
-8x both sides
20=-2x
Divide by -2
X=-10

You might be interested in

Convert 0.6 to a fraction to a percent and show a example how

11Alexandr11 [23.1K]

Answer:

6/10 | 60%

Step-by-step explanation:

0.6 is the same as 6/10 because it is six-tenths.

6/10 is the percentage 60% because you can multiply the top and bottom by ten to get 100 on the bottom.

5 0

2 years ago

Kristin goes to the mall and buys a pair of brand name sunglasses on sale for 13 off the regular price of $240.00. How much will

SIZIF [17.4K]

$13 (?) off the original price of 240 is 229. She will have to pay $229.00

7 0

3 years ago

Read 2 more answers

the owner of a zoo described to take a random sample of the zoos visitors to find out which exhibit was the visitor's favorite.

jok3333 [9.3K]

819 is the answer because thats how many visitors he got

7 0

2 years ago

In need help plz.....………

abruzzese [7]

Here you go, enjoy.

1. 8,8

2. 5,4

3. 4,4

7 0

3 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)

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

>> Therefore, your answers are x = -4 or x = 2.

Learn more about quadratic equations:

brainly.com/question/27638369

6 0

2 years ago

Read 2 more answers