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Ipatiy [6.2K]
2 years ago
9

4m^2-5m when m= -4 solve

Mathematics
1 answer:
yawa3891 [41]2 years ago
8 0

Answer:

84

Step-by-step explanation:

4 (-4)^2 - 5(-4)

4 × 16 - 5 × -4

64 + 20

84

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Nastasia [14]

Answer:

(3/2x)+14

Step-by-step explanation:


3 0
3 years ago
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OLEGan [10]
The answer is maybe C lol
4 0
3 years ago
Bob has some 10 lb weights and some 3 lb weights. Together, all his weights add up to 50 lbs. If x represents the number of 3 lb
olga2289 [7]

Answer:

3x = 50 - 10y

Step-by-step explanation:

Here, x represents the number of 3 lb weights and y represents the number of 10 lb weights,

So, total weight of 3 lb weights = 3x

And, the weight of 10 lb weights = 10y

Total weight of all his weights = 3x + 10y

According to the question,

Total weight = 50 lb,

⇒ 3x + 10y = 50

Or 3x = 50 - 10y

Which is the required equation.

Second option is correct.

3 0
3 years ago
5a+2=6-7a please help me fellows
stira [4]
<span>5a + 2 = 6 - 7a 
5a + 7a = 6 - 2
12a = 4
a = 4/12 = 1/3

</span>
6 0
3 years ago
Use distributive property to factor method 1. Use distributive property to expand method 2.
sasho [114]

Step-by-step explanation:

I guess method 1 means to deal with whole factors.

x + 5 = (x - 2)(x + 5)

for (x + 5) <> 0 we can divide both sides by this factor :

1 = x - 2

x = 3

for the second solution we deal with

x + 5 = 0

x = -5

so, for x = -5 and x = 3 both functions deliver the same output, and these are the intersection points.

method 2 : we multiply the expression out and solve it then

x + 5 = (x - 2)(x + 5)

x + 5 = x² + 5x - 2x - 10 = x² + 3x - 10

0 = x² + 2x - 15

the general solution to such a square equation is

x = (-b ± sqrt(b² - 4ac))/(2a)

in our case

a = 1

b = 2

c = -15

x = (-2 ± sqrt(2² - 4×1×-15))/(2×1) =

= (-2 ± sqrt(4 + 60))/2 = (-2 ± sqrt(64))/2 = (-2 ± 8)/2 =

= -1 ± 4

x1 = -1 + 4 = 3

x2 = -1 - 4 = -5

and you get the 2 solutions again. as expected, they are the same as with method 1, of course.

4 0
2 years ago
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