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Alborosie
3 years ago
7

A quadratic equation of the form 0 = ax2 + bx + c has a discriminant value of –16. How many real number solutions does the equat

ion have?
Mathematics
2 answers:
agasfer [191]3 years ago
6 0

The equation has 0 real solutions.

If an equation has a negative discriminant, than there are no real solutions. We know this because the quadratic formula requires you to take the square root of the discriminant and therefore you would have all imaginary numbers.

sergij07 [2.7K]3 years ago
3 0

Answer:

answer is 0 (b) on Ed

Step-by-step explanation:

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A cylinder has a radius of 5 feet and a height of 8 feet. Which of the following can be used to calculate the volume of a cone t
Lilit [14]
The answer is 1 over 3(3.14)(5²)(8) 

The volume of the cylinder (V1) is:
V1 = π · r² · h                (r - radius, h - height)

The volume of the cone (V2) is:
V2 = 1/3 π · r² · h         (r - radius, h - height)

It is given:
r = 5 feet
h = 8 feet
π = 3.14

Therefore, the volume of the cone is:
V2 = 1/3 · 3.14 · 5² · 8   which is the same as 1 over 3(3.14)(5²)(8). 

6 0
3 years ago
Read 2 more answers
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ValentinkaMS [17]

Answer:

-100

Step-by-step explanation:

8 0
3 years ago
(-10x-8)-70 (4x-11) + 46 -43(6x6) -54+ (8x5)
STatiana [176]
I feel like you just put random numbers in to waste someone time... but I did the problem any way.

7 0
3 years ago
If the mean weight of 4 backfield members on the football team is 221 lb and the mean weight of the 7 other players is 202 lb, w
MatroZZZ [7]

Let x_1, x_2, x_3, x_4, x_5, x_6, x_7, x_8, x_9, x_{10}, x_{11} be the weight of i-th player.

1. If the mean weight of 4 backfield members on the football team is 221 lb, then

\dfrac{x_1+x_2+x_3+x_4}{4}=221\ lb.

2. If the mean weight of the 7 other players is 202 lb, then

\dfrac{x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}}{7}=202\ lb.

3. From the previous statements you have that

x_1+x_2+x_3+x_4=221\cdot 4=884 \lb,\\ \\x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}=202\cdot 7=1414\ lb.

Add these two equalities and then divide by 11:

x_1+x_2+x_3+x_4+x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}=884+1414=2298\ lb,\\ \\\dfrac{x_1+x_2+x_3+x_4+x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}}{11}=\dfrac{2298}{11}=208\dfrac{10}{11}\ lb.

Answer: the mean weight of the 11-person team is 208\dfrac{10}{11}\ lb.

4 0
2 years ago
Which of the following, to the nearest tenth, is a solution of f(x) = g (x) if f (x) = in (x + 5 ) - 1 and g(x) =x^3 - 2x+1
Dovator [93]

Answer:

c

Step-by-step explanation:

5 0
3 years ago
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