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3 years ago

If 95% of students are coming to school tomorrow, what is the complement?

Mathematics

2 answers:

DaniilM [7]3 years ago

6 0

Answer:

the answer is d

Step-by-step explanation:

it only takes 5% to get to 100%

rjkz [21]3 years ago

3 0

Answer:

Step-by-step explanation:

can you help me with this question According to Frost in his poem "The Road Not Taken," what kind of difference did taking one road over another road make? A. It made a good difference, B. It made a bad diference, c. Frost does not say

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Equations and inequalities

DerKrebs [107]

Inequalities and equations use similar steps to solve, but inequalities give a range of numbers that can represent the variable while an equation will give a direct answer.

Example:

4+5x=24

First you would subtract 4 or add the opposite of 4 from 24.

4+5x=24
+(-4) +(-4)
—————
5x=20

Then divide each side by 5 or multiply each side by the reciprocal of 5 to isolate the variable.

1/5*5x=20*1/5 (reciprocal)
X= 4

You would use the same steps in an inequality but end up with a different sign.

8 0

3 years ago

Read 2 more answers

Sarah has 128 stickers. alex has 10 fewer stickers than sarah. how many stickers does Alex have.

nikitadnepr [17]

To get your answer you first subtract 10 from 128 which is 118.

4 0

3 years ago

1.) Determine the type of solutions for the function (Picture 1)

NNADVOKAT [17]

Answer:

1) 2 nonreal complex roots

2) 1 Real Solution

3) 16

4) Reflected, narrower by a factor of 2/5, slides right 4 units and slides up 6 (units)

Step-by-step explanation:

1) The graph does not intercept the x-axis, therefore, there are no real solutions at the point y = 0

We get;

y = a·x² + b·x + c

At y = 6, x = -2

Therefore;

6 = a·(-2)² - 2·b + c = 4·a - 2·b + c

6 = 4·a - 2·b + c...(1)

At y = 8, x = 0

8 = a·(0)² + b·0 + c

∴ c = 8...(2)

Similarly, we have;

At y = 8, x = -4

8 = a·(-4)² - 4·b + c = 16·a - 4·b + 8

16·a - 4·b = 0

∴ b = 16·a/4 = 4·a

b = 4·a...(3)

From equation (1), (2) and (3), we have;

6 = 4·a - 2·b + c

∴ 6 = b - 2·b + 8 = -b + 8

6 - 8 = -b

∴ -b = -2

b = 2

b = 4·a

∴ a = b/4 = 2/4 = 1/2

The equation is therefor;

y = (1/2)·x² + 2·x + 8

Solving we get;

x = (-2 ± √(2² - 4 × (1/2) × 8))/(2 × (1/2))

x =( -2 ± √(-12))/1 = -2 ± √(-12)

Therefore, we have;

2 nonreal complex roots

2) Give that the graph of the function touches the x-axis once, we have;

1 Real Solution

3) The given function is f(x) = 2·x² + 8·x + 6

The general form of the quadratic function is f(x) = a·x² + b·x + c

Comparing, we have;

a = 2, b = 8, c = 6

The discriminant of the function, D = b² - 4·a·c, therefore, for the function, we have;

D = 8² - 4 × 2 × 6 = 16

The discriminant of the function, D = 16

4.) The given function is g(x) = (-2/5)·(x - 4)² + 6

The parent function of a quadratic equation is y = x²

A vertical translation is given by the following equation;

y = f(x) + b

A horizontal to the right by 'a' translation is given by an equation of the form; y = f(x - a)

A vertical reflection is given by an equation of the form; y = -f(x) = -x²

A narrowing is given by an equation of the form; y = b·f(x), where b < 1

Therefore, the transformations of g(x) from the parent function are;

g(x) is a reflection of the parent function, with the graph of g(x) being narrower by 2/5 than the graph of the parent function. The graph of g(x) is shifted right by 4 units and is then slides up by 6 units.

7 0

2 years ago

A bacteria culture starts with 400 bacteria and grows at a rate proportional to its size. After 4 hours, there are 9000 bacteria

Kaylis [27]

Answer:

A) The expression for the number of bacteria is $P(t) = 400e^{0.7783t}$ .

B) After 5 hours there will be 19593 bacteria.

C) After 5.55 hours the population of bacteria will reach 30000.

Step-by-step explanation:

A) Here we have a problem with differential equations. Recall that we can interpret the rate of change of a magnitude as its derivative. So, as the rate change proportionally to the size of the population, we have

$P' = kP$

where $P$ stands for the population of bacteria.

Writing $P'$ as $\frac{dP}{dt}$ , we get

$\frac{dP}{dt} = kP$ .

Notice that this is a separable equation, so

$\frac{dP}{P} = kdt$ .

Then, integrating in both sides of the equality:

$\int\frac{dP}{P} = \int kdt$ .

We have,

$\ln P = kt+C$ .

Now, taking exponential

$P(t) = Ce^{kt}$ .

The next step is to find the value for the constant $C$ . We do this using the initial condition $P(0)=400$ . Recall that this is the initial population of bacteria. So,