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3 years ago

A binomial: a –1+4a–4a^2

Mathematics

1 answer:

Snowcat [4.5K]3 years ago

4 0

Answer:

-4a^2+5a-1

not a binomial but a trinomial/quadratic equation

Step-by-step explanation:

order of operation

Step 1: Commutative property

-4a^2+4a+a-1

Step 2: combine like terms

-4a^2+5a-1

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A number is multiplied by 5. If 12 is added to the product you get 37.

Korolek [52]

5x+12=37

5x=25

x=5

I hope this helps

5 0

4 years ago

Graph the line with slope 7 and y-intercept -7

poizon [28]

Given :-

To graph the line with slope 7 and y intercept -7.

To Find :-

The graph of the line .

Solution :-

Here since the slope of the line is 7 and y intercept is -7 , we can use the slope intercept form of the line to find the equation of the line . The slope intercept form of the line is ,

$\longrightarrow$ y = mx + c

On putting the respective values ,

$\longrightarrow$ y = 7(x) + (-7)

$\longrightarrow$ y = 7x - 7

For the graph see attachment .

6 0

2 years ago

What is the y-intercept of the following equation of y = 2x - 5?

ANTONII [103]

Answer:

-5

Step-by-step explanation:

The last number is always the y intercept. This number is -5 in this case.

7 0

3 years ago

A game is played with a spinner on a circle, like the minute hand on a clock. The circle is marked evenly from 0 to 100, so, for

zheka24 [161]

Answer:

The probability is 1/2

Step-by-step explanation:

The time a person is given corresponds to a uniform distribution with values between 0 and 100. The mean of this distribution is 0+100/2 = 50 and the variance is (100-0)²/12 = 833.3.

When we take 100 players we are taking 100 independent samples from this same random variable. The mean sample, lets call it X, has equal mean but the variance is equal to the variance divided by the length of the sample, hence it is 833.3/100 = 8.333.

As a consecuence of the Central Limit Theorem, the mean sample (taken from independant identically distributed random variables) has distribution Normal with parameters μ = 50, σ= 8.333. We take the standarization of X, calling it W, whose distribution is Normal Standard, in other words

$W = \frac{X - \mu}{\sigma} = \frac{X - 50}{8.333} \simeq N(0,1)$

The values of the cummulative distribution of the Standard Normal distribution, lets denote it $\phi$ , are tabulated and they can be found in the attached file, We want to know when X is above 50, we can solve that by using the standarization

$P(X > 50) = P(\frac{X-50}{8.33} > \frac{50-50}{8.33}) = P(W > 0) = \phi(0) = 1/2$

Download pdf

8 0

4 years ago

Integral of e^x (sinx +cosx)

disa [49]

integral of e^x (sinx +cosx)= e^x sin x + c

8 0

3 years ago