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3 years ago

What is the constant term in the expansion of the binomial (2x + 3)3?

Mathematics

1 answer:

guajiro [1.7K]3 years ago

4 0

The answer is 27 but I am not sure I hope my calculations ain't wrong

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What's 99 multiplied by 9

Fiesta28 [93]

9x99= 891
9x9=81
9x90=810

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3 years ago

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Please give me the correct answer.Please only answer if you're good at math.

lina2011 [118]

The rate of change is 2 euro per hour. Please mark me brainliest :)

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3 years ago

One of the roots of the quadratic equation dx^2+cx+p=0 is twice the other, find the relationship between d, c and p

scZoUnD [109]

Answer:

$c^2 = 9dp$

Step-by-step explanation:

Given

$dx^2 + cx + p = 0$

Let the roots be $\alpha$ and $\beta$

So:

$\alpha = 2\beta$

Required

Determine the relationship between d, c and p

$dx^2 + cx + p = 0$

Divide through by d

$\frac{dx^2}{d} + \frac{cx}{d} + \frac{p}{d} = 0$

$x^2 + \frac{c}{d}x + \frac{p}{d} = 0$

A quadratic equation has the form:

$x^2 - (\alpha + \beta)x + \alpha \beta = 0$

So:

$x^2 - (2\beta+ \beta)x + \beta*\beta = 0$

$x^2 - (3\beta)x + \beta^2 = 0$

So, we have:

$\frac{c}{d} = -3\beta$ -- (1)

and

$\frac{p}{d} = \beta^2$ -- (2)

Make $\beta$ the subject in (1)

$\frac{c}{d} = -3\beta$

$\beta = -\frac{c}{3d}$

Substitute $\beta = -\frac{c}{3d}$ in (2)

$\frac{p}{d} = (-\frac{c}{3d})^2$

$\frac{p}{d} = \frac{c^2}{9d^2}$

Multiply both sides by d

$d * \frac{p}{d} = \frac{c^2}{9d^2}*d$

$p = \frac{c^2}{9d}$

Cross Multiply

$9dp = c^2$

$c^2 = 9dp$

Hence, the relationship between d, c and p is: $c^2 = 9dp$

8 0

3 years ago

The ratio of girls to boys in Liza’s classroom is 5

alexira [117]

There are 15 girls in the classroom

6 0

3 years ago

You have to write 0.33 as a fraction.how do you choose the denominator?

yaroslaw [1]

100 is the easiest because 1/100 =0.01
33/100=0.33

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3 years ago

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