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3 years ago

(2x+2)(3x+1) Multiply the binomials Please show Step-by-Step

Mathematics

2 answers:

enyata [817]3 years ago

4 0

Answer:

2(x+1)(3x+1)

Step-by-step explanation:

(2x+2)(3x+1)

Apply the distributive property by multiplying each term of 2x+2 by each term of 3x+1.

6x^2+2x+6x+2

Combine 2x and 6x to get 8x.

6x^2+8x+2

Marysya12 [62]3 years ago

3 0

Answer:

6x²+8x+2

Step-by-step explanation:

multiply each ferm in the first set of parenthess by each term in the second parentheses (FOIL)

2x times 3x +2x+ 2x times 3x+2

6x+2x+2 times 3x +2

6x²+2x+6x+2

now collect like terms

6x²+8x+2

You might be interested in

An engineer wishes to determine the width of a particular electronic component. If she knows that the standard deviation is 1.6

nexus9112 [7]

Answer:

n=1705

Step-by-step explanation:

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Assuming the X follows a normal distribution

$X \sim N(\mu, \sigma=1.6)$

And the distribution for $\bar X$ is:

$\bar X \sim N(\mu, \frac{1.6}{\sqrt{n}})$

We know that the margin of error for a confidence interval is given by:

$Me=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

The next step would be find the value of $\z_{\alpha/2}$ , $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$

Using the normal standard table, excel or a calculator we see that:

$z_{\alpha/2}=\pm 2.58$

If we solve for n from formula (1) we got:

$\sqrt{n}=\frac{z_{\alpha/2} \sigma}{Me}$

$n=(\frac{z_{\alpha/2} \sigma}{Me})^2$

And we have everything to replace into the formula:

$n=(\frac{2.58(1.6)}{0.1})^2 =1704.03$

And if we round up the answer we see that the value of n to ensure the margin of error required $\pm=0.1$ mm is n=1705.

6 0

3 years ago

How to differentiate ?

Bas_tet [7]

Use the power, product, and chain rules:

$y = x^2 (3x-1)^3$

• product rule

$\dfrac{\mathrm dy}{\mathrm dx} = \dfrac{\mathrm d(x^2)}{\mathrm dx}\times(3x-1)^3 + x^2\times\dfrac{\mathrm d(3x-1)^3}{\mathrm dx}$

• power rule for the first term, and power/chain rules for the second term:

$\dfrac{\mathrm dy}{\mathrm dx} = 2x\times(3x-1)^3 + x^2\times3(x-1)^2\times\dfrac{\mathrm d(3x-1)}{\mathrm dx}$

• power rule

$\dfrac{\mathrm dy}{\mathrm dx} = 2x\times(3x-1)^3 + x^2\times3(3x-1)^2\times3$

Now simplify.

$\dfrac{\mathrm dy}{\mathrm dx} = 2x(3x-1)^3 + 9x^2(3x-1)^2 \\\\ \dfrac{\mathrm dy}{\mathrm dx} = x(3x-1)^2 \times (2(3x-1) + 9x) \\\\ \boxed{\dfrac{\mathrm dy}{\mathrm dx} = x(3x-1)^2(15x-2)}$

You could also use logarithmic differentiation, which involves taking logarithms of both sides and differentiating with the chain rule.

On the right side, the logarithm of a product can be expanded as a sum of logarithms. Then use other properties of logarithms to simplify

$\ln(y) = \ln\left(x^2(3x-1)^3\right) \\\\ \ln(y) = \ln\left(x^2\right) + \ln\left((3x-1)^3\right) \\\\ \ln(y) = 2\ln(x) + 3\ln(3x-1)$

Differentiate both sides and you end up with the same derivative:

$\dfrac1y\dfrac{\mathrm dy}{\mathrm dx} = \dfrac2x + \dfrac9{3x-1} \\\\ \dfrac1y\dfrac{\mathrm dy}{\mathrm dx} = \dfrac{15x-2}{x(3x-1)} \\\\ \dfrac{\mathrm dy}{\mathrm dx} = \dfrac{15x-2}{x(3x-1)} \times x^2(3x-1)^3 \\\\ \dfrac{\mathrm dy}{\mathrm dx} = x(15x-2)(3x-1)^2$