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3 years ago

Factors of 4x^3-14x^2+12x

Mathematics

1 answer:

IRISSAK [1]3 years ago

7 0

Step-by-step explanation:

4x³ - 14x² + 12x

= 2x[2x² - 7x + 6]

= 2x[(2x-3)(x-2)]

= 2x(2x-3)(x-2)

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Hi can someone help me with this pls

Mama L [17]

Answer:

Step-by-step explanation:

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3 years ago

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Find the sum of the first 47 terms of the following series, to the nearest integer.

topjm [15]

Answer:

The sum of the first 47 terms of the given series = 6016

Step-by-step explanation:

Given the sequence

13, 18, 23, ...

An arithmetic sequence has a constant difference 'd' and is defined by

$a_n=a_1+\left(n-1\right)d$

$18-13=5,\:\quad \:23-18=5$

As the difference between all the adjacent terms is the same.

$d=5$

$a_1=13$

Arithmetic sequence sum formula

$n\left(a_1+\frac{d\left(n-1\right)}{2}\right)$

Put the values

$d=5$

$a_1=13$

$n=47$

$=47\left(13+\frac{5\left(47-1\right)}{2}\right)$

$=47\left(13+115\right)$

$=47\cdot \:128$

$=6016$

Thus, the sum of the first 47 terms of the given series = 6016

8 0

3 years ago

Cos 0=9/15 find tan 0

Liula [17]

Answer:

I HAVE NO CLUE BUT HAVE A NICE DAY....

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2 years ago

What are the real zeros of y=(x+3)^3+10

aivan3 [116]

<span> y - x3 - 9x2 - 27x - 37 = 0 is this what you were looking for?</span>

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3 years ago

Write an equation for a circle with a diameter that has endpoints at (–4, –7) and (–2, –5). Round to the nearest tenth if necess

Zinaida [17]

since we know the endpoints of the circle, we know then that distance from one to another is really the diameter, and half of that is its radius.

we can also find the midpoint of those two endpoints and we'll be landing right on the center of the circle.

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-4}~,~\stackrel{y_1}{-7})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-5})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{diameter}{d}=\sqrt{[-2-(-4)]^2+[-5-(-7)]^2}\implies d=\sqrt{(-2+4)^2+(-5+7)^2} \\\\\\ d=\sqrt{2^2+2^2}\implies d=\sqrt{2\cdot 2^2}\implies d=2\sqrt{2}~\hfill \stackrel{~\hfill radius}{\cfrac{2\sqrt{2}}{2}\implies\boxed{ \sqrt{2}}} \\\\[-0.35em] ~\dotfill$

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{-4}~,~\stackrel{y_1}{-7})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-5})\qquad \qquad \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{-2-4}{2}~~,~~\cfrac{-5-7}{2} \right)\implies \left( \cfrac{-6}{2}~,~\cfrac{-12}{2} \right)\implies \stackrel{center}{\boxed{(-3,-6)}} \\\\[-0.35em] ~\dotfill$

$\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{-3}{ h},\stackrel{-6}{ k})\qquad \qquad radius=\stackrel{\sqrt{2}}{ r} \\[2em] [x-(-3)]^2+[y-(-6)]^2=(\sqrt{2})^2\implies (x+3)^2+(y+6)^2=2$

4 0

3 years ago

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