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3 years ago

What is the factorization of 3x^2-8x+5

Mathematics

1 answer:

VARVARA [1.3K]3 years ago

4 0

Answer:

(x - 1)(3x - 5)

Step-by-step explanation:

Consider the factors of the product of the coefficient of the x² term and the constant term which sum to give the coefficient of the x-term.

product = 3 × 5 = 15 and sum = - 8

The factors are - 3 and - 5

use these factors to split the middle term

3x² - 3x - 5x + 5 ( factor the first/second and third/fourth terms )

= 3x(x - 1) - 5(x - 1) ( take out the common factor (x - 1) )

= (x - 1)(3x - 5)

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Divide -5/12 and -5/8

kiruha [24]

Answer:

2/3

Step-by-step explanation:

-5/12 ÷ -5/8

Copy dot flip

-5/12 * -8/5

The 5's cancel

-8/-12

Divide the top and bottom by -4

2/3

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3 years ago

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I need help on this question

sasho [114]

Answer:

The shape is a kite and x=26

Step-by-step explanation:

The shape is a kite because the two adjacent sides at the right match and the two sides on the left match

To find x:

125+125=250

360-250=110

(3x-11)+(2x-9)=110

Combine like terms: 5x-20=110

Add 20 to both sides: 5x-20+20=110+20

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Divide both sides by 5: 5x/5=130/5

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3 years ago

What is the value of x? 0.9(x+1.4)−2.3+0.1x=1.6 Help quick

Lyrx [107]

The answer is x is equal to 2.64

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3 years ago

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Please help with this calculus problem!

ELEN [110]

$\dfrac{\mathrm dQ}{\mathrm dt}=\dfrac{a(1-5bt^2)}{(1+bt^2)^4}$
$Q(t)=\displaystyle\int\frac{a(1-5bt^2)}{(1+bt^2)^4}\,\mathrm dt$

Let $t=\dfrac1{\sqrt b}\tan u$ , so that $\mathrm dt=\dfrac1{\sqrt b}\sec^2u\,\mathrm du$ . Then the integral becomes

$\displaystyle\frac a{\sqrt b}\int\frac{1-5b\left(\frac1{\sqrt b}\tan u\right)^2}{\left(1+b\left(\frac1{\sqrtb}\tan u\right)^2\right)^4}\sec^2u\,\mathrm du$
$=\displaystyle\frac a{\sqrt b}\int\frac{1-5\tan^2u}{(1+\tan^2u)^4}\sec^2u\,\mathrm du$
$=\displaystyle\frac a{\sqrt b}\int\frac{1-5\tan^2u}{\sec^6u}\,\mathrm du$
$=\displaystyle\frac a{\sqrt b}\int(\cos^6u-5\sin^2u\cos^4u)\,\mathrm du$
$=\displaystyle\frac a{\sqrt b}\int(6\cos^2u-5)\cos^4u\,\mathrm du$

One way to proceed from here is to use the power reduction formula for cosine:

$\cos^{2n}x=\left(\dfrac{1+\cos2x}2\right)^n$

You'll end up with

$=\displaystyle\frac a{16\sqrt b}\int(5\cos2u+8\cos4u+3\cos6u)\,\mathrm du$
$=\displaystyle\frac a{16\sqrt b}\left(\frac52\sin2u+2\sin4u+\frac12\sin6u\right)+C$
$=\dfrac a{32\sqrt b}(5\sin2u+4\sin4u+\sin6u)+C$
$Q(t)=\dfrac{at}{(1+bt^2)^3}+C$

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3 years ago

(Tot

Andrej [43]

Answer:

0,07 0,7 0,704 0,74 0,744

Step-by-step explanation:

When u make it 100 times bigger, its clear.

7 70 704 740 744

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3 years ago