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3 years ago

(2x3 + 7x2 + x) + 2(x2 - 2x - 6) splifly

Mathematics

1 answer:

zimovet [89]3 years ago

8 0

Answer:

2x³ + 3(3x - 4)(x + 1)

Step-by-step explanation:

(2x³ + 7x² + x) + 2(x² - 2x - 6)

2x³ + 7x² + x + 2x² - 4x - 12

2x³ + 9x² - 3x - 12

2x³ + 3(3x² - x - 4)

2x³ +3(3x - 4)(3x + 3)/3

2x³ + (3x - 4)(3x + 3)

2x³ + 3(3x - 4)(x + 1)

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HELP ME PLZZZZZZZZ ASAP

Alex_Xolod [135]

Answer:

find the common factor

Step-by-step explanation:

1)13

2) 13(1+5)

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Abigail's car uses 3 2/3 gallons of gas every hour on a long distance trip. How many gallons does the car use in 8 1/2 hours?

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3 years ago

Which equation can be used to find the surface area of the cylinder?

sergij07 [2.7K]

Answer:

S = 2π(36) + 2π(6)(8)

S = 2(3.14)(36) + 2(3.14)(6)(8)

Step-by-step explanation:

The surface area of a cylinder is the sum of the area of the base and the lateral area. A cylinder has 2 circular base .

Surface area = 2πr² + 2πrh

surface area = 2πr(r + h)

where

r = radius

h = height

surface area = 2πr(r + h)

r = 6 cm

h = 8 cm

The equation to find the surface area is represented below.

Surface area = 2πr² + 2πrh

S = 2π (6²) + 2π (6)(8)

S = 2π(36) + 2π (6)(8)

Assuming π is given as 3.14 the equation will be as follows

S = 2(3.14)(36) + 2(3.14)(6)(8)

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3 years ago

16x+28=16x+28. help please?

castortr0y [4]

Answer:

All real numbers could equal x.

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4 years ago

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Find the center that eliminates the linear terms in the translation of 4x^2 - y^2 + 24x + 4y + 28 = 0.(-3, 2)(-3,- 2)(4, 0)

baherus [9]

Step 1

Given;

$4x^2-y^2+24x+4y+28=0$

Required; To find the center that eliminates the linear terms

Step 2

$\begin{gathered} 4x^2-y^2+24x+4y=-28 \\ 4x^2+24x-y^2+4y=-28 \\ Complete\text{ the square }; \\ 4x^2+24x \\ \text{use the form ax}^2+bx\text{ +c} \\ \text{where} \\ a=4 \\ b=24 \\ c=0 \end{gathered}$ $\begin{gathered} consider\text{ the vertex }form\text{ of a }parabola \\ a(x+d)^2+e \\ d=\frac{b}{2a} \\ d=\frac{24}{2\times4} \\ d=\frac{24}{8} \\ d=3 \end{gathered}$ $\begin{gathered} Find\text{ the value of e using }e=c-\frac{b^2}{4a} \\ e=0-\frac{24^2}{4\times4} \\ e=0-\frac{576}{16}=-36 \end{gathered}$

Step 3

Substitute a,d,e into the vertex form

$\begin{gathered} a(x+d)^2+e \\ 4(x+_{}3)^2-36 \end{gathered}$ $\begin{gathered} 4(x+3)^2-36-y^2+4y=-28 \\ 4(x+3)^2-y^2+4y=\text{ -28+36} \\ \\ \end{gathered}$

Step 4

Completing the square for -y²+4y

$\begin{gathered} \text{use the form ax}^2+bx\text{ +c} \\ \text{where} \\ a=-1 \\ b=4 \\ c=0 \end{gathered}$ $\begin{gathered} consider\text{ the vertex }form\text{ of a }parabola \\ a(x+d)^2+e \\ d=\frac{b}{2a} \\ d=\text{ }\frac{4}{2\times-1} \\ d=\frac{4}{-2} \\ d=-2 \end{gathered}$ $\begin{gathered} Find\text{ the value of e using }e=c-\frac{b^2}{4a} \\ e=0-\frac{4^2}{4\times(-1)} \\ \\ e=0-\frac{16}{-4} \\ e=4 \end{gathered}$

Step 5

Substitute a,d,e into the vertex form

$\begin{gathered} a(y+d)^2+e \\ =-1(y+(-2))^2+4 \\ =-(y-2)^2+4 \end{gathered}$

Step 6

$\begin{gathered} 4(x+3)^2-y^2+4y=\text{ -28+36} \\ 4(x+3)^2-(y-2)^2+4=-28+36 \\ 4(x+3)^2-(y-2)^2=-28+36-4 \\ 4(x+3)^2-(y-2)^2=4 \\ \frac{4(x+3)^2}{4}-\frac{(y-2)^2}{4}=\frac{4}{4} \\ (x+3)^2-\frac{(y-2)^2}{2^2}=1 \end{gathered}$

Step 7

$\begin{gathered} \frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1 \\ \text{This is the }form\text{ of a hyperbola.} \\ \text{From here } \\ a=1 \\ b=2 \\ k=2 \\ h=-3 \end{gathered}$

Hence the answer is (-3,2)

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1 year ago