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musickatia [10]

4 years ago

5

Simplify (3x2 - x + 7) + (6x2-3)

2 answers:

nydimaria [60]4 years ago

5 0

The simplified version of the equation is 9x^2-x+4

Novosadov [1.4K]4 years ago

4 0

Answer: $9x^2-x+4$

<u>Combine Like Terms</u>

$3x^2+-x+7+6x^2+-3\\(3x^2+6x^2)+(-x)+(7+-3)\\9x^2+-x+4$

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Divide 190 to the ratio 0.2:0.3:0.5

Lostsunrise [7]

0.2 + 0.3 + 0.5 = 1
190 x 0.2/1 = 38
190 x 0.3/1 = 57
190 x 0.5/1 = 95
190
0.2: 0.3: 0.5
38: 57: 95

3 0

3 years ago

A train travels 146 miles in 2 hours. At this rate, how many miles will it travel in 3.5 hours?

Svetllana [295]

Answer:

255.5 miles

Step-by-step explanation:

146 miles - 2 hours

x miles - 3.5 hours

Cross multiply

2x = 146 x 3.5

2x = 511

Divide both sides by 2

x = 511/2

x = 255.5 miles

7 0

3 years ago

Read 2 more answers

What is the expansion of (3+x)^4

Vlad1618 [11]

Answer:

$\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81$

Step-by-step explanation:

Considering the expression

$\left(3+x\right)^4$

Lets determine the expansion of the expression

$\left(3+x\right)^4$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=3,\:\:b=x$

$=\sum _{i=0}^4\binom{4}{i}\cdot \:3^{\left(4-i\right)}x^i$

Expanding summation

$\binom{n}{i}=\frac{n!}{i!\left(n-i\right)!}$

$i=0\quad :\quad \frac{4!}{0!\left(4-0\right)!}3^4x^0$

$i=1\quad :\quad \frac{4!}{1!\left(4-1\right)!}3^3x^1$

$i=2\quad :\quad \frac{4!}{2!\left(4-2\right)!}3^2x^2$

$i=3\quad :\quad \frac{4!}{3!\left(4-3\right)!}3^1x^3$

$i=4\quad :\quad \frac{4!}{4!\left(4-4\right)!}3^0x^4$

$=\frac{4!}{0!\left(4-0\right)!}\cdot \:3^4x^0+\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1+\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2+\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3+\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4$

$=\frac{4!}{0!\left(4-0\right)!}\cdot \:3^4x^0+\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1+\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2+\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3+\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4$

as

$\frac{4!}{0!\left(4-0\right)!}\cdot \:\:3^4x^0:\:\:\:\:\:\:81$

$\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1:\quad 108x$

$\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2:\quad 54x^2$

$\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3:\quad 12x^3$

$\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4:\quad x^4$

so equation becomes

$=81+108x+54x^2+12x^3+x^4$

$=x^4+12x^3+54x^2+108x+81$

Therefore,

$\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81$

6 0

4 years ago

What is the distance between the points H(-5,9) and J(1,6)?

Ahat [919]

Answer:The answer would be b

Step-by-step explanation:

8 0

3 years ago

Can someone please help me with this?

Fantom [35]

Answer:

step 1: 100

step 2: is 100

Step 3: 5,600

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers