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damaskus [11]
3 years ago
11

What is the result when 6x^3+13x^2+27x+14 is divided by 3x+2

Mathematics
1 answer:
aliya0001 [1]3 years ago
4 0
6x^3+13x^2+16 is the result.
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A.
Veronika [31]

Answer:

a

Step-by-step explanation:

y= 1/2 ^(x+4)

7 0
2 years ago
Solve the problem.
marta [7]

Answer:

There will be 90 ways to reach Greenup from Charleston.

Step-by-step explanation:

<em>Option C: 90 is correct.</em>

Let's name all the ways and try to visualize the roads.

C = Charleston

M = Mattoon

T = Toledo

G = Greenup

Task = Charleston to Greenup. How many different ways to reach?

    1            1          

    2           2           1

C   3     M   3   T      2   G

     4           4           3

     5           5

                   6

So, Refer to this above diagram.

If we Start from C then go to 1 and then go to M and then go to 1 and then go to T and then go to 1 and then go G.

If you notice, in this single possibility we have 3 ways: C to 1 to M, M to 1 to T, T to 1 to G.

It means we will have: 5 x 6 x 3 = 90 number of ways to reach greenup from Charleston.

8 0
3 years ago
Two trains leave the station at the same time, one heading east and the other west. The eastbound train travels at 75 miles per
Alisiya [41]

Let the time = x

Distance = Speed x time

One train is 75x, the other train 85x

Add together to equal total distance:

75x + 85x = 384

Simplify:

160x = 384

Divide both sides by 160:

x = 384 / 160

x = 2.4 hours

8 0
2 years ago
Evaluate 11011-101 base 2
choli [55]

11011 - 101 = 10110 (putting the zero in front does not matter)

Base two only accommodate 1's and 0's. Base two are called binary.

Check the picture below

Note 10 - 1 in binary equals 1

when we borrow to 1, that zero becomes 10, then 10 minus 1 equals 1.

For similar question check the link : brainly.com/question/17099589?referrer=searchResults

7 0
2 years ago
Read 2 more answers
Solve irrational equation pls
rusak2 [61]
\hbox{Domain:}\\&#10;x^2+x-2\geq0 \wedge x^2-4x+3\geq0 \wedge x^2-1\geq0\\&#10;x^2-x+2x-2\geq0 \wedge x^2-x-3x+3\geq0 \wedge x^2\geq1\\&#10;x(x-1)+2(x-1)\geq 0 \wedge x(x-1)-3(x-1)\geq0 \wedge (x\geq 1 \vee x\leq-1)\\&#10;(x+2)(x-1)\geq0 \wedge (x-3)(x-1)\geq0\wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\&#10;x\in(-\infty,-2\rangle\cup\langle1,\infty) \wedge x\in(-\infty,1\rangle \cup\langle3,\infty) \wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\&#10;x\in(-\infty,-2\rangle\cup\langle3,\infty)


&#10;\sqrt{x^2+x-2}+\sqrt{x^2-4x+3}=\sqrt{x^2-1}\\&#10;x^2-1=x^2+x-2+2\sqrt{(x^2+x-2)(x^2-4x+3)}+x^2-4x+3\\&#10;2\sqrt{(x^2+x-2)(x^2-4x+3)}=-x^2+3x-2\\&#10;\sqrt{(x^2+x-2)(x^2-4x+3)}=\dfrac{-x^2+3x-2}{2}\\&#10;(x^2+x-2)(x^2-4x+3)=\left(\dfrac{-x^2+3x-2}{2}\right)^2\\&#10;(x+2)(x-1)(x-3)(x-1)=\left(\dfrac{-x^2+x+2x-2}{2}\right)^2\\&#10;(x+2)(x-3)(x-1)^2=\left(\dfrac{-x(x-1)+2(x-1)}{2}\right)^2\\&#10;(x+2)(x-3)(x-1)^2=\left(\dfrac{-(x-2)(x-1)}{2}\right)^2\\&#10;(x+2)(x-3)(x-1)^2=\dfrac{(x-2)^2(x-1)^2}{4}\\&#10;4(x+2)(x-3)(x-1)^2=(x-2)^2(x-1)^2\\
&#10;4(x+2)(x-3)(x-1)^2-(x-2)^2(x-1)^2=0\\&#10;(x-1)^2(4(x+2)(x-3)-(x-2)^2)=0\\&#10;(x-1)^2(4(x^2-3x+2x-6)-(x^2-4x+4))=0\\&#10;(x-1)^2(4x^2-4x-24-x^2+4x-4)=0\\&#10;(x-1)^2(3x^2-28)=0\\&#10;x-1=0 \vee 3x^2-28=0\\&#10;x=1 \vee 3x^2=28\\&#10;x=1 \vee x^2=\dfrac{28}{3}\\&#10;x=1 \vee x=\sqrt{\dfrac{28}{3}} \vee x=-\sqrt{\dfrac{28}{3}}\\

There's one more condition I forgot about
-(x-2)(x-1)\geq0\\&#10;x\in\langle1,2\rangle\\

Finally
x\in(-\infty,-2\rangle\cup\langle3,\infty) \wedge x\in\langle1,2\rangle \wedge x=\{1,\sqrt{\dfrac{28}{3}}, -\sqrt{\dfrac{28}{3}}\}\\&#10;\boxed{\boxed{x=1}}
3 0
2 years ago
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