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nataly862011 [7]
3 years ago
10

Select the correct answer from each drop-down menu.

Mathematics
1 answer:
Harman [31]3 years ago
3 0

Answer:

q(x)= 2x+5

r(x)= 6

Step-by-step explanation:

Given:

2x^2+13x+26 /x+4

Re-writing in the form of q(x) + r(x)/d(x)

Let

Polynomial P(x)= 2x^2+13x+26

Polynomial d(x)= x+4

Using remainder theorem, putting -4 in P(x) to find r(x)

P(-4)= 2(-4)^2+13(-4)+26

       =6

i.e r(x)= 6

Finding q(x) by long division

q(x)= 2x+5

Hence given expression in the form q(x) + r(x)/d(x)

= (2x+5) + 6/x+4 !

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When you divide a whole number by a decimal less than 1, the quotient is greater than the whole number. Why?
Setler [38]
For this type of problem you would have to move the decimal as many places to the right as needed to make the divisor a whole number the you would move the dividend the same amount of times even if it is a whole number . Then you use what you have to divide normally. Hope this helps!
3 0
4 years ago
Can someone please help me out here? :)
Marizza181 [45]

Answer:

i believe it's 133.13

3 0
3 years ago
PLEASE HELP!!!
soldi70 [24.7K]

Answer:

See below

Step-by-step explanation:

4(x + 5) = 9x + 4x − 34

4x + 20 = 9x + 4x − 34   [Distributive]

4x + 20 =  13x − 34         {Combine Like Terms]

4x - 13x + 20 = 13x - 13x − 34  [Subtractive:  -13x both sides]

-5x + 20 = - 34            [Combine Like Terms]

-5x + 20 - 20 = -34 - 20     [Subtractive:  - 20 both sides]

-5x  = -54               [Combine Like Terms]

x = -54/-5             [Division Property:  divide both sides by -5]

x =  - 10 4/5            

5 0
2 years ago
Find the intersection points using substitution or elimination for each system of equations:
natka813 [3]

Answer:

Step-by-step explanation:

Okay, so I think I know what the equations are, but I might have misinterpreted them because of the syntax- I think when you ask a question you can use the symbols tool to input it in a more clear way, otherwise you can use parentheses and such.

Problem 1:

(x²)/4 +y²= 1

y= x+1

*substitute for y*

Now we have a one-variable equation we can solve-

x²/4 + (x+1)² = 1

x²/4 + (x+1)(x+1)= 1

x²/4 + x²+2x+1= 1

*subtract 1 from both sides to set equal to 0*

x²/4 +x^2+2x=0

x²/4 can also be 1/4 * x²

1/4 * x² +1*x² +2x = 0

*combine like terms*

5/4 * x^2+2x+ 0 =0

now, you can use the quadratic equation to solve for x

a= 5/4

b= 2

c=0

the syntax on this will be rough, but I'll do my best...

x= (-b ± √(b²-4ac))/(2a)

x= (-2 ±√(2²-4*(5/4)*(0))/(2*(5/4))

x= (-2 ±√(4-0))/(2.5)

x= (-2±2)/2.5

x will have 2 answers because of ±

x= 0 or x= 1.6

now plug that back into one of the equations and solve.

y= 0+1 = 1

y= 1.6+1= 2.6

Hopefully this explanation was enough to help you solve problem 2.

Problem 2:

x² + y² -16y +39= 0

y²- x² -9= 0

6 0
3 years ago
Model the pair of situations with exponential functions f and g. Find the approximate value of x that makes f(x)= g(x).
Law Incorporation [45]
I like the cat on your pfp LOL sorry i couldn’t help tho
6 0
3 years ago
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