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Papessa [141]
3 years ago
11

I’ve looked at this for a while, can someone clear up the answers for me? I can’t figure it out.

Mathematics
1 answer:
navik [9.2K]3 years ago
4 0
What answers? I don't see any
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The quotient of a number and 5 is less than 4
cupoosta [38]

Answer:

n ÷ 5 < 4

Step-by-step explanation:

Quotient is to divide.

Let n be the unknown number.

Now you have the equation needed.

8 0
2 years ago
miles has 25 coins In his pocket. Some are Nichols and some are dimes. He has a total of $1.65 how many of each type of coin doe
Leni [432]

▪ Answer:

n = 17

d = 8

▪ Step-by-step explanation:

Hi there !

1 nickel = 5c

1 dime = 10c

$1.65 = 165c

n + d = 25 => n = 25 - d

5n + 10d = 165

replace n

5(25 - d) + 10d = 165

125 - 5d + 10d = 165

5d = 165 - 125

5d = 40

d = 8 (dimes)

n = 25 - d

n = 25 - 8

n = 17 (nickels)

Good luck !

7 0
3 years ago
Find all real solutions and show work: <br> 2x^3-3x^2+18x-27=0
Amiraneli [1.4K]

Given:

The given equation is

2x^3-3x^2+18x-27=0

To find:

All the real solutions.

Solution:

We have,

2x^3-3x^2+18x-27=0

It can be written as

x^2(2x-3)+9(2x-3)=0

(2x-3)(x^2+9)=0

Using zero product property, we get

(2x-3)=0 and (x^2+9)=0

2x=3 and x^2=-9

x=\dfrac{3}{2} and x=\pm\sqrt{-9}

We know that x=\pm\sqrt{-9} is an imaginary number because there is a negative sign under the square root.

Therefore, x=\dfrac{3}{2} is the only real solution of the given equation.

5 0
3 years ago
After being rearranged and simplified, which of the following equations could be solved using the quadratic formula?
Alecsey [184]

For this case we must indicate which of the equations shown can be solved using the quadratic formula.

By definition, the quadratic formula is applied to equations of the second degree, of the form:

ax ^ 2+ bx+ c = 0

Option A:

2x ^ 2-3x +10 = 2x + 21

Rewriting we have:

2x ^ 2-3x-2x+ 10-21 = 0\\2x ^ 2-5x-11 = 0

This equation can be solved using the quadratic formula

Option B:

2x ^ 2-6x-7 = 2x ^ 2

Rewriting we have:

2x ^ 2-2x ^ 2-6x-7 = 0\\-6x-7 = 0

It can not be solved with the quadratic formula.

Option C:

5x ^ 2 + 2x-4 = 2x ^ 2

Rewriting we have:

5x ^ 2-2x ^ 2 + 2x-4 = 0\\3x ^ 2 + 2x-4 = 0

This equation can be solved using the quadratic formula

Option D:

5x ^ 3-3x + 10 = 2x ^ 2

Rewriting we have:

5x ^ 3-2x ^ 2-3x + 10 = 0

It can not be solved with the quadratic formula.

Answer:

A and C

4 0
3 years ago
Read 2 more answers
Jake tutors students. He charges $10 for the first session and $5 for each additional session. Let x
SIZIF [17.4K]

Answer: y = 5x + 10

Step-by-step explanation:

y= 5x + 10

THINK ABOUT IT! if 5 is for EACH additional session and x is equal to the NUMBER OF SESSIONS...... then 5 times x will give you the cost for each additional session, but NOT INCLUDING THE 10 FOR THE FIRST SESSION. so simply just do 5x + 10 and y is the total cost so

y = 5x + 10

8 0
2 years ago
Read 2 more answers
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