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RoseWind [281]
3 years ago
10

sharon earns $25 per item that she sells plus a base salary of $150 per week. write and solve an inequality to find how many ite

ms she must sell to earn at least $750 per week.
Mathematics
1 answer:
lozanna [386]3 years ago
7 0
She has to sell 24 items
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Add polynomials (intro) Add. Your answer should be an expanded polynomial in standard form. (3x^3 + 4x²) + (3x^3 – 4x^2 – 9x) =
Vlad1618 [11]

(3x^3 + 4x²) + (3x^3 – 4x^2 – 9x) =​

We have to combine like terms, terms that have x raised to the same power:

First, eliminate the parenthesis:

3x^3 + 4x² + 3x^3 – 4x^2 – 9x =​

We have 2 terms raised to the second power (4x^2 and -4 x^2) simply add and subtract.

3x^3+3x^3+4x^2-4x^2-9x

6x^3-9x

7 0
1 year ago
place the following steps in order to complete the square and solve the quadratic equation, x^2-6x+7=0
MatroZZZ [7]
We have that
x²<span>-6x+7=0
</span>Group terms that contain the same variable
(x²-6x)+7=0
Complete the square  Remember to balance the equation
(x²-6x+9-9)+7=0
Rewrite as perfect squares
(x-3)²+7-9=0
(x-3)²-2=0
(x-3)²=2
(x-3)=(+/-)√2
x=(+/-)√2+3

the solutions are
x=√2+3
x=-√2+3



4 0
4 years ago
Which inequality is true?
LenaWriter [7]

Answer:

its B the answer is b

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
PLS HELP! WILL GIVE BRAINLIEST
Leviafan [203]

Answer:

  • The center (2, 2.5), radius \sqrt{37} / 2

Step-by-step explanation:

<u>The standard form of the equation of a circle is: </u>

  • ( x - h )^2 + ( y - k )^2 = r^2, where ( h, k ) is the center and r is the radius

<u>Rewrite the given equation in the standard form:</u>

  • 2x^2 + 2y^2 - 8x + 10y + 2 = 0
  • x^2 - 4x + y^2 + 5y = -1
  • x^2 - 4x + 2^2 + y^2 + 5y + (5/2)^2 = -1 + 4 + 25/4
  • (x - 2)^2 + (y + 2.5)^2 = 37/4

<u>The center is:</u>

  • (2, 2.5)

<u>And radius is:</u>

  • <u />\sqrt{37} / 2

4 0
3 years ago
Consider the following statement. If r is any rational number and s is any irrational number, then r s is irrational. (a) Which
olganol [36]

Answer:

(a) If r is any rational number and s is any irrational number, then r/s is rational

(b) The statement is false when r is 0

Step-by-step explanation:

Given

r \to rational number

s \to irrational number

\frac{r}{s} \to irrational number

Solving (a): The negation

To get the negation of a statement, we only need to negate the end result

In other words, the number type of r and s will remain the same, but r/s will be negated.

So, the negation is:

r \to rational number

s \to irrational number

\frac{r}{s} \to rational number

Solving (b): When r/s is irrational is false

Given that:

\frac{r}{s} \to irrational number

Set r to 0

So:

\frac{r}{s} = \frac{0}{s}

\frac{r}{s} = 0 -- rational

<em>Hence, the statement is false when r is 0</em>

8 0
3 years ago
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