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Marat540 [252]
3 years ago
13

What happens when the ordered pair (x,y) moves to the right?

Mathematics
1 answer:
UNO [17]3 years ago
4 0
Question is unclear
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The store sells walnuts for $2.95 per pound. Corey bought 2.4 pounds of walnuts. Before tax, how much will the walnuts cost?
stepladder [879]

Answer:

$7.18

Step-by-step explanation:

Multiply the number of pounds bought by the cost of one pound.

 (2.4 pounds)($2.99 per pound) = $7.176 ≅ $7.18

7 0
2 years ago
Read 2 more answers
When Sam gave Clarissa 105 chocolates, Clarissa had 5 times as many chocolates as Sam. They had 504 chocolates altogether. How m
IgorLugansk [536]

Answer:

315 chocolate

Step-by-step explanation:

Let Sam's chocolate be S

Let Clariss's chocolate be C

When Sam gave Clarissa 105 chocolates, Clarissa had 5 times as many chocolates as Sam.

This can be written as:

C = 5S

The sum of their chocolate is 504 i.e

S + C = 504

Now, let us determine the chocolate of Clarissa after receiving 105 chocolate from Sam. This can be obtained as follow:

S + C = 504

But: C = 5S

S + 5S = 504

6S = 504

Divide both side by 6

S = 504/6

S = 84.

C = 5S = 5 x 84 = 420

Therefore, Clarissa have 420 chocolate after receiving 105 chocolate from Sam.

Now, to know the amount of chocolate that Clarissa has at first, we simply subtract 105 from the present amount that Clarissa have. This is illustrated below:

Amount of chocolate that Clarissa has a first = 420 – 105 = 315

Therefore, Clarissa had 315 chocolate at first.

3 0
3 years ago
Can you please help me?​
Musya8 [376]

Answer:

Option (a)

Step-by-step explanation:

Slope of a line passing through two points (x_1,y_1) and (x_2,y_2) is given by,

Slope (m) = \frac{y_2-y_1}{x_2-x_1}

Slope of the line passing through (-4, 3) and (6, 3) will be,

m = \frac{3-3}{6+4}

m = 0

In other words, slope of any line parallel to x-axis is zero.

Therefore, Option (a) is the answer.

6 0
2 years ago
I am less than 10 I am not a multiple of 2 I am a composite number
Rama09 [41]

Answer:

9

Step-by-step explanation:

Composite Numbers before 10: 4, 6, 8, and 9

The only one of those 4 that is NOT a multiple of 2: 9

4 0
2 years ago
1) Determine the discriminant of the 2nd degree equation below:
Aleksandr-060686 [28]

\LARGE{ \boxed{ \mathbb{ \color{purple}{SOLUTION:}}}}

We have, Discriminant formula for finding roots:

\large{ \boxed{ \rm{x =  \frac{  - b \pm \:  \sqrt{ {b}^{2}  - 4ac} }{2a} }}}

Here,

  • x is the root of the equation.
  • a is the coefficient of x^2
  • b is the coefficient of x
  • c is the constant term

1) Given,

3x^2 - 2x - 1

Finding the discriminant,

➝ D = b^2 - 4ac

➝ D = (-2)^2 - 4 × 3 × (-1)

➝ D = 4 - (-12)

➝ D = 4 + 12

➝ D = 16

2) Solving by using Bhaskar formula,

❒ p(x) = x^2 + 5x + 6 = 0

\large{ \rm{ \longrightarrow \: x =  \dfrac{ - 5\pm  \sqrt{( - 5) {}^{2} - 4 \times 1 \times 6 }} {2 \times 1}}}

\large{ \rm{ \longrightarrow \: x =  \dfrac{ - 5  \pm  \sqrt{25 - 24} }{2 \times 1} }}

\large{ \rm{ \longrightarrow \: x =  \dfrac{ - 5 \pm 1}{2} }}

So here,

\large{\boxed{ \rm{ \longrightarrow \: x =  - 2 \: or  - 3}}}

❒ p(x) = x^2 + 2x + 1 = 0

\large{ \rm{ \longrightarrow \: x =  \dfrac{  - 2 \pm  \sqrt{ {2}^{2}  - 4 \times 1 \times 1} }{2 \times 1} }}

\large{ \rm{ \longrightarrow \: x =  \dfrac{ - 2 \pm \sqrt{4 - 4} }{2} }}

\large{ \rm{ \longrightarrow \: x =  \dfrac{ - 2 \pm 0}{2} }}

So here,

\large{\boxed{ \rm{ \longrightarrow \: x =  - 1 \: or \:  - 1}}}

❒ p(x) = x^2 - x - 20 = 0

\large{ \rm{ \longrightarrow \: x =  \dfrac{ - ( - 1) \pm  \sqrt{( - 1) {}^{2} - 4 \times 1 \times ( - 20) } }{2 \times 1} }}

\large{ \rm{ \longrightarrow \: x =  \dfrac{ 1 \pm \sqrt{1 + 80} }{2} }}

\large{ \rm{ \longrightarrow \: x =  \dfrac{1 \pm 9}{2} }}

So here,

\large{\boxed{ \rm{ \longrightarrow \: x = 5 \: or \:  - 4}}}

❒ p(x) = x^2 - 3x - 4 = 0

\large{ \rm{ \longrightarrow \: x =   \dfrac{  - ( - 3) \pm \sqrt{( - 3) {}^{2} - 4 \times 1 \times ( - 4) } }{2 \times 1} }}

\large{ \rm{ \longrightarrow \: x =  \dfrac{3 \pm \sqrt{9  + 16} }{2 \times 1} }}

\large{ \rm{ \longrightarrow \: x =  \dfrac{3  \pm 5}{2} }}

So here,

\large{\boxed{ \rm{ \longrightarrow \: x = 4 \: or \:  - 1}}}

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5 0
2 years ago
Read 2 more answers
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