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3 years ago

Please help me, Thanks.

Mathematics

1 answer:

Lady_Fox [76]3 years ago

3 0

Answer:

6.9

Step-by-step explanation:

$6 \frac{4}{5} = 6.8$

$6.9 > 6.8$

$6.9 < 7$

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Two lines, A and B, are represented by the following equations:

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Answer:

Step-by-step explanation:

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Describe the graph of the equation y = 6x.

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y = 6x (8.9)

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When using the inspection method, the number you would add to (and subtract from) the constant term of the numerator so the poly

krok68 [10]

Question:

Consider the expression $\frac{x^2 + x -10}{x + 3}$

When using the inspection method the number you would add to (and subtract from) the constant term of the numerator so the polynomial in the numerator will have (x + 3) as a factor is?

Answer:

The constant to add is 4

Step-by-step explanation:

Given

$\frac{x^2 + x -10}{x + 3}$

First, we need to get an expression that has x + 3 has its factor.

Represent this expression with: $(x + 3)(x + k)$

Expand

$x^2 + 3x + kx + 3k$

Group like terms

$x^2 + (3 + k)x + 3k$

Compare the above expression to: $x^2 + x - 10$

$(3 + k)x = x$

$3k = -10$

However, we only consider solving $(3 + k)x = x$ for k

$(3 + k)x = x$

$3 + k = 1$

Subtract 3 from both sides

$3 - 3 + k = 1 - 3$

$k = 1 - 3$

$k= -2$

Substitute -2 for k in $(x + 3)(x + k)$

$(x + 3)(x + k) = (x + 3)(x -2)$

$(x + 3)(x + k) = x^2 + 3x - 2x - 6$

$(x + 3)(x + k) = x^2 + x - 6$

So, the expression that has a factor of x + 3 is $x^2 + x - 6$

To get the constant term to add/subtract, we have:

$Constant = (x^2 + x - 6) - (x^2 + x - 10)$

Open brackets

$Constant = x^2 + x - 6 - x^2 - x + 10$

Collect Like Terms

$Constant = x^2 - x^2+ x - x - 6+ 10$

$Constant = - 6+ 10$

$Constant = 4$

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3 years ago

Write an expression for the net change in charge if two electrons are added to each of six atoms in a group. Also, write an expr

dusya [7]

Given:

Two electrons are added to each of six atoms in a group.

To find:

The expression for the net change in charge.

Solution:

Charge of atoms = Number protons - Number of electrons.

Let number of atoms be x.

6 atoms = 2 electrons added

1 atom = $\dfrac{2}{6}$ electrons added

1 atom = $\dfrac{1}{3}$ electrons added

x atoms = $\dfrac{1}{3}x$ electrons added

Therefore, the net change in charge is represented by the expression $-\dfrac{1}{3}x$ because we subtract number of electrons to find atom charge, where, x is the number of atoms.

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1 atom = $\dfrac{2}{6}$ protons removed

1 atom = $\dfrac{1}{3}$ protons removed

x atoms = $\dfrac{1}{3}x$ protons removed

Therefore, the net change in charge is represented by the expression $-\dfrac{1}{3}x$ , where, x is the number of atoms.

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