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3 years ago

What is the value of the expression?

Mathematics

1 answer:

serious [3.7K]3 years ago

7 0

B. -10.6

5 + (-11.3) Is the same thing as 5 - 11.3, which equals -6.3
-6.3 + (-4.3) = -10.6

Hope I helped!

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what is the degree of the polynomial ? what is the leading coefficient ? describe the end behavior of each function. f(x)=x^4-2x

Verizon [17]

Answer:

4th degree polynomial with leading coefficient of 1.

As x goes to negative or positive infinity, y goes to positive infinity in both cases.

Step-by-step explanation:

The degree of a polynomial is the highest exponent on the variable. Here it is 4.

The leading coefficient is the coefficient on the the term with the highest degree, Here there is none so it is 1.

The end behavior is how x and y behave at negative and positive infinity. When graphed, this equation has a W shape. This means at each end y goes to positive infinity.

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3 years ago

If f(x) is an even function and (6, 8) is one the points on the graph of f(x), which reason explains why (–6, 8) must also be a

Vikki [24]

If $f(x)$ is an even function, then $f(x)=f(-x)$ . In this case, $f(6)=8=f(-6)$ .

This is because an even function is symmetric about the $y-axis$ . Hence the outputs of a negative x-value and a positive x-value are the same.

The correct option is the first option.

Since the graph is even, the outputs of a negative x-value and a positive x-value are the same.

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3 years ago

Find the midpoint of the segment between (6,7) and (6,-5)

Norma-Jean [14]

Answer:

(6,1)

Step-by-step explanation:

midpoint = ((x2+x1),(y2+y1)/2)

= (6+6)/2, (-5+7)/2

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= 6,1

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3 years ago

Read 2 more answers

Maggie graphed the image of a 90 counterclockwise rotation about vertex A of . Coordinates B and C of are (2, 6) and (4, 3) and

Lemur [1.5K]

Answer:

A(2,2)

Step-by-step explanation:

Let the vertex A has coordinates $(x_A,y_A)$

Vectors AB and AB' are perpendicular, then

$\overrightarrow {AB}=(2-x_A,6-y_A)\\ \\\overrightarrow {AB'}=(-2-x_A,2-y_A)\\ \\\overrightarrow {AB}\perp\overrightarrow {AB'}\Rightarrow \overrightarrow {AB}\cdot \overrightarrow {AB'}=0\Rightarrow (2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0$

Vectors AC and AC' are perpendicular, then

$\overrightarrow {AC}=(4-x_A,3-y_A)\\ \\\overrightarrow {AC'}=(1-x_A,4-y_A)\\ \\\overrightarrow {AC}\perp\overrightarrow {AC'}\Rightarrow \overrightarrow {AC}\cdot \overrightarrow {AC'}=0\Rightarrow (4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0$

Now, solve the system of two equations:

$\left\{\begin{array}{l}(2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0\\ \\(4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0\end{array}\right.\\ \\\left\{\begin{array}{l}-4-2x_A+2x_A+x_A^2+12-6y_A-2y_A+y^2_A=0\\ \\4-4x_A-x_A+x_A^2+12-3y_A-4y_A+y_A^2=0\end{array}\right.\\ \\\left\{\begin{array}{l}x_A^2+y_A^2-8y_A+8=0\\ \\x_A^2+y_A^2-5x_A-7y_A+16=0\end{array}\right.$

Subtract these two equations:

$5x_A-y_A-8=0\Rightarrow y_A=5x_A-8$

Substitute it into the first equation:

$x_A^2+(5x_A-8)^2-8(5x_A-8)+8=0\\ \\x_A^2+25x_A^2-80x_A+64-40x_A+64+8=0\\ \\26x_A^2-120x_A+136=0\\ \\13x_A^2-60x_A+68=0\\ \\D=(-60)^2-4\cdot 13\cdot 68=3600-3536=64\\ \\x_{A_{1,2}}=\dfrac{60\pm8}{2\cdot 13}=\dfrac{34}{13},2$