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2 years ago

Solve the following inequality.

Mathematics

2 answers:

Zanzabum2 years ago

7 0

Answer:

Diagnostic PreTest

Which graph shows the correct solution?

A number line going from 27 to 37. An open circle is at 32. Everything to the right of the circle is shaded.

Step-by-step explanation:

san4es73 [151]2 years ago

5 0

Answer:

graph 1, i took the test

Step-by-step explanation:

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Help please thanks :)

Gre4nikov [31]

Answer:

the third one down, there is no gap in the data set

Step-by-step explanation:

ill be here all day

8 0

2 years ago

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:

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

Then

$y_{A_{1,2}}=5\cdot \dfrac{34}{13}-8 \text{ or } 5\cdot 2-8\\ \\=\dfrac{66}{13}\text{ or } 2$

Rotation by 90° counterclockwise about A(2,2) gives image points B' and C' (see attached diagram)

8 0

2 years ago

Someone please tell me what this is, the question is what is the slope of this line ?

AnnyKZ [126]

Answer:2/3

Step-by-step explanation:

rise/run. y=2 x=3

7 0

2 years ago

Read 2 more answers

If there is a parallelogram with one pair of opposite sides that are congruent and another pair that is parallel, how do I prove

dybincka [34]

Simply state that it is a parallelogram according to the parallelogram theorm. If you need the segments here's an example (A ≈ B and C ≈ D) thats it!

3 0

2 years ago

For which product or quotient is this expression the simplest form? (image attached)! please help :((

mars1129 [50]

Answer:

Step-by-step explanation:

For simplify the work we can start to factorise all the possibles expressions:

2x + 8.

8 is multiple of 2, so it can became

2(x+4)

x^2 - 16 this is a difference of two squares, so it can be rewritten as:

(x+4)(x-4)

x^2 + 8x + 16

we have to find two numbers whose sum is 8 and whose product is 16

the two number are 4 and 4

it becames:

(x+4)(x+4)

x+ 4 can‘t be simplified

if we look at the expression, we can find that x-4 appears at the numerator so

x^2 - 16 must be at numerator

but the second factor (x+4) doesn’t appear, so has been simplified. This situation can be possible only in the D option

in fact

(x+4)(x-4)/2(x+4) * (x+4)/(x+4)(x+4)

it became

(x+4)(x-4)/2 * 1/(x+4)(x+4)

(x-4)/2(x+4)

4 0

2 years ago