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3 years ago

CAN SOMEONE PLEASE SHOW WORK FOR THESE PROBLEMS WILL GIVE BAINLIEST!!!!!!!!!EMERGENCY

Mathematics

1 answer:

azamat3 years ago

5 0

Answer:

Part 1) $x\geq10$

Part 2) $m\leq -9$

Part 3) $p\geq 5$

Part 4) $x$

Part 5) $b$

Part 6) $n$

Part 7) $n$

Part 8) $r\leq 4$

Part 9) $x\geq 7$

Part 10) $p\leq 0$

Part 11) $x$

Part 12) $a > 24$

Step-by-step explanation:

Part 1) $2x+4\geq24$

Subtract 4 both sides

$2x\geq24-4$

$2x\geq20$

Divide by 2 both sides

$x\geq10$

the solution is the interval ------> [10,∞)

The solution is the shaded area to the right of the solid line at number 10 (closed circle).

see the attached figure

Part 2) $\frac{m}{3}-3\leq -6$

Adds 3 both sides

$\frac{m}{3}\leq -6+3$

$\frac{m}{3}\leq -3$

Multiply by 3 both sides

$m\leq -9$

the solution is the interval ------> (-∞,-9]

The solution is the shaded area to the left of the solid line at number -9 (closed circle).

see the attached figure

Part 3) $-3(p+1)\leq -18$

applying the distributive property left side

$-3p-3\leq -18$

adds 3 both sides

$-3p\leq -18+3$

$-3p\leq -15$

Multiply by -1 both sides

$3p\geq 15$

Divide by 3 both sides

$p\geq 5$

the solution is the interval ------> [5,∞)

The solution is the shaded area to the right of the solid line at number 5 (closed circle).

see the attached figure

Part 4) $-4(-4+x)>56$

applying the distributive property left side

$16-4x>56$

Subtract 16 both sides

$-4x>56-16$

$-4x>40$

Multiply by -1 both sides

$4x$

Divide by 4 both sides

$x$

the solution is the interval ------> (-∞,-10)

The solution is the shaded area to the left of the dashed line at number -10 (open circle).

see the attached figure

Part 5) $-b-2>8$

adds 2 both sides

$-b>8+2$

$-b>10$

Multiply by -1 both sides

$b$

the solution is the interval ------> (-∞,-10)

The solution is the shaded area to the left of the dashed line at number -10 (open circle).

Part 6) $-4(3+n)>-32$

applying the distributive property left side

$-12-4n>-32$

adds 12 both sides

$-4n>-32+12$

$-4n>-20$

multiply by -1 both sides

$4n$

divide by 4 both sides

$n$

the solution is the interval ------> (-∞,5)

The solution is the shaded area to the left of the dashed line at number 5 (open circle).

see the attached figure

Part 7) $4+\frac{n}{3}$

Subtract 4 both sides

$\frac{n}{3}$

Multiply by 3 both sides

$n$

the solution is the interval ------> (-∞,6)

The solution is the shaded area to the left of the dashed line at number 6 (open circle).

see the attached figure

Part 8) $-3(r-4)\geq 0$

applying the distributive property left side

$-3r+12\geq 0$

subtract 12 both sides

$-3r\geq -12$

divide by -1 both sides

$3r\leq 12$

divide by 3 both sides

$r\leq 4$

the solution is the interval ------> (-∞,4]

The solution is the shaded area to the left of the solid line at number 4 (closed circle).

see the attached figure

Part 9) $-7x-7\leq -56$

Adds 7 both sides

$-7x\leq -56+7$

$-7x\leq -49$

Multiply by -1 both sides

$7x\geq 49$

Divide by 7 both sides

$x\geq 7$

the solution is the interval ------> [7,∞)

The solution is the shaded area to the right of the solid line at number 7 (closed circle).

see the attached figure

Part 10) $-3(p-7)\geq 21$

applying the distributive property left side

$-3p+21\geq 21$

subtract 21 both sides

$-3p\geq 21-21$

$-3p\geq 0$

Multiply by -1 both sides

$3p\leq 0$

$p\leq 0$

the solution is the interval ------> (-∞,0]

The solution is the shaded area to the left of the solid line at number 0 (closed circle).

see the attached figure

Part 11) $-11x-4> -15$

Adds 4 both sides

$-11x> -15+4$

$-11x> -11$

Multiply by -1 both sides

$11x$

Divide by 11 both sides

$x$

the solution is the interval ------> (-∞,1)

The solution is the shaded area to the left of the dashed line at number 1 (open circle).

see the attached figure

Part 12) $\frac{-9+a}{15}>1$

Multiply by 15 both sides

$-9+a > 15$

Adds 9 both sides

$a > 15+9$

$a > 24$

the solution is the interval ------> (24,∞)

The solution is the shaded area to the right of the dashed line at number 24 (open circle).

see the attached figure

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ANSWER: 1.211 or 13.211

Step-by-step explanation:

STEP

1.1 Evaluate : (x-6)2 = x2-12x+36

Trying to factor by splitting the middle term

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The first term is, x2 its coefficient is 1 .

The middle term is, -12x its coefficient is -12 .

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Conclusion : Trinomial can not be factored

Equation at the end of step

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STEP

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2.1 Find the Vertex of y = x2-12x-16

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Each parabola has a vertical line of symmetry that passes through its vertex. Because of this symmetry, the line of symmetry would, for example, pass through the midpoint of the two x -intercepts (roots or solutions) of the parabola. That is, if the parabola has indeed two real solutions.

Parabolas can model many real life situations, such as the height above ground, of an object thrown upward, after some period of time. The vertex of the parabola can provide us with information, such as the maximum height that object, thrown upwards, can reach. For this reason we want to be able to find the coordinates of the vertex.

For any parabola,Ax2+Bx+C,the x -coordinate of the vertex is given by -B/(2A) . In our case the x coordinate is 6.0000

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2.2 Solving x2-12x-16 = 0 by Completing The Square .

Add 16 to both side of the equation :

x2-12x = 16

Now the clever bit: Take the coefficient of x , which is 12 , divide by two, giving 6 , and finally square it giving 36

Add 36 to both sides of the equation :

On the right hand side we have :

16 + 36 or, (16/1)+(36/1)

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Things which are equal to the same thing are also equal to one another. Since

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x2-12x+36 = (x-6)2

then, according to the law of transitivity,

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The Square Root Principle says that When two things are equal, their square roots are equal.

Note that the square root of

(x-6)2 is

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(x-6)1 =

x-6

Now, applying the Square Root Principle to Eq. #2.2.1 we get:

x-6 = √ 52

Add 6 to both sides to obtain:

x = 6 + √ 52

Since a square root has two values, one positive and the other negative

x2 - 12x - 16 = 0

has two solutions:

x = 6 + √ 52

x = 6 - √ 52

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According to the Quadratic Formula, x , the solution for Ax2+Bx+C = 0 , where A, B and C are numbers, often called coefficients, is given by :

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