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3 years ago

Select the correct difference. -3z 5 - (-7z 5) -10z5 -4z5 4z5 4z

Mathematics

2 answers:

bezimeni [28]3 years ago

8 0

-10z5 is the answer.

Alinara [238K]3 years ago

3 0

<span>-10z5 is the answer. </span>

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What is the graph of y = 2f(x)

Licemer1 [7]

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2 years ago

What is the positive root of the equation x 2 + 5x = 150? a0

Radda [10]

Solution for x^2+5x=150 equation:
<span>Simplifying x2 + 5x = 150 Reorder the terms: 5x + x2 = 150 Solving 5x + x2 = 150 Solving for variable 'x'. Reorder the terms: -150 + 5x + x2 = 150 + -150 Combine like terms: 150 + -150 = 0 -150 + 5x + x2 = 0 Factor a trinomial. (-15 + -1x)(10 + -1x) = 0 Subproblem 1Set the factor '(-15 + -1x)' equal to zero and attempt to solve: Simplifying -15 + -1x = 0 Solving -15 + -1x = 0 Move all terms containing x to the left, all other terms to the right. Add '15' to each side of the equation. -15 + 15 + -1x = 0 + 15 Combine like terms: -15 + 15 = 0 0 + -1x = 0 + 15 -1x = 0 + 15 Combine like terms: 0 + 15 = 15 -1x = 15 Divide each side by '-1'. x = -15 Simplifying x = -15 Subproblem 2Set the factor '(10 + -1x)' equal to zero and attempt to solve: Simplifying 10 + -1x = 0 Solving 10 + -1x = 0 Move all terms containing x to the left, all other terms to the right. Add '-10' to each side of the equation. 10 + -10 + -1x = 0 + -10 Combine like terms: 10 + -10 = 0 0 + -1x = 0 + -10 -1x = 0 + -10 Combine like terms: 0 + -10 = -10 -1x = -10 Divide each side by '-1'. x = 10 Simplifying x = 10Solutionx = {-15, 10}</span>

5 0

2 years ago

I’m stuck can some one help me

abruzzese [7]

Answer:

3) Midpoint is (-4,0.5)

Option A is correct.

4) Midpoint is (2.5,0)

Option B is correct.

5) The factors are (x+4)(x-7)

Option C is correct.

6) The factors are (x+4)(x+2)

Option A is correct.

Step-by-step explanation:

Question 3

Find midpoint of the following:

(2,-7), (-10,8)

The formula used to find midpoint is: $Midpoint=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2} )$

We have $x_1=2, y_1=-7, x_2=-10,y_2=8$

Putting values and finding midpoint

$Midpoint=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2} )\\Midpoint=(\frac{2-10}{2},\frac{-7+8}{2} )\\Midpoint=(\frac{-8}{2},\frac{1}{2} )\\Midpoint=(-4,0.5 )$

So, Midpoint is (-4,0.5)

Option A is correct.

Question 4

Find midpoint of the following:

(2,-10), (3,10)

The formula used to find midpoint is: $Midpoint=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2} )$

We have $x_1=2, y_1=-10, x_2=3,y_2=10$

Putting values and finding midpoint

$Midpoint=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2} )\\Midpoint=(\frac{2+3}{2},\frac{-10+10}{2} )\\Midpoint=(\frac{5}{2},\frac{0}{2} )\\Midpoint=(2.5,0 )$

So, Midpoint is (2.5,0)

Option B is correct.

Question 5

Factor each completely

$x^2-3x-28$

We will break the middle term and find factors

$x^2-3x-28\\=x^2-7x+4x-28\\Taking\:common\\=x(x-7)+4(x-7)\\=(x+4)(x-7)$

So, the factors are (x+4)(x-7)

Option C is correct.

Question 6

Factor each completely

$x^2+6x+8$

We will break the middle term and find factors

$x^2+6x+8\\=x^2+4x+2x+8\\=x(x+4)+2(x+4)\\=(x+4)(x+2)$

So, the factors are (x+4)(x+2)

Option A is correct.

7 0

2 years ago

What is the distace tweenn (4,-7) and (-5,-7)

love history [14]

The distance between them is 9.

5 0

3 years ago

Read 2 more answers

the volume v of a right circular cylinder of radius r and heigh h is V = pi r^2 h 1. how is dV/dt related to dr/dt if h is const

laiz [17]

In general, the volume

$V=\pi r^2h$

has total derivative

$\dfrac{\mathrm dV}{\mathrm dt}=\pi\left(2rh\dfrac{\mathrm dr}{\mathrm dt}+r^2\dfrac{\mathrm dh}{\mathrm dt}\right)$

If the cylinder's height is kept constant, then $\dfrac{\mathrm dh}{\mathrm dt}=0$ and we have

$\dfrac{\mathrm dV}{\mathrm dt}=2\pi rh\dfrac{\mathrm dt}{\mathrm dt}$

which is to say, $\dfrac{\mathrm dV}{\mathrm dt}$ and $\dfrac{\mathrm dr}{\mathrm dt}$ are directly proportional by a factor equivalent to the lateral surface area of the cylinder ( $2\pi r h$ ).

Meanwhile, if the cylinder's radius is kept fixed, then

$\dfrac{\mathrm dV}{\mathrm dt}=\pi r^2\dfrac{\mathrm dh}{\mathrm dt}$

since $\dfrac{\mathrm dr}{\mathrm dt}=0$ . In other words, $\dfrac{\mathrm dV}{\mathrm dt}$ and $\dfrac{\mathrm dh}{\mathrm dt}$ are directly proportional by a factor of the surface area of the cylinder's circular face ( $\pi r^2$ ).

Finally, the general case ( $r$ and $h$ not constant), you can see from the total derivative that $\dfrac{\mathrm dV}{\mathrm dt}$ is affected by both $\dfrac{\mathrm dh}{\mathrm dt}$ and $\dfrac{\mathrm dr}{\mathrm dt}$ in combination.

8 0

2 years ago