I need word form and decimal

(I already shaded it in so that is all I need​

Answer:

  0 and 1

Step-by-step explanation:

The scale of the graph is not marked, so we assume each grid line represents 1 unit. The leftmost crossing of the x-axis is between 0 and 1.

_____

Additional comment

The other real zero is between 2 and 3. If this is a polynomial function, it will have additional complex zeros.

Answer:

The graph moved down 5 units

Step-by-step explanation:

Answer:

Step-by-step explanation:

The best way to get an answer to this question is to study the graph you get when you study the result from Desmos, which is a free  graphing program.

Notice the way the graph is laid out. If the graph was simply y = x^2 - 6x  then the y value for the minimum value would be x = 3 (just like the 2 values you do get) and y = -9

But y = x^2 - 6x + 3 has you moving up 3 units to y = - 6 and y = x^2 - 6x + 8 moves you to y = - 1. Which way are you going? The difference is 5 units and you are moving down because you start  at y = x^2 - 6x + 8 and move down to x^2 - 6x + 3.

Without finding minimums (which is a cumbersome process which needs to be done twice) there's no easy way to do this except graphing.

The answer is B.

Equivalent expressions are simply expressions that have the same value, irrespective of their form.

The equivalent expression of [tex]\mathbf{6x(3x + 11)}[/tex] is [tex]\mathbf{18x^2 + 66x}[/tex]

The expression is given as:

[tex]\mathbf{6x(3x + 11)}[/tex]

Start by opening the brackets; i.e. apply distributive property

[tex]\mathbf{6x(3x + 11) = 6x \times 3x + 6x \tiimes 11}[/tex]

Evaluate all products

[tex]\mathbf{6x(3x + 11) = 18x^2 + 66x}[/tex]

The expression cannot be further simplified.

Hence, the equivalent of [tex]\mathbf{6x(3x + 11)}[/tex] is [tex]\mathbf{18x^2 + 66x}[/tex]

Read more about equivalent expressions at:

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Answer: The both involve writing a rate

I hope this helps i would appreciate brainliest if possible :)

Answer:

I and II

Step-by-step explanation:

All trapezoids have a midsegment. The base angles of isosceles trapezoids are always congruent. Because the top and bottom side are parallel, the same side interior angles are supplementary not complementary.

Answer:

D. 604.8

Step-by-step explanation:

Answer: these are correct

1,750

14,000,000

1,120,000

604.8

Answer:

7

Step-by-step explanation:

Working backwards:

19 + 9 = 28

28/ 4 = 7

Checking work:

7 x 4 = 28

28 - 9 = 19

Answer:

Well I’m not sure what Answers do you want

Step-by-step explanation:

Answer:

[tex]3\dfrac{1}{4} \ \ or \ \ 3,25[/tex]

Step-by-step explanation:

[tex]\displaystyle12 \frac{3}{8} -9\frac{1}{8} =12-9+\frac{3}{8} -\frac{1}{8} =3+\frac{2}{8} =3\frac{1}{4} \ \ or \ \ 3,25[/tex]

Answer:

Step-by-step explanation:

here you go m = (y2-y1)/(x2-x1)

  = (11-(-9))/(3-(-1))

  = 20/4

  = 5

Answer:

d .  37/12 unit^2.

Step-by-step explanation:

d) First find the points of intersection of the the 2 graphs, by solving them simultaneously:

x^3 = x^2 + 2x

x^3 - x^2 - 2x) = 0

x(x^2 - x - 2) = 0

x(x - 2)(x + 1) = 0

So they intersect at x = -1, x = 0 and x = 2.

Take the area from x = 0 to x = 2)

           2                        2

Area =  ∫ x^2 + 2x dx -   ∫  x^3 dx

            0                       0

   2

=  [ x^3/3 + x^2 ) - x^4/4]

   0

=   ((8/3 + 4) - 0) - (16/4 - 0) -

=   8/3 unit^2

Now calculate the area  from x = -1 to x = 0.

   0

=    (x^3/3 + x^2 ) - x^4/4 )

   -1

=   (( 0 +1/3 - 1) ) - (0 - 1/4)

=  5/12 unit^2

So the total area of the region =  8/3 + 5/12 = 37/12 unit^2.

Answer:

Parallel lines: Line B and C

Perpendicular lines: Line A and B; Line A and C

Step-by-step explanation:

To answer this question, we only need to find the slope of each equation. The formula for slope is attached in a picture.

LINE A

Substituting the first line into the formula, we will get...

(2 - 5)/(0 -(-4))

-3/4

LINE B

Substituting the second line's points into the formula we will get...

(2 - 6)/(3 - 6)

-4/-3

4/3

LINE C

Substituting the third line's points into the formula we will get...

(3 -(-1)/(0 -(-3))

4/3

So, parallel lines have the same slope, right? Since line b and line c have the same slope, they are parallel.

And, perpendicular lines have negative reciprocal slopes, right? (reciprocal means x/y -> y/x) Since -3/4 and 4/3 are negative reciprocals of each other, they are perpendicular.

Therefore...

Parallel lines: Line B and C

Perpendicular lines: Line A and B; Line A and C

Answer:

5

Step-by-step explanation:

set up point slope form

y + 8 = 2(x-1)

set y to zero

0+8=2(x-1)

8=2x-2

10= 2x

x = 5

Answer:

I think the answer would be A

Answer:

5. is III. Since there is a vertical asymptote at x= -3 and hole at x=0

Answer:

8

Step-by-step explanation:

11-3=8

11-3=8 must subtract for anser

Answer:

The final velocity, V, of an object under constant acceleration can be found using the formula V2=v2+2as , where v is the initial velocity in meters per second, a is acceleration in meters per second, and s is the distance in meters.

[tex]y = 2x -2 ~~~~.....(i) \\\\y = x^2 -4 x +3 ~~~~ .....(ii) \\\\\\x^2-4x +3 = 2x -2\\\\\implies x^2 -4x -2x +3 +2 =0\\\\\implies x^2 -6x +5 =0\\\\\implies x^2 -5x -x +5 =0\\\\\implies x(x-5) -(x-5) =0\\\\\implies(x-5)(x-1) =0\\\\\implies x = 5 ~~ \text{or}~~ x = 1 \\\\\\\text{Substitute x = 1 in equation (i):}\\\\y = 2 -2 =0\\\\ \text{Substitute x = 5 in equation (i):}\\\\\\y = 2(5) -2 = 10 -2 =8 \\\\\text{Hence,}~~ (x,y) = (1,0) ~ \text{and} ~ (x,y) = (5,8)[/tex]

Answer:

that have no answer so sorry

Answer:

[tex]11[/tex]

Step-by-step explanation:

[tex]\sf{Use\: the\: distance\: formula\: to\: determine\: the\: distance\: between\: two\: points.[/tex]

[tex]\sf{Distance=\sqrt{(x_2 - x_1)^2 + (y_2-y_1)^2}[/tex]

[tex]\mathrm{Plug\:the\:points}[/tex]

[tex]\sqrt{\left(4-\left(-7\right)\right)^2+\left(5-5\right)^2}[/tex]

[tex](4+7)^2=11^2\\(5-5)^2=0[/tex]

[tex]\sqrt{11^2+0}[/tex]

[tex]11^2+0=11^2\\\sqrt{11^2}[/tex]

[tex]\mathrm{Apply\:radical\:rule:\:}\sqrt[n]{a^n}=a,\:\quad \mathrm{\:assuming\:}a\:\mathrm{is\ greater\ than \:or\ equal\: to \:0}[/tex]

[tex]=11[/tex]

Answer:

161oz = 10.0625 (then you round)

Step-by-step explanation:

Answer: Y=7x+19

Step-by-step explanation:

got no time for explain hope this helper

9514 1404 393

Answer:

  y -5 = 7(x +2)

Step-by-step explanation:

Values for the slope and a point are given. That makes it convenient to use the point-slope form of the equation for a line:

  y -k = m(x -h) . . . . . . line with slope m through point (h, k)

  y -5 = 7(x +2) . . . . . . line with slope 7 through point (-2, 5)

__

Sometimes, you want the slope-intercept form of the equation. That can be found by solving for y

  y -5 = 7x +14 . . . . . . . . eliminate parentheses

  y = 7x +19 . . . . . . . . . . add 5 to both sides

__

In standard form, the same line has equation ...

  7x -y = -19

Answer:

75 miles

Step-by-step explanation:

We know that she has driven 120 miles before the first stop.

Then she drives for 1/4 of the the distance she has already driven (120 miles). One fourth of 120 is 30.

So now, she's driven for 120 + 30 miles, or 150 miles.

She drives for another 2/3 of the distance she's traveled (150 miles). Two thirds of 150 is 100.

So now, she's driven for 150 + 100 miles, or 250 miles.

What we need to find out now is how much of the trip has she traveled.

We know that the entire distance she has driven is 55 miles more than 1 ⅔ of the remaining trip. So, in other words, the distance she has traveled minus 55 is equal to 1 2/3 of the trip.

To put this in an equation:

250 - 55 = 1 2/3t

t = the whole trip

Now, we solve for t.

250 - 55 = 195

1 2/3 = 5/3

so, 195 = 5/3t

t = 325

We subtract what she has already driven to the total to get the remaining distance.

325 - 250 = 75

[tex]2x^2 -5x-8 =0\\\\\implies x = \dfrac{ -(-5) \pm \sqrt{(-5)^2 - 4\cdot 2 \cdot (-8)}}{2 \cdot 2}\\\\\\\implies x = \dfrac{5 \pm \sqrt{89}}{4}\\\\\\\text{Hence}~ x = \dfrac{5 + \sqrt{89}}{4},~~ x = \dfrac{5 - \sqrt{89}}{4}[/tex]

Answer:

[tex]\sf \boxed{\sf x=\frac{5+\sqrt{89}}{4}}[/tex]

[tex]\sf \boxed{\sf x=\frac{5-\sqrt{89}}{4}}[/tex]

[tex]\sf 2x^2-5x-8=0[/tex]

☁ We'll solve this equation using The Quadratic Formula:

[tex]\boxed{\sf \cfrac{-b\pm \sqrt{b^2-4ac}}{2a}}[/tex]

A= 2, b= -5, c= -8

[tex]\sf x=\cfrac{-\left(-5\right)\pm \sqrt{\left(-5\right)^2-4\times \:2\left(-8\right)}}{2\times \:2}[/tex]

[tex]\mapsto \sf \sqrt{\left(-5\right)^2-4\times \:2\left(-8\right)}[/tex]

[tex]**\sf \left(-5\right)^2=5^2[/tex]

[tex]\mapsto \sf \sqrt{5^2+4\times \:2\times \:8}[/tex]

[tex]\mapsto \sf \sqrt{5^2+64}[/tex]

[tex]\mapsto \sf \sqrt{89}[/tex]

______________

[tex]\sf x=\cfrac{-\left(-5\right)\pm \sqrt{89}}{2\times \:2}[/tex]

☁ Now, we'll Separate solutions, first, we'll solve the equation when ± is plus:

[tex]\mapsto \sf \cfrac{-\left(-5\right)+\sqrt{89}}{2\times \:2}[/tex]

☁ Multiply 2*2 =4

[tex]\boxed{\sf \cfrac{5+\sqrt{89}}{4}}[/tex]

_________________

☁ Now solve the equation when ± is minus:

[tex]\mapsto \sf \cfrac{-\left(-5\right)-\sqrt{89}}{2\times \:2}[/tex]

☁  Multiply 2*2 =4

[tex]\boxed{\sf \cfrac{5-\sqrt{89}}{4}}[/tex]

[tex]\sf Solution \:1:\boxed{\sf x=\frac{5+\sqrt{89}}{4}}[/tex]

[tex]\sf Solution \:2:\boxed{\sf x=\frac{5-\sqrt{89}}{4}}[/tex]

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Answer:

Y=5 y=-3

Step-by-step explanation:

Two real solutions:

y =(2+√64)/2=1+4= 5.000

or:

y =(2-√64)/2=1-4= -3.000

Answer:

y = 9/10x - 7/5

Step-by-step explanation:

Slope = 9/10

Point (6,4)

y-intercept = 4 - (9/10)(6) = 4 - 54/10

= - 7/5

Answer:

x=6

Step-by-step explanation:

Answer:

$11

Step-by-step explanation:

This is because If you subtract 19 from 30 your answer will be 19.