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2 years ago

Which inequality has an open circle when it is graphed on a number line?

Mathematics

1 answer:

strojnjashka [21]2 years ago

5 0

Answer:

x > 3/5

Step-by-step explanation:

x > 3/5 has an open circle because 3/5 is not included

if it was x ≥ 3/5 then that would include 3/5 and the circle would be closed

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Alex learning algebra 2 quiz

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Subtract that number down so its a

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Jana planted 30 tulips. A total of 2/3 of those tulips have bloomed. How many of the tulips that Jana planted have bloomed?

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20 tulips. 2/3=20/30 when you multiply both the denominator and numerator by 10.

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Formula to find volume of a right triangular prism

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3 years ago

The function h(t) = −5t2 + 20t shown in the graph models the curvature of a satellite dish: graph of parabola starting at zero c

ipn [44]

Answer:

Part 1) Option B: 0 ≤ x ≤ 4

Part 2) Option C: (5x − 1)(2x − 3)

Part 3) Option C: The crop yield decreased by 10 pounds per acre from year 10 to year 20.

Part 4) Option B: f(x) = (2x + 1)(2x − 1)

Part 5) Option A: The company earned 85 billion dollars in 2008

Part 6) Option C: (3, 0) and (5, 0)

Step-by-step explanation:

Part 1) we have

$h(t)=-5t^{2}+20t$

This is a vertical parabola open downward

The vertex is a maximum

The x-intercepts are the points (0,0) and (4,0)

The vertex is the point (2,20) ---> The x-coordinate of the vertex is the midpoint of the x-intercepts

The domain is the interval ----> [0,4]

$0\leq t\leq 4$

The range is the interval ----> [0,20]

$0\leq h(t)\leq 20$

Part 2) we have

$V=10x^{2} -17x+13$

where

x is the time in minutes

V is the volume in gallons

we know that

When the trough is empty, the volume is equal to zero

For V=0

$10x^{2} -17x+3=0$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$10x^{2} -17x+3=0$

$a=10\\b=-17\\c=3$

substitute in the formula

$x=\frac{-(-17)(+/-)\sqrt{-17^{2}-4(10)(3)}} {2(10)}$

$x=\frac{17(+/-)\sqrt{169}} {20}$

$x=\frac{17(+/-)13} {20}$

$x=\frac{17(+)13} {20}=\frac{3}{2}$

$x=\frac{17(-)13} {20}=\frac{1}{5}$

$10x^{2} -17x+3=10(x-\frac{3}{2})(x-\frac{1}{5})$

simplify

$10(x-\frac{3}{2})(x-\frac{1}{5})=(2x-3)(5x-1)$

Part 3) we have

$f(x)=-x^{2}+20x+100$

we know that

To find the average rate of change, we divide the change in the output value by the change in the input value

the average rate of change is equal to

$\frac{f(b)-f(a)}{b-a}$

In this problem we have

$f(a)=f(10)=-(10)^{2}+20(10)+100=200$

$f(b)=f(20)=-(20)^{2}+20(20)+100=100$

$a=10$

$b=20$

Substitute

$\frac{100-200}{20-10}$

$-\frac{100}{10}=-10$ ----> is a decreasing function

therefore

The crop yield decreased by 10 pounds per acre from year 10 to year 20.

Part 4) we have

$f(x)=4x^{2} -1$

<em>Find out the x-intercepts</em>

Remember that the x-intercepts are the values of x when the value of the function is equal to zero

For f(x)=0

$4x^{2} -1=0$

Solve for x

$4x^{2}=1$

$x^{2}=\frac{1}{4}$

take square root both sides

$x=(+/-)\frac{1}{2}$

$f(x)=4x^{2} -1=4(x-\frac{1}{2})(x+\frac{1}{2})=(2x-1)(2x+1)$

Part 5) we have

$t^{2}+14t+85$

Remember that

The y-intercept of a function is the value of the function (output value) when the value of x (input value) is equal to zero

In this problem

The expression represent a company's net sales, in billions, from 2008 to 2018

t is the number of years since the end of 2008

For t=0 -----> represent the end of 2008

$(0)^{2}+14(0)+85=85\ billion\ dollars$

therefore

The company earned 85 billion dollars in 2008.

Part 6) What are the x-intercepts of the parabola?

we have the points (2,3),(4,-1),(6,3)

This is a vertical parabola open upward

The vertex represent a minimum

In this problem the vertex is the point (4,-1)

The equation of a vertical parabola in vertex form is equal to

$y=a(x-h)^2+k$

where

a is a coefficient

(h,k) is the vertex

we have

(h,k)=(4,-1)

substitute

$y=a(x-4)^2-1$

take the point (2,3) and substitute in the quadratic equation to solve for a

For x=2, y=3

$3=a(2-4)^2-1$

$3=4a-1$

$4a=4$

$a=1$

The quadratic equation is

$y=(x-4)^2-1$

Find out the x-intercepts

Remember that the x-intercepts are the values of x when the value of the function is equal to zero

For y=0

$(x-4)^2-1=0$

Solve for x

$(x-4)^2=1$

take square root both sides

$(x-4)=(+/-)1$

$x=4(+/-)1$

$x=4+1=5\\x=4-1=3$

therefore

The x-intercepts are the points (3,0) and (5,0)

7 0

3 years ago

Read 2 more answers

Graph the image of the given triangle, translated 4 units to the right and 5 units down.

Goryan [66]

Answer- Follow the attachment attached herewith.

Solution-

As we know if we are moving in left or right, we have to deal with the x - direction or x co-ordinate and if we are moving in up and down , we have to deal with y - direction or y co-ordinate .

If we are moving in right ,we have to add in x co-ordinate and if we are moving left we have subtract from x co-ordinate. Likewise, if we are moving up, we have to add in y co-ordinate and if we are moving down, then we have to subtract from y co-ordinate.

As given in the question, we have move the triangle 4 units right and 5 units down,

The co-ordinate of given triangles'

A = (-1,3)

B = (-5,-3)

C = (2,-1)

∴Then the new co-ordinate will be

A' = (-1+4 , 3-5) = (3,-2)

B' = (-5+4 , -3-5) = (-1,-8)

C' = (2+4 , -1-5) = (6,-6)

7 0

2 years ago