The cost of a ticket t will be no more than $26

Can you please help me with these problems, Thank you

Kaylis [27]

A) A mark dwon greater than 99% and less than 100%

Mark down 99.1% of x

Original price: x

Final price: original price - mark down = x - 0.991 x = 0.009x

=> If orginal price is 1000 and mark down is 99.1% , final price = 0.009*1000 = 9

You can check that the mark down is 99.1% of 1000 = 991.

b) A mark down that is less than 1%

Mark down: 0.1%

Original price: 10,000

Mark down: 10,000 * 0.1% = 10

Final price: 10,000 - 10 = 9,990

c) A mark up that is more than 200%

Mark up: 300%

Original price: 100

=> final price = 100 + 100*300% = 100 + 300 = 400

7 0

2 years ago

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Find the maximum value of

noname [10]

Answer:

(3, 1)

Step-by-step explanation:

Given the following constraints:

2x + 4y ≤ 10

x + 9y ≤ 12

x ≥ 0

y ≥ 0

We have to first graph the above constraints and secondly, we find the boundary points for the feasible region. This is done by plotting the graphs of 2x + 4y ≤ 10 and x + 9y ≤ 12.

From the graph, the following points satisfy the constraints:

(0,0) , (3,1) , (0,4/3) , (5,0)

Given that:

P = x + 6y

For (0, 0): P = x + 6y = 0 + 6(0) = 0

For (3, 1): P = x + 6y = 3 + 6(1) = 9

For (0, 4/3) P = x + 6y = 0 + 6(4/3) = 8

For (5, 0) P = x + 6y = 5 + 6(0) = 5

6 0

2 years ago

Kuri spent a total of $68 on trading cards. His collection contains two types of cards. A pack of trading card A cost $4, and a

devlian [24]

4a+7b=68 go to mathpapa

8 0

3 years ago

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the frame of a picture is 28 cm by 32 cm outside and is of uniform width.What is the width of the frame if 192 cm2 of the pictur

natita [175]

Answer:

The width of the frame is equal to $8\ cm$

Step-by-step explanation:

Let

x-------> the width of the frame picture

we know that

$(28-2x)(32-2x)=192$

$896-56x-64x+4x^{2} =192$

$4x^{2}-120x+704=0$

Simplify

$x^{2}-30x+176=0$

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

$x^{2}-30x+176=0$

so

$a=1\\b=-30\\c=176$

substitute in the formula

$x=\frac{30(+/-)\sqrt{(-30)^{2}-4(1)(176)}} {2(1)}$

$x=\frac{30(+/-)\sqrt{196}} {2}$

$x=\frac{30(+/-)14} {2}$

$x=\frac{30+14} {2}=22\ cm$ ----> this solution is not logic

$x=\frac{30-14} {2}=8\ cm$

The width of the frame is equal to $8\ cm$

8 0

3 years ago

(I need the answer right now please) ΔA'B'C' was constructed using ΔABC and line segment EH. 2 triangles are shown. Line E H is

irina1246 [14]

Answer:

The true statements are

1. BD = DB'

3. m∠EFA = 90°

4. The line of reflection, EH, is the perpendicular bisector of BB', AA', and

CC'

Step-by-step explanation:

* Lets explain how to solve the problem

- Reflection is flipping an object over the line of reflection.

- The object and its image have the same shape and size, but the

figures are in opposite directions from the line of reflection

- The objects appear as if they are mirror reflections, with right and left

reversed

- The line of reflection is a perpendicular bisector for all lines joining

points on the figure with their corresponding images

- Look to the attached figure for more understand

* Lets solve the problem

- ΔA'B'C' was constructed using ΔABC and line segment EH, where

EH is the reflection line

- D is the mid-point of BB'

- F is the mid-point of AA'

- G is the mid-point of CC'

* Lets find from the answer the true statements

1. BD = DB'

∵ D is the mid point of BB'

- Point D divides BB' into two equal parts

∴ BD = DB' ⇒ True

2. DF = FG

- It depends on the size of the sides and angles of the triangle

∵ We can't prove that

∴ DF = FG ⇒ Not true

3. m∠EFA = 90°

∵ The line of reflection ⊥ the lines joining the points with their

corresponding images

∴ EH ⊥ AA' and bisect it at F

∴ m∠EFA = 90° ⇒ True

4. The line of reflection, EH, is the perpendicular bisector of BB',

AA', and CC'

- Yes the line of reflection is perpendicular bisectors of them

∴ The line of reflection, EH, is the perpendicular bisector of BB',

AA', and CC' ⇒ True

5. ΔABC is not congruent to ΔA'B'C'

∵ In reflection the object and its image have the same shape and size

∴ Δ ABC is congruent to Δ A'B'C'

∴ ΔABC is not congruent to ΔA'B'C' ⇒ Not true

4 0

3 years ago

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