Huan runs a business selling sweets. She is designing a container to hold the sweets. She uses rectangular pieces of cardboard measuring 20 cm by 30 cm. The mass of 1 square metre of the cardboard is 210 grams. She cuts a square from each corner of a piece of cardboard. She then folds the cardboard to make an open box. By changing the size of the square she cuts out, Huan can change the volume of the box.
(a) Work out the mass of the box made when she cuts a square of side 5cm from each corner.
(b) Huan cuts a square of side x from each corner of the cardboard.
(i) Explain why the volume of the box she makes is given by x(20-2x)(30-2x).
(ii) Explain why 0<x<10
(c) Huan wants to make a box with volume of at least 900cm3. She wants the mass of the box to be as small as possible.
By drawing a suitable graph, work out the mass of the box she will make ​

Answer:

answer is 121/9

Step-by-step explanation:

13.4444444444 (decimal form)

121/9 (improper fraction form)

13 4/9 (mixed fraction form)

Answer:we smoking fredoooooooooo

[tex]8 \: cars = 120 \: dollars \\ 1 \: car = \frac{120}{8} = 15 \: dollars[/tex]

[tex]m(c) = 15c[/tex]

Answer:

d: m=15c

Step-by-step explanation:

Usually (not always), variable y depends on variable x.

Since in this question, the amount of money Paul can earn changes based on the number of cars he washes, y is the amount of money, and x is the number of cars he washes.

This can be represented as: 120 = 8 times ___

___ is the amount of money Paul earns by cleaning 1 car.

___ is 120 divided by 8, which is 15.

--> Paul earns $15 by cleaning each car.

The new equation will be m = 15c

Answer:

(D). 6.

Step-by-step explanation:

The number of parts = 1 + 10 = 11.

So 10/11 of the alloy is copper and 1/11 is aluminium.

So the answer is 1/11 * 66

= 66/11= 6 pound,s

The bottom edge of the window is 8 inches above the ground, computed using the Pythagoras Theorem.

According to the Pythagoras Theorem, in a right triangle, the square of the hypotenuse, that is, the side opposite to the right angle is equal to the sum of the square of the legs, that is, the two other sides.

If in a right triangle, c is the hypotenuse, and a and b are the two legs, then by the Pythagoras Theorem, we can write:

a² + b² = c².

In the question, we are informed that a building meets the ground at a right angle. The top of a 10-foot ladder is placed against the bottom edge of a window in the building, and the base of the ladder is placed 6 feet from where the building meets the ground.

We are asked to draw a diagram that represents the situation and find the height of the bottom edge of the window from the ground.

We assume the building to be DB, meeting the ground BC at a right angle. At point C, the ladder meets the ground and meets the bottom edge of the window at point A.

The diagram is attached.

Now, we get a right triangle ABC, right-angled at B.

By Pythagoras' Theorem, we know that:

AB² + BC² = AC²,

or, AB² + 6² = 10²,

or, AB² = 10² - 6² sq. inches,

or, AB² = 100 - 36 sq. inches,

or, AB = √64 inches,

or, AB = 8 inches.

Thus, the bottom edge of the window is 8 inches above the ground, computed using the Pythagoras Theorem.

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The provided question is incomplete. The complete question is:

"A building meets the ground at a right angle. The top of a 10-foot ladder is placed against the bottom edge of a window in the building, and the base of the ladder is placed 6 feet from where the building meets the ground. Draw a diagram that represents this situation. How far up from the ground is the bottom edge of the window?"

Answer:

1/2

1/3

Step-by-step explanation:

x=1/2

x=-1/3

Answer:

[tex]x_{1} =-\frac{1}{3} \\x_{2} =\frac{1}{2}[/tex]

Step-by-step explanation:

[tex]6x^{2} -x-1=0[/tex]

Factoring the equation:

[tex](3x+1)(2x-1)=0[/tex]

Extract the roots:

First factor

[tex]3x+1=0\\3x=-1\\x_{1} =-\frac{1}{3}[/tex]

Second factor

[tex]2x-1=0\\2x=1\\x_{2} =\frac{1}{2}[/tex]

Hope this helps

The Polygon ABCD was reflected across the x axis, followed by a reflection across the line y = x to form polygon A'B'CD'.

Transformation is the movement of a point from its initial location to a new location. Types of transformation are reflection, translation, rotation and dilation.

The Polygon ABCD was reflected across the x axis, followed by a reflection across the line y = x to form polygon A'B'CD'.

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Answer:

1.5, -1.5

Step-by-step explanation:

Using the quadratic formula, you should have a plus, and minus solution. These should be your answers.

The solution of the differential equation dy/dx=[tex]2xy/(x^{2} -1)[/tex] is y=[tex]-x^{2} +1[/tex].

Given a differential equation dy/dx=[tex]2xy/(x^{2} -1)[/tex].

We are required to find the solution of the differential equation when y(0)=1.

dy/dx=[tex]2xy/(x^{2} -1)[/tex]

Taking y to left side and dx to right side.

1/y dy=2x/[tex](x^{2} -1)[/tex] dx

Integrating both sides.

[tex]\int\limits {1/y} \, dy[/tex]=[tex]\int\limits {2x/(x^{2} -1)} \, dx[/tex]----------1

log y=[tex]\int\limits {2x/(x^{2} -1)} \, dx[/tex]

Solving right side.

[tex]\int\limits {2x/(x^{2} -1)} \, dx[/tex]

let [tex]x^{2} -1[/tex]=z

differentiating both sides with respect to x.

2x=dz/dx

2x dx=dz

Put 2x dx=dz in 1.

[tex]\int\limits {1/y} \, dy[/tex]=[tex]\int\limits {1/z } \, dz[/tex]

log y=log z+log c

Put z=[tex]x^{2} -1[/tex]

log y=log([tex]x^{2} -1[/tex])+log c

log y=log [{[tex]x^{2} -1[/tex])c]

y=([tex]x^{2} -1[/tex])c------------2

Put x=0 and y=1

1=(0-1)c

1=-c

c=-1

Put c=-1 in 2 to get the solution.

y=-1([tex]x^{2} -1[/tex])

y=[tex]-x^{2} +1[/tex]

Hence the solution of the differential equation dy/dx=[tex]2xy/(x^{2} -1)[/tex] is

y=[tex]-x^{2} +1[/tex].

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The length of the road that connects the service station with Cray is 46.2 mi.

We need to apply the principle of similar triangles to obtain the length of the road that will connect the service station with Cray as shown in the image attached.

Hence;

The distance between Alba and Blare = 130 mi

The distance between Alba and Cray = 120 mi

Road connecting the service station with Cray =x

And;

120/130 = x/50

x =  46.2 mi

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Answer:

Θ ≈ 37° , x ≈ 13 cm

Step-by-step explanation:

(a)

using the cosine ratio in the right triangle

cosΘ = [tex]\frac{adjacent}{hypotenuse}[/tex] = [tex]\frac{BC}{AC}[/tex] = [tex]\frac{8}{10}[/tex] , then

Θ = [tex]cos^{-1}[/tex] ([tex]\frac{8}{10}[/tex] ) ≈ 37° ( to the nearest whole number )

(b)

using the sine ratio in the right triangle and the exact value

sin60° = [tex]\frac{\sqrt{3} }{2}[/tex] , then

sin60° = [tex]\frac{opposite}{hypotenuse}[/tex] = [tex]\frac{x}{15}[/tex] = [tex]\frac{\sqrt{3} }{2}[/tex] ( cross- multiply )

2x = 15[tex]\sqrt{3}[/tex] ( divide both sides by 2 )

x = 7.5[tex]\sqrt{3}[/tex] ≈ 13 cm ( to the nearest whole number )

3a) The coordinate of R after dilation is; R' = (-6, 21)

3b) The coordinate of T after dilation is; T' = (1, 3)

5) A dilation by a scale factor of 2 followed by a translation right 1 up 3.

3a) We are given the coordinates of the rectangle as;

Q(-2, -1), R(-2, 7), S(4, 7), T(4, -1)

Now, we are told that R(-2, 7) is dilated about the origin with a scale factor of 3. Thus;

R' = (-2 *3), (3 * 7)

R' = (-6, 21)

3b) We are now told that T(4, -1) is dilated by a scale factor of 1/2 with R as center of origin. R has a coordinate of (-2, 7). Thus;

Coordinate of T' = (-2, 7) + ¹/₂((4, -1) - (-2, 7))

Coordinate of T' = (-2, 7) + ¹/₂(6, -8)

Coordinate of T' = (-2, 7) + (3, -4)

Coordinate of T' =  (1, 3)

5) We can see that LN is 2 units while L'N' is 4 units and so we can say scale factor in the transformation of ΔLMN to ΔL'M'N' is 2.

Now, if we apply scale factor of 2 to coordinates of N before transformation, we will have; 2(3, 1) = (6, 2)

Comparing to N' which is (5, 5) we can say that it was shifted up by 3 units and right by 1 unit.

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The value of the given expression; f(x) = 4 - 2x, for f(3r -1) as required in the task content is; f(3r-1) = 6 - 6r.

Since it follows from the task content that the given function is defined by the expression;

f(x) = 4 - 2x.

To evaluate the function f(x) = 4 - 2x, for f(3r -1); we must substitute 3r -1 for x in the equation as follows;

f(3r -1) = 4 - 2(3r-1)

f(3r-1) = 6 - 6r.

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Answer:

138

Step-by-step explanation:

There is 60 seconds in a minute, so if the heart beats 23 times in 10 seconds, it is 138 bpm.

23 * 6 = 138

The cost of one gizmo is $18.

The cost of one thingamajigs is $9

3g + 6t = 102 equation 1

5g + 11t = 179  equation 2

Where:

Multiply equation 1 by 5 and equation 2 by 3

15g + 30t = 510 equation 3

15g + 33t = 537 equation 4

Subtract equation 3 from equation 4

3t = 27

t = 27 / 3

t = $9

Substitute for t in equation 1

3g + 6(9) = 102

3g + 54 = 102

3g = 102 - 54

3g = 48

g = 48 / 3

g = $18

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When the balloon, rising at 10 ft/sec, rises to 100 feet, the rate of change of the angle of elevation is 0.05 rad/sec

The rate at which the balloon rises = 10 ft/sec

Horizontal distance of the balloon from the observer, d = 100 feet

Therefore;

[tex] \frac{dy}{dt} = 10[/tex]

The angle of elevation can be presented as follows;

[tex] tan(\theta) = \frac{y}{100} [/tex]

[tex] y = 100 \times tan(\theta) [/tex]

Therefore;

[tex] \frac{dy}{dt} = \mathbf{ 100\cdot sec^2 \frac{d \theta}{dt}} [/tex]

Which gives;

[tex] \frac{d \theta}{dt}= \frac{ \frac{d y}{dt}}{100} \times cos^2 \theta [/tex]

[tex] \mathbf{ \frac{d \theta}{dt}} = \frac{ 10}{100} \times cos^2 \theta [/tex]

When the balloon is 100 ft. from the ground, we have;

Therefore;

[tex] \frac{d \theta}{dt}= \frac{ 10}{100} \times \left(\frac{1}{\sqrt{2}} \right)^2 [/tex]

[tex] \frac{ 10}{100} \times \left(\frac{1}{\sqrt{2}} \right)^2 = \frac{ 1}{20} [/tex]

Which gives;

[tex] \frac{d \theta}{dt}=\frac{ 1}{20} = 0.05 [/tex]

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The correct option is D: G(x) = -5x².

The equation of G(x) will be -5x² after vertically stretching and flipping.

A function change called a stretch or compression makes a graph bigger or narrower without moving it horizontally or vertically.

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The complete question is -

The graph of F(x) can be stretched vertically and flipped over the x-axis to produce the graph of G(x). If F(x) = x², which of the following could be the equation of G(x)?

A: G(x) = -1/5x²

B: G(x) = 5x²

C: G(x) = 1/5x²

D: G(x) = -5x²

Answer:

16. C

17. C

Step-by-step explanation:

16.

this is just a multiplication of the 2 functions.

that means we multiply the functional expressions, and the result is the new functional expression of the multiplied functions.

2(3x - 5) × (-2x + 1) = (6x - 10) × (-2x + 1) =

= 6x×-2x + 6x×1 + -10×-2x + -10×1 =

= -12x² + 6x + 20x - 10 = -12x² + 26x - 10

17.

the domain of a function is the interval or set of all valid x values (or input values).

the range is the interval or set of all valid y or functional result values.

so, for the domain we need to see what values x may have for a 4th root.

for this we can imagine that a 4th root is the square root of the square root of x.

this is directly related to x⁴ being the square of the square of x.

the valid input values for a square root are all positive numbers (incl. 0). the result will be again a positive number (incl. 0), which is again a valid input number for the second square root.

therefore, the valid values of x for a 4th root are also the valid values for a square root. and vice versa.

and that is

0 <= x < infinity

FYI : we can never say x = infinity, because infinity is not a number, just a generation concept. so, infinity can never be included in any interval, it can only be listed as an excluded limit concept.

Answer:

Step-by-step explanation:

B is the answer

Answer:

[tex]\sf 1\dfrac{2}{5} \ cm[/tex]

Step-by-step explanation:

    If two sides of the triangle are equal, then the triangle is called isosceles triangle.

Let the two equal sides = x cm

Perimeter of the triangle = [tex]\sf 4\dfrac{2}{15}[/tex] cm

[tex]\sf x + x + \dfrac{4}{3}=4\dfrac{2}{15}\\\\[/tex]

      [tex]\sf 2x +\dfrac{4}{3}=\dfrac{62}{15}[/tex]

             [tex]\sf 2x = \dfrac{62}{15}-\dfrac{4}{3} \ [\text{\bf LCM of 15 , 3 = 15}]\\\\2x = \dfrac{62}{15}-\dfrac{4*5}{3*5}\\\\2x = \dfrac{62}{15}-\dfrac{20}{15}\\\\2x = \dfrac{42}{15} \ [\text{\bf Divide both sides by 2}]\\\\ x = \dfrac{42}{15*2}\\\\ x = \dfrac{7}{5}\\\\ x = 1\dfrac{2}{5}[/tex]

[tex]\sf \boxed{\text{Equal sides of isosceles triangle = $1\dfrac{2}{5}$ cm}}[/tex]

Answer:

y =sin(x)

Step-by-step explanation:

small brain

Answer:

B: k pi for any integer k

Step-by-step explanation:

I took the test

Answer:

  P = 160/A

Step-by-step explanation:

For P to vary inversely with A, we must have P be proportional to the reciprocal of A. The constant of proportionality is given.

Two quantities are directly proportional if there is a constant of proportionality (k) that multiplies one of them to give the value of the other:

  y = kx

They are inversely proportional if the inverse of one of them can be multiplied by a constant of proportionality to give the value of the other:

  y = k(1/x) = k/x

In this problem, the variables P and A are said to be inversely proportional. That means the equation that relates them will be of the form ...

  P = k/A

You are given such a form with k=160, which is an appropriate value for the relation given by the problem. All you need to do is fill in the variable:

  P = 160/A

Answer:

  √5 is irrational

Step-by-step explanation:

A rational number is one that can be written exactly as an integer or ratio of integers. Written as a decimal number, it will have a finite number of digits, or a repeating decimal fraction.

Usually, a number that can only be expressed exactly using a symbol will be irrational. For square roots, any root of an integer other than a perfect square will be irrational.

The integer 5 is not a perfect square. It is between the squares 2²=4 and 3²=9. The square root of 5 is irrational.

__

Additional comment

A reduced fraction whose denominator has factors other than 2 or 5 will translate to a repeating decimal. The number of repeating digits may be as many as 1 less than the denominator. For example, 1/19 has an 18-digit repeating decimal equivalent.

Answer:

f(2) = 3

Step-by-step explanation:

Plug in 2 for x

You did not provide directions. What do you want us to do with your 8th degree polynomial?

Answer:

The volume of the cylinder is 2,400pi cubic units

Step-by-step explanation:

I hope this helps and have a good day!

The expression which shows the total cost is 2.50+0.4x and the cost of riding 20 miles is $10.50.

Given a taxi charges $2.50 for the first mile and $0.40 for each mile

We are required to make an expression showing total cost and total cost of a 20 mile ride.

let the miles ride by tax be x.

Total cost of the product or service=Number of units*price of one units.

We have been given that $2.50 is fixed so we will not multiply our variable charge with $2.50 and multiply 0.40 with x.

So,the expression showing looks like 2.50+0.40x.

Now to find the cost when taxi travels 20 miles, we have to put x=20,

Cost=2.50+0.40*20

=2.50+(40*20)/100

=2.50+800/100

=2.50+8

=$10.50

Hence the expression which shows the total cost is 2.50+0.4x and the cost of riding 20 miles is $10.50.

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Question 2

1) [tex]\overline{PR} \perp\overline{QS}, \overline{PQ} \cong \overline{PS}[/tex] (given)

2) [tex]\overline{PR} \cong \overline{PR}[/tex] (reflexive property)

3) [tex]\angle PRQ, \angle PRS[/tex] are right angles (perpendicular lines form right angles)

4) [tex]\triangle PRQ, \triangle PRS[/tex] are right triangles (a triangle with a right angle is a right triangle)

5) [tex]\triangle PRQ \cong \triangle PRS[/tex] (HL)

Question 3

1) [tex]\angle C[/tex] is a right angle, [tex]\overline{AC} \cong \overline{AE}, \overline{DE} \perp \overline{AB}[/tex] (given)

2) [tex]\angle DEA[/tex] is a right angle (perpendicular lines form right angles)

3) [tex]\triangle ACD, \triangle DAE[/tex] are right triangles (a triangle with a right triangle is a right angle)

4) [tex]\overline{AD} \cong \overline{AD}[/tex] (reflexive property)

5) [tex]\triangle ACD \cong \triangle AED[/tex] (HL)

6) [tex]\angle CAD \cong \angle DAE[/tex] (CPCTC)

7) [tex]\overline{AD}[/tex] bisects [tex]\angle BAC[/tex] (if a segment splits an angle into two congruent parts, it is an angle bisector)