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2 years ago

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A food company originally sells cereal in boxes with dimensions 25 cm by 14 cm by 10 cm. To make more profit, the company decrea

ses each dimension of the box by x centimeters but keeps the price the same. If the new volume is 2,208 cm3 by how much was each dimension decreased

1 answer:

hjlf2 years ago

3 0

Answer:

1.75 cm

Step-by-step explanation:

We are given that the dimensions of the box as 25 cm by 14 cm by 10 cm.

Now, to increase the profit, the dimensions of the box by 'x' cm.

Thus, the new dimensions are (25-x) cm by (14-x) cm by (10-x) cm.

Further, it is given that the volume of the new box is 2,208 $cm^{3}$

As, Volume of the box = Length × Breadth × Height

i.e. 2208 = (25-x) × (14-x) × (10-x)

i.e. $2208 = (25-x) \times (140-24x-x^{2})$

i.e. $2208=x^{3}-x^{2}-740x+3500$

i.e. $x^{3}-x^{2}-740x+1292=0$

On solving this cubic equations, we get that, the solutions are x = 27.6, x = 1.75 and x = 26.8.

Since, the dimensions 25 cm by 14 cm by 10 cm are reduced by x cm.

Thus, if x = 27.6 or x = 26.8, then (10-x) = -17.6 or -16.8, which cannot be possible. So, x = 1.75 cm.

Hence, the dimensions were decreased by 1.75 cm.

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Answer: 0.05

Step-by-step explanation:

Let M = Event of getting an A in Marketing class.

S = Event of getting an A in Spanish class,

i.e. P(M) = 0.80 , P(S) = 0.60 and P(M∩S)=0.45

Required probability = P(neither M nor S)

= P(M'∩S')

= P(M∪S)' [∵P(A'∩B')=P(A∪B)']

=1- P(M∪S) [∵P(A')=1-P(A)]

= 1- (P(M)+P(S)- P(M∩S)) [∵P(A∪B)=P(A)+P(B)-P(A∩B)]

= 1- (0.80+0.60-0.45)

= 1- 0.95

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hence, the probability that Helen does not get an A in either class= 0.05

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2 years ago

The table shows y as a function of x. Suppose a point is added to this table. Which choice gives a point that preserves the func

Hunter-Best [27]

Answer:

<em>A) (-5,7)</em>

Step-by-step explanation:

<u>Functions and Relations</u>

A set of values A can have a relation with another set B as long as at least one element of A has at least one image in B. Functions are special relations where each element of A (the domain of the function) has one and only one image on B (the range of the function).

By looking at the options, we can see that x=9, x=-8, and x=-1 already have defined values in Y, so if we define another value for any of them the relation will stop being a function. The only possible choice to preserve the function is the option

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If x is the midpoint of ZW, then ZX=XW

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True.

Step-by-step explanation:

By definition of a mid point, it is located in the middle of two connected points. This means that the split is equal on both sides.

IF X is the midpoint of ZW, then ZX = XW.

~

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2 years ago

Which statements are true regarding triangle LMN? Check all that apply.

dimaraw [331]

Answer:

$NM = x$

$LM = x\sqrt{2}$

$tan (45) = 1$

Step-by-step explanation:

Step 1: Pythagoras Theorem

Pythagoras theorem relates the three sides of the triangle in such a way that the sum of the square of base and perpendicular is equal to hypotenuse, such as:

$LM^{2} =LN^{2} +NM^{2}$

Step 2: Trigonometric Functions

Only for a right angle triangle following three trigonometric relations are valid

$sin (\theta) = \frac{opposite}{hypotenuse}$

$cos (\theta) = \frac{adjacent}{hypotenuse}$

$tan (\theta)=\frac{sin (\theta)}{cos (\theta)} = \frac{opposite}{adjacent}$

Step 3: Verifying all the possible answers

A: Since, LN = x and using $tan (45) =1$

we can calculate

$tan (\theta)= \frac{opposite}{adjacent}$

$tan (45)= \frac{NM}{x} =1$

therefore, NM = x (true)

B: As NM = x therefore it can not be equal to $x\sqrt{2\\}$ .

C: Using Pythagoras Theorem

$LM^{2} =LN^{2} +NM^{2}$

$LM^{2} =x^{2} +x^{2}$

$LM^{2} =2x^{2}$

$LM = \sqrt{2x^{2}} = x\sqrt{2}$

It can also be proved using trigonometric relation

$cos (45) = \frac{x}{LM}$

$LM = \frac{x}{cos (45)}$

As, $\frac{1}{cos (45)}= \sqrt{2}$

Therefore

$LM = x\sqrt{2}$

D and E:

Using same approach similar to part A

Since, LN = x and NM = x

we can calculate

$tan (\theta)= \frac{opposite}{adjacent}$

$tan (45)= \frac{x}{x} =1$

Therefore, $tan (45) = 1$ and not equal to $\frac{\sqrt{2} }{2}$

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Please see the attached picture to understand this.

You will find the area of the total space created by the new triangle and subtract the space represented by the original triangle to find the area of the exclusion zone.

(1/2 x 6 x 5) - (1/2 x 4 x 3)
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The exclusion space is 9 km².

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