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MakcuM [25]
3 years ago
5

Suppose you have a rectangular shipping box that measures 30 cm × 15 cm × 20 cm. What is the longest length of a 1 cm diameter w

ooden dowel that could fit inside the box? (round to the nearest whole centimeter)
PLEASE HURRY
Mathematics
2 answers:
balandron [24]3 years ago
5 0

the box is cuboid in shape.

Longest wooden dowel that could fit the box is equal to length of diagonal of cuboid  

length of diagonal of cuboid  = sqrt(l^2+b^2+h^2)= sqrt(1525)

ngest wooden dowel that could fit the box is equal = 39.05 = 39 cm



mr_godi [17]3 years ago
3 0
See what happened was you failed         .               
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Solve the equation by completing the square. Round to the nearest tenth, if necessary. Simplify your solutions and enter them fr
Slav-nsk [51]

Answer:

-1.1, 6.1

Step-by-step explanation:

-2X^2+10X+14=0

(-2X^2+10X+14)/-2=0

X^2-5X-7=0

Then use the quadratic formula because you can't factor

Fill in the equations and solve. One should have the plus and the other should have the minus and that's how you get the 2 different answers.

A=1

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4 0
2 years ago
Please help, need help on problem will mark BRAINLIEST AND 5 STARS
horsena [70]

Answer: positive, linear association

Step-by-step explanation: that is the answer, because the dots are rising up in a slope

5 0
2 years ago
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Order pair of x=y+2 and 2y=x-1
EastWind [94]
<span>x=y+2
2y=x-1
substitute </span>x=y+2  into 2y=x-1

2y=x-1
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<span>x=y+2
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x=3

answer
x = 3 and y = 1</span>
5 0
3 years ago
Suppose an airline policy states that all baggage must be box shaped with a sum of length, width, and height not exceeding 174 i
Leya [2.2K]

Answer:

The square-based box with the greatest volume under the condition that the sum of length, width, and heigth does not exceed 174 in is a cube with each edge of 58 in and a volume of 195112 in^{3}

Step-by-step explanation:

For this problem we have two constraints, that are as follows:

1) Sum of length, width, and heigth not exceeding 174 in

2) Lenght and width have the same measure (square-based box)

We know that volume is equal to the product of all three edges, and with the two conditions into account we have the next function:

V=(w^{2})(174-2w)\\V=174w^{2}-2w^{3}

The interval of interest of the objective function is [0, 87]

This problem requieres that we maximize the function that defines the volume. We start calculating the derivative of the function, wich is:

V'=348w-6w^{2} \\V'=(348-6w)(w)

We need to remember that the derivative of a function represents the slope of said function at a given point. The maximum value of the function will have a slope equal to zero.

So we find the value in wich the derivative equals zero:

0=(348-6w)(w)\\w_1=0\\w_2=348/6=58

The first value (w=0) will leave us with a 'height-only box', so the answer must be w=58 in

The value is between the interval of interest.

And, once we solve for the constraints, we have that:

Lenght = Width = Heigth = 58 in

Volume = 195112 in^{2}

8 0
3 years ago
A farmer grows crops with an average height of 5 feet. He breeds his tallest plants, which are each 8 feet tall. They produce of
irinina [24]

Answer:

Qa: The heritability for tallness in these crops is 0.08, while Qb: Genes slightly influence the trait for height.

Step-by-step explanation:

Qa: The average height of regular crops grown by the farmer was 5 feet. The amazingly tall breeds have a 3feet increment than the usual crops. This progeny of the farmer's choice breeds has a 0.25 feet increment in height than what is expected in other normal breeds. Therefore, the heritability of the tall trait by these progeny was 0.25/3, which is 0.08333.....Approximately 0.08.

Qb: This low heritability index showed that the trait for tallness is slightly influenced by gene, because if otherwise, the index would have been a far way higher than this. However, we cannot conclude that it is the environment that determines the tall phenotype to be expected in the crops; this renders other options wrong.

7 0
3 years ago
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