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3 years ago

Just look at the question on the picture

Mathematics

1 answer:

snow_tiger [21]3 years ago

4 0

Answer:

A = ½bh + lw

Step-by-step explanation:

Your composite figure is a triangle on top of a rectangle.

The formula for the area of a triangle is A = ½bh.

The formula for the area of a rectangle is A = lw.

The formula for the total area is

A = ½bh + lw.

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Helpppppppp meeeeeeeeeeeeeeeeeeee

Alex73 [517]

Answer:

The length of the shorter piece is $5\textrm{ ft}$ .

The length of the longer piece is $11\textrm{ ft}$ .

Step-by-step explanation:

The five-step method is GUESS method.

Given:

Length of steel beam, $L=16\textrm{ ft}$

Length of longer piece is 1 foot longer than twice the shorter one.

Unknown:

Length of longer and shorter pieces of beam.

Let the length of shorter piece be $x$ feet.

Therefore, the length of longer beam is equal to $1+2x$

Evaluate:

Sum of lengths of longer and shorter pieces of beam is equal to the total length of beam. So,

$x+1+2x=16\\3x+1=16\\3x=16-1\\3x=15\\x=\frac{15}{3}=5$

Substitute and Solve:

Now, we have, $x=5$ .

Substitute 5 for $x$ for the lengths of shorter and longer pieces and solve for each. This gives,

Length of shorter piece = $x=5\textrm{ ft}$

Length of longer piece = $1+2x=1+2(5)=1+10=11\textrm{ ft}$

Therefore, the lengths of shorter and longer pieces are $5\textrm{ ft}$ and $11\textrm{ ft}$ respectively.

6 0

4 years ago

How many ways can the letters of the word snow be arranged horizontally so that the letters n and o are always immediately next

vitfil [10]

Answer:

6 times

Step-by-step explanation:

nosw

swno

wsno

nows

snow

wnos

7 0

3 years ago

Can someone help me pls

jeka94

The answer for this is C

8 0

3 years ago

Find the dimensions of the open rectangular box of maximum volume that can be made from a sheet of cardboard 21 in. by 12 in. by

a_sh-v [17]

Answer:

Dimension of the box is $16.1\times 7.1\times 2.45$

The volume of the box is 280.05 in³.

Step-by-step explanation:

Given : The open rectangular box of maximum volume that can be made from a sheet of cardboard 21 in. by 12 in. by cutting congruent squares from the corners and folding up the sides.

To find : The dimensions and the volume of the box?

Solution :

Let h be the height of the box which is the side length of a corner square.

According to question,

A sheet of cardboard 21 in. by 12 in. by cutting congruent squares from the corners and folding up the sides.

The length of the box is $L=21-2h$

The width of the box is $W=12-2h$

The volume of the box is $V=L\times W\times H$

$V=(21-2h)\times (12-2h)\times h$

$V=(21-2h)\times (12h-2h^2)$

$V=252h-42h^2-24h^2+4h^3$

$V=4h^3-66h^2+252h$

To maximize the volume we find derivative of volume and put it to zero.

$V'=12h^2-132h+252$

$0=12h^2-132h+252$

Solving by quadratic formula,

$h=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$h=\frac{-(-132)\pm\sqrt{132^2-4(12)(252)}}{2(12)}$

$h=\frac{132\pm72.99}{24}$

$h=2.45,8.54$

Now, substitute the value of h in the volume,

$V=4h^3-66h^2+252h$

When, h=2.45

$V=4(2.45)^3-66(2.45)^2+252(2.45)$

$V\approx 280.05$

When, h=8.54

$V=4(8.54)^3-66(8.54)^2+252(8.54)$

$V\approx -170.06$

Rejecting the negative volume as it is not possible.

Therefore, The volume of the box is 280.05 in³.

The dimension of the box is

The height of the box is h=2.45

The length of the box is $L=21-2(2.45)=16.1$

The width of the box is $W=12-2(2.45)=7.1$

So, Dimension of the box is $16.1\times 7.1\times 2.45$

6 0

4 years ago

What is the volume of this sphere?

PIT_PIT [208]

12 cubic feet!! this may be it

7 0

3 years ago

Just look at the question on the picture​

Just look at the question on the picture