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KiRa [710]
3 years ago
12

Could someone help me with these questions? I can’t understand it

Mathematics
1 answer:
vampirchik [111]3 years ago
6 0
First question: So, we know a fence that is 3 sections long, requires 4 posts and 6 rails.

Now, we have to use this kind of fence (3 sections, 4 posts, 6 rails) to fence off a yard. The yard is rectangular, and is 6 sections long and 3 sections wide.

Since we know the fence we are using is 3 sections wide, we have to count how many of them we’ll be using for the yard.
We know all four sides of the rectangle - two sides are 3 sections long, and the other two are 6 sections long.

Our fence is 3 sections, so we can already use 2 of those for the two sides of 3 sections. We multiply the number of rails and posts by two (since we are using two of these fences) and get 8 posts and 12 rails. We’ll keep this in mind.

Now, for the other two sides: they are both 6 sections long. We know that 3+3=6, thus one of those sides would be 2 fences long, or 3 sections times 2. We have two of these 6 section sides, so we actually would be using 12 sections (6*2), or in other words, we’d use this fence 4 times (4*3). We multiply the rails and posts by 4, and get 16 posts and 24 rails.

We use the 8 posts and 12 rails from earlier and add the number of the posts and rails we got just now: 16p and 24r, and get a total of 24 posts and 36 rails.

As for the second question, even I’m having a tough time. I’ll get back to you if I can solve it.







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K is between H and J, HK = x + 4, KJ = 2x - 1, and HJ = 5x - 3. What is the value of x?
Tanya [424]

HK + KJ = HJ\\x + 4 + 2x - 1 = 5x - 3\\x + 2x - 5x = -3 - 4 + 1\\-2x = -6\\x = 3

5 0
2 years ago
Solve the system by elimination. $x+y=12$ $3x=2y+6$
Andrei [34K]

Answer:

(6,6)

Step-by-step explanation:

3 0
2 years ago
Paul Kim estimates that he will drive about 10,000 miles during the year and will have $3,460 in annual fixed costs. He projects
Mama L [17]

Answer:

The maximum annual variable cost he can have to reach his projection is $1,940

Step-by-step explanation:

Given;

Number of miles drive per year N = 10,000 miles

Total annual Fixed cost F = $3,460

cost per mile(rate) r = $0.54 or less

Total cost = fixed cost + variable cost

Total cost = cost per mile × number of miles

Total cost = r × N = $0.54 × 10,000 = $5,400

Let V represent the total variable cost per year;

F + V ≤ r × N

Substituting the values;

3,460 + V ≤ 5,400

V ≤ 5,400 - 3,460

V ≤ 1,940

The maximum annual variable cost he can have to reach his projection is $1,940

5 0
3 years ago
A square pyramid has side length 9 in. and volume = 270 in^3. find the height.
aleksandrvk [35]

If the square pyramid has side length 9 inches and volume of 270 inches cube then by putting the values within the formula of volume height are going to be 10 inches.

Given Length and volume of a square pyramid are 9 inches and 270 inches cube. Square pyramid is the pyramid having square base. It is a right square pyramid.

We know that the amount of square pyramid is 1/3 *l^{2} *h

Put the values in formula to induce the values oh h.

Volume=1/3*9^{2}*h

270=1/3* 81*h

h=(270*3)/81

h=810/81

h=10

Hence the peak of the square pyramid having length 9 inches and volume of 270 inches cube is 10 inches.

Learn more about volume at brainly.com/question/1972490

#SPJ10

7 0
2 years ago
A tank in the shape of a hemisphere with a radius r = 13 ft is full of water to a depth of 11 ft. Set up the integral that would
Jobisdone [24]

Answer:

(a). The integral is W=62.5\pi\int_{0}^{11}(169-y^2)ydy

(b). The integral is W=62.5\pi\int_{0}^{11}{48+22y-y^2(11-y)}dy

Step-by-step explanation:

Given that,

Radius = 13 ft

Depth = 11 ft

Weight of water = 62.5 lb/ft³

(a). The origin is at the top of the tank with positive y variables going down.

We need to calculate the volume of the strip

Using formula of volume

dV=\pi r^2dy

Put the value into the formula

dV=\pi(\sqrt{13^2-y^2})^2dy

dV=\pi({169-y^2})dy

We need to calculate the mass of the strip

Using formula of mass

W=\rho dVgy

Put the value into the formula

W=\pi({169-y^2})dy\times62.5\ y

We need to calculate the work done to pump the water out of the spout

Using formula of work done

W= \int_{0}^{11}(\pi({169-y^2})dy\times62.5\ y)

W=62.5\pi\int_{0}^{11}(169-y^2)ydy

(b). The origin is at the bottom of the tank with positive y-values going up

We need to calculate the volume of the strip

Using formula of volume

dV=\pi r^2dy

Put the value into the formula

dV=\pi(\sqrt{13^2-(11-y)^2})^2dy

dV=\pi(169-(121-22y+y^2))dy

dV=\pi(169-121+22y-y^2)dy

dV=\pi(48+22y-y^2)dy

We need to calculate the mass of the strip

Using formula of mass

W=\rho Vg

Put the value into the formula

W=dV\rho g(11-y)

W=\pi(48+22y-y^2)dy\rho g(11-y)

W=62.5\pi(48+22y-y^2)(11-y)dy

We need to calculate the work done to pump the water out of the spout

Using formula of work done

W=\int_{0}^{11}{62.5\pi(48+22y-y^2)(11-y)dy}

W=62.5\pi\int_{0}^{11}{48+22y-y^2(11-y)}dy

Hence, (a). The integral is W=62.5\pi\int_{0}^{11}(169-y^2)ydy

(b). The integral is W=62.5\pi\int_{0}^{11}{48+22y-y^2(11-y)}dy

5 0
3 years ago
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