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Mrac [35]
4 years ago
13

Molly bought two heads of cabbage for $1.80. How many heads of cabbage can Willie by if he has $28.00

Mathematics
1 answer:
enot [183]4 years ago
5 0
The closest you can get without going over exactly 28 is 15
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Distance (feet)
Advocard [28]

The area of a 2D form is the amount of space within its perimeter. The area of the garden is 72 feet² and the perimeter of the garden is 36 feet.

<h3>What is an area?</h3>

The area of a 2D form is the amount of space within its perimeter. It is measured in square units such as cm2, m2, and so on. To find the area of a square formula or another quadrilateral, multiply its length by its width.

The diagram for the given garden is given below.

1.) The area of the garden is,

The area of the garden = Area of rectangle + Area of triangle

                                       = (6 x 8) + (0.5 x 6 x 8)

                                       = 48 + 24

                                       = 72 feet²

2.) The perimeter of the garden is,

The perimeter of the triangle = 12 + 8 + 6 + 10 = 36 feet

Hence, the area of the garden is 72 feet² and the perimeter of the garden is 36 feet.

Learn more about the Area:

brainly.com/question/1631786

#SPJ1

4 0
2 years ago
Given f (X) = 4x +6 find f (x) = 54
Nikitich [7]

Answer:

x = 12

Step-by-step explanation:

Step 1: Define

f(x) = 4x + 6

f(x) = 54

Step 2: Substitute variables

54 = 4x + 6

Step 3: Solve for <em>x</em>

<u>Subtract 6 on both sides:</u> 48 = 4x

<u>Divide both sides by 4:</u> 12 = x

Step 4: Check

<em>Plug in x to verify it is a solution.</em>

f(12) = 4(12) + 6

f(12) = 48 + 6

f(12) = 54

3 0
4 years ago
Read 2 more answers
Evaluate the integral. (sec2(t) i t(t2 1)8 j t7 ln(t) k) dt
polet [3.4K]

If you're just integrating a vector-valued function, you just integrate each component:

\displaystyle\int(\sec^2t\,\hat\imath+t(t^2-1)^8\,\hat\jmath+t^7\ln t\,\hat k)\,\mathrm dt

=\displaystyle\left(\int\sec^2t\,\mathrm dt\right)\hat\imath+\left(\int t(t^2-1)^8\,\mathrm dt\right)\hat\jmath+\left(\int t^7\ln t\,\mathrm dt\right)\hat k

The first integral is trivial since (\tan t)'=\sec^2t.

The second can be done by substituting u=t^2-1:

u=t^2-1\implies\mathrm du=2t\,\mathrm dt\implies\displaystyle\frac12\int u^8\,\mathrm du=\frac1{18}(t^2-1)^9+C

The third can be found by integrating by parts:

u=\ln t\implies\mathrm du=\dfrac{\mathrm dt}t

\mathrm dv=t^7\,\mathrm dt\implies v=\dfrac18t^8

\displaystyle\int t^7\ln t\,\mathrm dt=\frac18t^8\ln t-\frac18\int t^7\,\mathrm dt=\frac18t^8\ln t-\frac1{64}t^8+C

8 0
4 years ago
The graph below shows the hours students spent studying and their science test scores.
scoray [572]

Answer:

4.5

Step-by-step explanation:

Just took the Test

7 1
3 years ago
Read 2 more answers
What is the product of 3 and 57 explain your thinking
kirza4 [7]

Answer:

3*57=171

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
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