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Mrrafil [7]
3 years ago
9

The ratio of the length of the rectangular field to its width 5:2. The length of the field is 30yd. Find the perimeter and area

Mathematics
1 answer:
almond37 [142]3 years ago
5 0

Answer:

Perimeter= 84

Area= 360

Step-by-step explanation:

We know that this field has portions split up by 7. The length is 5/7 of that, and the width is 2/7.

We know that the length is 30yd. 30/42 is the equivalent of 5/7. That means that 12/42 is the equivalent of 2/7.

Thus, the width of the field is equal to 12.

To find the perimeter we will add all sides. 30 + 30 + 12 + 12 = 84

We will multiply them to find area. 30 * 12 = 360

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Factor the polynomial. 48w^7 30w^4 a) 6w^4(8w^3 5) b) 6w^3(8w^4 5w) c) w^4(48w^3 30) d) 6(8w^7 5w^4)
Oliga [24]
48w^7 + 30w^4 = 6w^4(8w^3 + 5)
8 0
3 years ago
MUTIPLE CHOICE AND GIVE A BRIEF EXPLANATION PLEASE!!!
zepelin [54]

Answer:

bottom one

Step-by-step explanation:

Because it's tall and if it was the same as the circumference it would have been wide

8 0
2 years ago
What step is incorrect
tatyana61 [14]

Answer:

3

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5 0
2 years ago
Read 2 more answers
The area of the shaded region
antiseptic1488 [7]

Answer:

125.66cm²

Step-by-step explanation:

so you need to find the area of the big circle and subtract it from the area of the small circle

area of a circle= π r²

small circle: substitute in what you know

a= π x 3²

a= 28.2743

big circle: substitute in what you know

a= π x 7²

a= 153.9380

now do 153.9380-28.2743=125.6637

now round this number to the nearest hundredth:

125.66cm²

4 0
3 years ago
This is an ADDMATHS question!! It's a kinematics problem. Please help me for both a(i) and a(ii).
Kitty [74]
a(i). Since you are given a velocity v. time graph, the distance will be represented by:
\vec{s}(t) = \int_{a}^{b}\left \| \vec{v}(t) \right \|

In this case, however, we can just use simple geometry to evaluate the area under the graph v(t). I split it up into 2 trapezoids, and 1 rectangle. So, the area will be as follows:
A_t = A_{t1}+A_{t2}+A_r
A_{t1} = \frac{30+15}{2}(10) = 225m
A_{t2} = \frac{15+25}{2}(15) = 300m
A_r = 25(30) = 750m
A_t = 750+300+225 = 1275m

So, the particle traveled a total of 1275m assuming it never turned back (because it says to calculate distance).

a(iii). Deceleration is a word for negative acceleration. Acceleration is the first derivative of velocity, and so deceleration is too. So, we just need to find the slope of the line that passes through t = 30 because it has a linear slope (meaning the slope doesn't change). So, we can just use simple algebra instead of calculus to figure this out. Recall from algebra that slope (m):
m = \frac{y_2-y_1}{x_2-x_1}

So, let's just pick values. I'm going to pick (25, 30) and (35, 15). Let's plug and chug:
m = \frac{15-30}{35-25} = -\frac{15}{10} = -\frac{3}{2}

Since it's a negative value, this means that acceleration is negative but deceleration is positive (because deceleration is negative acceleration). So, your answer is: The deceleration of the particle at t = 30s is 3/2 or 1.5.
5 0
3 years ago
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