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SCORPION-xisa [38]

3 years ago

7

your neighbor is mowing their 6 yards by 8 yards rectangular lawn diagonally. How far did your neighbor mow the diagonal from co

rner to corner?

1 answer:

masya89 [10]3 years ago

7 0

Answer:

48?

Explanation:

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1. A rabbit can hop 6 feet in 2 seconds and 12 feet in 4 seconds. If the number of feet varies directly with the time the rabbit

kykrilka [37]

Answer: The rabbit is hopping at 3 feet/ sec in speed

Step-by-step explanation:

Step 1

Let the The number of feet of the rabbit be expressed as N

AND time rabbit spends hopping can be expressed as t

From the question, The number of feet varies directly with the time rabbit spends hopping can be expressed as

N α t

Let us introduce a constant of proportionality ,S which represents speed in this case, we now have that

N = St

STEP 2 -- SOLVING

If the rabbit hop 6 feet in 2 seconds, Then speed =

N = St

6= S X2

S = 6/ 2

S = 3

also

If the rabbit hop 12 feet in 4 seconds, Then speed =

N = St

12= S X 4

S = 12/ 4

S = 3

Also recall

that Speed = Distance / time

Pluging the values in the first case,

speed = 6 feet/ 2 secconds = 3 feet/ seconds

in the second case

Speed= 12 feet/ 4 second s= 3 feet/ secs.

5 0

3 years ago

Which relationships describe the angle pair x° and 50°

worty [1.4K]

Answer:

I don't know because I would need a visual picture of the diagram you are most likely looking at, it would be helpful if you include the picture somehow.

Step-by-step explanation:

7 0

3 years ago

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PJ is Decorating Pupcakes for Rocky´s Birthday Party. Her Color Scheme is Black, Purple, and Green. She Will use This Spinner to

Veseljchak [2.6K]

Answer:

173

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

The base of a triangle is seven less than twice its height. If the area of the triangle is 102 cm2, find the length of the base.

Shalnov [3]

Answer:

195

Step-by-step explanation:

5 0

3 years ago

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Use Simpson's Rule with n = 10 to approximate the area of the surface obtained by rotating the curve about the x-axis. Compare y

DiKsa [7]

The area of the surface is given exactly by the integral,

$\displaystyle\pi\int_0^5\sqrt{1+(y'(x))^2}\,\mathrm dx$

We have

$y(x)=\dfrac15x^5\implies y'(x)=x^4$

so the area is

$\displaystyle\pi\int_0^5\sqrt{1+x^8}\,\mathrm dx$

We split up the domain of integration into 10 subintervals,

[0, 1/2], [1/2, 1], [1, 3/2], ..., [4, 9/2], [9/2, 5]

where the left and right endpoints for the $i$ -th subinterval are, respectively,

$\ell_i=\dfrac{5-0}{10}(i-1)=\dfrac{i-1}2$

$r_i=\dfrac{5-0}{10}i=\dfrac i2$

with midpoint

$m_i=\dfrac{\ell_i+r_i}2=\dfrac{2i-1}4$

with $1\le i\le10$ .

Over each subinterval, we interpolate $f(x)=\sqrt{1+x^8}$ with the quadratic polynomial,

$p_i(x)=f(\ell_i)\dfrac{(x-m_i)(x-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+f(m_i)\dfrac{(x-\ell_i)(x-r_i)}{(m_i-\ell_i)(m_i-r_i)}+f(r_i)\dfrac{(x-\ell_i)(x-m_i)}{(r_i-\ell_i)(r_i-m_i)}$

Then

$\displaystyle\int_0^5f(x)\,\mathrm dx\approx\sum_{i=1}^{10}\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx$

It turns out that the latter integral reduces significantly to

$\displaystyle\int_0^5f(x)\,\mathrm dx\approx\frac56\left(f(0)+4f\left(\frac{0+5}2\right)+f(5)\right)=\frac56\left(1+\sqrt{390,626}+\dfrac{\sqrt{390,881}}4\right)$

which is about 651.918, so that the area is approximately $651.918\pi\approx\boxed{2048}$ .

Compare this to actual value of the integral, which is closer to 1967.

4 0

3 years ago