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Schach [20]
2 years ago
6

Three consecutive even integers that add up to -6. What are these integers?

Mathematics
1 answer:
Gekata [30.6K]2 years ago
7 0

Answer:

-1, -2, -3 Are all consecutive

Step-by-step explanation:

-1 + -2 + -3 = -6

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A rectangle has an area of 96 cm2 The length of the rectangle is 4 cm longer than the width. Work out the length and width of th
Svetach [21]
Area of a rectangle: 2(L+W)
length: 4+W

Area: 2(4+W) + 2W = 96
8+2W+2W = 96
8+4W = 96
4W = 88
W= 22cm

Calculate for length: 2L + 2(22) = 96
2L + 44 = 96
2L = 52
L = 26cm

• The length is 26cm and width is 22 cm.
3 0
3 years ago
Is 1/6 greater than 1/3
____ [38]
No, 1/3 is greater than 1/6. 
1/3 converts to 2/6. 
2/6 > 1/6
4 0
3 years ago
Read 2 more answers
NASA launches a rocket at t = 0 seconds. Its height, in meters above sea-level, as a function of time is given by h ( t ) = − 4.
Aleks [24]

Answer:

A.) 24.08 seconds

B.) 825.42 metres

Step-by-step explanation:

function of time is given as

h ( t ) = − 4.9 t 2 + 118 t + 115 .

Where a = -4.9, b = 118, c = 115

Let's assume that the trajectory of the rocket is a perfect parabola.

The time t the rocket will reach its maximum height will be at the symmetry of the parabola.

t = -b/2a

Substitute b and a into the formula

t = -118/-2(4.9)

t = 118/9.8

t = 12.041 seconds

Since NASA launches the rocket at t = 0 seconds, the time it will splash down into the ocean will be 2t.

2t = 2 × 12.041 = 24.08 seconds

Therefore, the rocket splashes down after 24.08 seconds.

B.) At maximum height, time t = 12.041s

Substitute t for 12.041 in the function

h ( t ) = − 4.9 t 2 + 118 t + 115

h(t) = -4.9(12.041)^2 + 118(12.041) + 115

h(t) = -4.9(144.98) + 118(12.041) + 115

h(t) = -710.402 + 1420.82 + 115

h(t) = 825.42 metres

Therefore, the rocket get to the peak at 825.42 metres

6 0
3 years ago
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
Rationalize the denominator in the expression. <br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B4%7D%7B%5Csqrt%7B2%7D%20%7D%20
Dovator [93]

Answer:

2\sqrt{2}

Step-by-step explanation:

See the photo for the steps. :)

7 0
3 years ago
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