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4 years ago

Is 0.02468 greater than 0.024

1 answer:

3 0

Use zeroes as place holders
0.02468
0.02400
so clearly 0.02468 is greater than 0.02400
so yes, 0.02468 is greater than 0.024
but remember don't use those extra zeroes in ur actual answer.

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Can someone help answer the first two?

Murljashka [212]

Answer:

below given!

Step-by-step explanation:

1) mean = (7.4 + 5.8 + 7.3 + 7 + 8.9 + 9.4 + 8.3 + 9.3 + 6.9 + 7.5 + 9 + 5.8 + 5.5 + 8.6 + 9.3 +3.8) / 16

mean = 7.47

5 0

3 years ago

The length of a garden is 23 feet. The width of the garden is 14 feet. Which is the correct unit of measurement to use when figu

Free_Kalibri [48]

The fence should be measured in just feet
the answer is B. ft.

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3 years ago

Read 2 more answers

Integration questions .

dlinn [17]

$\\\\\ \textbf{a)}\\\\~~~\displaystyle \int (6x- \sin 3x) ~ dx\\\\=6\displaystyle \int x ~ dx - \displaystyle \int \sin 3x ~ dx\\\\=6 \cdot \dfrac{x^2}2 - \dfrac 13 (- \cos 3x) +C~~~~~~~~~~~;\left[\displaystyle \int x^n~ dx = \dfrac{x^{n+1}}{n+1}+C,~~~n \neq -1\right]\\\\ =3x^2 +\dfrac{\cos 3x}3 +C~~~~~~~~~~~~~~~~~~~~;\left[\displaystyle \int \sin (mx) ~dx = -\dfrac 1m ~ (\cos mx)+C \right]\\$

$\textbf{b)}\\\\~~~~\displaystyle \int(3e^{-2x} +\cos (0.5 x)) dx\\\\=3\displaystyle \int e^{-2x} ~dx+ \displaystyle \int \cos(0.5 x) ~dx\\\\\\=-\dfrac 32 e^{-2x} + \dfrac 1{0.5} \sin (0.5 x) +C~~~~~~~~~~~~~~;\left[\displaystyle \int e^{mx}~dx = \dfrac 1m e^{mx} +C \right]\\\\\\=-\dfrac 32 e^{-2x} + 2 \sin(0.5 x) +C~~~~~~~~~~~~~~~~~;\left[\displaystyle \int \cos(mx)~ dx = \dfrac 1m \sin(mx) +C\right]\\\\\\=-1.5e^{-2x} +2\sin(0.5x) +C$

$\textbf{a)}\\\\y = \displaystyle \int \cos(x+5) ~ dx\\\\\text{Let,}\\\\~~~~~~~u = x+5\\\\\implies \dfrac{du}{dx} = 1+0~~~~~~;[\text{Differentiate both sides.}]\\\\\implies \dfrac{du}{dx} = 1\\\\\implies du = dx\\\\\text{Now,}\\\\y= \displaystyle \int \cos u ~ du\\\\~~~= \sin u +C\\\\~~~=\sin(x+5) + C$

$\textbf{b)}\\\\y = \displaystyle \int 2(5x-3)^4 dx\\\\\text{Let,}\\~~~~~~~~u = 5x-3\\\\\implies \dfrac{du}{dx} = 5~~~~~~~~~~;[\text{Differentiate both sides}]\\\\\implies dx = \dfrac{du}5\\\\\text{Now,}\\\\y = 2\cdot \dfrac 1 5 \displaystyle \int u^4 ~ du\\\\\\~~=\dfrac 25 \cdot \dfrac{u^{4+1}}{4+1} +C\\\\\\~~=\dfrac 25 \cdot \dfrac{u^5}5+C\\\\\\~~=\dfrac{2u^5}{25}+C\\\\\\~~=\dfrac{2(5x-3)^5}{25}+C$

$\textbf{a)}\\\\y = \displaystyle \int xe^{3x} dx\\\\\text{We know that,}\\\\ \displaystyle \int (uv) ~dx = u \displaystyle \int v ~ dx - \displaystyle \int \left[ \dfrac{du}{dx} \displaystyle \int ~ v ~ dx \right]~ dx\\\\\text{Let}, u =x~ \text{and}~ v=e^{3x} .\\\\y= \displaystyle \int xe^{3x} ~dx\\\\\\~~= x\displaystyle \int e^{3x} ~ dx - \displaystyle \int \left[\dfrac{d}{dx}(x) \displaystyle \int e^{3x}~ dx \right]~ dx\\\\\\$

$=x\displaystyle \int e^{3x}~ dx - \displaystyle \dfrac 13 \int \left(e^{3x} \right)~ dx\\\\\\=\dfrac{xe^{3x}}3 - \dfrac 13 \cdot \dfrac{ e^{3x}}3+C\\\\\\= \dfrac{xe^{3x}}{3}- \dfrac{e^{3x}}{9}+C\\\\\\=\dfrac{3xe^{3x}}{9}- \dfrac{e^{3x}}9 + C\\\\\\= \dfrac 19e^{3x}(3x-1)+C$

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8 0

2 years ago

We have $10,000 to invest for 44 months. How much money will we have if we put the money into an account that has an annual inte

polet [3.4K]

A = p(1 + r/n)^nt
p = principal
r = rate, change to a decimal
n = number of times it is compounded per year
t = time in years
A = 10,000(1 + .055/12)^(12*(44/12))
Since the time is in months and the formulas is in years, I put the (44/12) to represent the time. This will actually simplify to 44.
A = 10,000(1 + .055/12)^(44)
A = $ 12,228.77

3 0

3 years ago

Find the x-intercept of the line 5× - y =10

LenKa [72]

Answer:

(2,0)

Step-by-step explanation:

To find the x intercept, set y =0 and solve for x

5x-y =10

5x -0 = 10

5x=10

Divide by 5

5x/5=10/5

x=2

(2,0)

4 0

4 years ago