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grandymaker [24]
3 years ago
5

Is 1/7 greater than -5​

Mathematics
2 answers:
gizmo_the_mogwai [7]3 years ago
7 0
Yes.
positives are greater than negatives.
Montano1993 [528]3 years ago
4 0

Answer:

no

Step-by-step explanation:

1/7 is positive, -5 is negative. Negative is less than positive

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Consider the graphs below. What are the explanatory variables?​
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A

Step-by-step explanation:

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The weekly ad for a local grocery store advertises a 5 pound bag of organic apples for $13.95. Round each rate to the nearest hu
beks73 [17]

Answer:

$2.79 per pound of apples.

0.36 pounds per dollar.

The unit rate representing cost of per pound apples is typically used.

Step-by-step explanation:

We have been given that the weekly ad for a local grocery store advertises a 5 pound bag of organic apples for $13.95.

Since we know when rates are expressed as a quantity of 1, such as 100 meter per second or 5 miles per hour, they are called unit rates. We need to have 1 in our denominator to express two quantities as unit rate.

Let us find the unit rate in terms of cost of per pound bag of apples.

\text{Unit rate as the cost of organic apples per pound}=\frac{\$13.95}{5 \text{ pounds of organic apples}}

\text{Unit rate as the cost of organic apples per pound}=\frac{\$2.79} {\text{ pound of organic apples}}

Therefore, our unit rate will be $2.79 per pound of apples.

Let us find unit rate of pounds of apples per dollar.

\text{Unit rate as the pounds of organic apples per dollar}=\frac{5\text{ pounds of organic apples}}{\$13.95}

\text{Unit rate as the pounds of organic apples per dollar}=\frac{0.3584 \text{ pounds of organic apples}}{\$}

\text{Unit rate as the pounds of organic apples per dollar}\approx \frac{0.36 \text{ pounds of organic apples}}{\$}

Therefore, our another unit rate will be 0.36 pounds per dollar.

Since, we purchase apples or other items according to their price per piece or their price per pound in our daily life, therefore, the unit rate representing cost of per pound apples is typically used.

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Find the length of the curve. R(t) = cos(8t) i + sin(8t) j + 8 ln cos t k, 0 ≤ t ≤ π/4
arsen [322]

we are given

R(t)=cos(8t)i+sin(8t)j+8ln(cos(t))k

now, we can find x , y and z components

x=cos(8t),y=sin(8t),z=8ln(cos(t))

Arc length calculation:

we can use formula

L=\int\limits^a_b {\sqrt{(x')^2+(y')^2+(z')^2} } \, dt

x'=-8sin(8t),y=8cos(8t),z=-8tan(t)

now, we can plug these values

L=\int _0^{\frac{\pi }{4}}\sqrt{(-8sin(8t))^2+(8cos(8t))^2+(-8tan(t))^2} dt

now, we can simplify it

L=\int _0^{\frac{\pi }{4}}\sqrt{64+64tan^2(t)} dt

L=\int _0^{\frac{\pi }{4}}8\sqrt{1+tan^2(t)} dt

L=\int _0^{\frac{\pi }{4}}8\sqrt{sec^2(t)} dt

L=\int _0^{\frac{\pi }{4}}8sec(t) dt

now, we can solve integral

\int \:8\sec \left(t\right)dt

=8\ln \left|\tan \left(t\right)+\sec \left(t\right)\right|

now, we can plug bounds

and we get

=8\ln \left(\sqrt{2}+1\right)-0

so,

L=8\ln \left(1+\sqrt{2}\right)..............Answer

5 0
3 years ago
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