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lesya692 [45]
3 years ago
8

What is the waist to hip ratio? Calculate the WHR of a male with a waist measurement of 40 inches and hip measurement 2 inches l

ess.
Mathematics
1 answer:
aev [14]3 years ago
8 0
Waist to Hip Ratio (WHR)  =  Waist Measurement / Hip Measurement

Waist Measurement = 40 inches
Hip Measurement = 2 inches less =  (40 -2) = 38 inches.

Waist to Hip Ratio (WHR)  =  40/38 =  1.0526

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2x-2<-12or2x+3>7 what is the solution to this compound inequality
cluponka [151]

2x-2 < -12\qquad|+2\\\\2x < -10\qquad|:2\\\\x < -5

-----------------------------------------------------------------------------------------------

2x+3 > 7\qquad|-3\\\\2x > 4\qquad|:2\\\\x > 2


Answer:\ x < -5\ or\ x > 2

8 0
3 years ago
What is the simplified version of 3\sqrt{135}
Aleonysh [2.5K]

Answer:

The simplified version of \sqrt[3]{135} is 3\sqrt[3]{5}.

Step-by-step explanation:

The given expression is

\sqrt[3]{135}

According to the property of radical expression.

\sqrt[n]{x}=(x)^{\frac{1}{n}}

Using this property we get

\sqrt[3]{135}=(135)^{\frac{1}{3}}

\sqrt[3]{135}=(27\times 5)^{\frac{1}{3}}

\sqrt[3]{135}=(3^3\times 5)^{\frac{1}{3}}

\sqrt[3]{135}=(3^3)^{\frac{1}{3}}\times (5)^{\frac{1}{3}}      [\because (ab)^x=a^xb^x]

\sqrt[3]{135}=3\times \sqrt[3]{5}     [\because \sqrt[n]{x}=(x)^{\frac{1}{n}}]

\sqrt[3]{135}=3\sqrt[3]{5}

Therefore the simplified version of \sqrt[3]{135} is 3\sqrt[3]{5}.

7 0
3 years ago
Read 2 more answers
Jesse served 13 pints of orange juice at her party how many quarts of orange juice did she serve
Alexus [3.1K]

Answer:

26 pints

Step-by-step explanation:

1 quart = 2 pints

13*2=26


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3 years ago
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Answer:

c hope this helps

Step-by-step explanation:

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2 years ago
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Let’s look at another one of Homer’s rocket launches. It was launched from ground level with an initial velocity of 208 feet per
kozerog [31]

Answer:

Smax = 676 ft

the maximum altitude (height) the rocket will attain during its flight is 676 ft

Step-by-step explanation:

Given;

The height function S(t) of the rocket as;

S(t) = -16t2 + 208t

The maximum altitude Smax, will occur at dS/dt = 0

differentiating S(t);

dS/dt = -32t + 208 = 0

-32t +208 = 0

32t = 208

t = 208/32

t = 6.5 seconds.

The maximum altitude Smax is;

Substituting t = 6.5 s

Smax = -16(6.5)^2 + 208(6.5)

Smax = 676 ft

the maximum altitude (height) the rocket will attain during its flight is 676 ft

8 0
3 years ago
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