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3 years ago

Which set of numbers CANNOT represent the lengths of the sides of a triangle?

Mathematics

1 answer:

posledela3 years ago

4 0

Answer:

Step-by-step explanation:

The general rule is that any 2 must exceed the third in order for a triangle to form. If any one is an exception, it won't work.

A will form a triangle. Any 2 exceeds the third.

B also forms a triangle. Same rule.

C does not work. 7 + 11 = 18 which is the value of the third leg.

D words Any two exceeds the 3rd.

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a carpenter has a board that is 8 feet long he cuts of 2 pieces one piece is 3 and a half feet long the other is 2 and one feet

kakasveta [241]

3 1/2 I think
8-3 1/2=5 1/2
5 1/2-2=3 1/2

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3 years ago

Read 2 more answers

A tank contains 30 lb of salt dissolved in 300 gallons of water. a brine solution is pumped into the tank at a rate of 3 gal/min

sesenic [268]

$A'(t)=(\text{flow rate in})(\text{inflow concentration})-(\text{flow rate out})(\text{outflow concentration})$
$\implies A'(t)=\dfrac{3\text{ gal}}{1\text{ min}}\cdot\left(2+\sin\dfrac t4\right)\dfrac{\text{lb}}{\text{gal}}-\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{300+(3-3)t\text{ gal}}$
$A'(t)+\dfrac1{100}A(t)=6+3\sin\dfrac t4$

We're given that $A(0)=30$ . Multiply both sides by the integrating factor $e^{t/100}$ , then

$e^{t/100}A'(t)+\dfrac1{100}e^{t/100}A(t)=6e^{t/100}+3e^{t/100}\sin\dfrac t4$
$\left(e^{t/100}A(t)\right)'=6e^{t/100}+3e^{t/100}\sin\dfrac t4$
$e^{t/100}A(t)=600e^{t/100}-\dfrac{150}{313}e^{t/100}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+C$
$A(t)=600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+Ce^{-t/100}$

Given that $A(0)=30$ , we have

$30=600-\dfrac{150}{313}\cdot25+C\implies C=-\dfrac{174660}{313}\approx-558.02$

so the amount of salt in the tank at time $t$ is

$A(t)\approx600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)-558.02e^{-t/100}$

3 0

3 years ago

A 2 kg metal cylinder is supplied with 1600J of energy to heat it from 5°C to 13°C. What is the specific heat capacity of the me

Neporo4naja [7]

Answer:

$c = 100\,\frac{J}{kg\cdot ^{\circ}C}$

Step-by-step explanation:

According to the First Law of Thermodynamics, the heat received by the metal cylinder is equal to the change in the internal energy. That is:

$Q = \Delta U$

$Q = m \cdot c \cdot \Delta T$

The specific heat is clear in the previous expression and finally computed:

$c = \frac{Q}{m\cdot \Delta T}$

$c = \frac{1600\,J}{(2\,kg)\cdot (13^{\circ}C-5^{\circ}C)}$

$c = 100\,\frac{J}{kg\cdot ^{\circ}C}$

6 0

3 years ago

Earl runs 60 meters in 40 seconds. How many meters per second does Earl run?

Yuki888 [10]

Earl runs 1.5 meters per second.................

5 0

3 years ago

You are given the following information obtained from a random sample of 5 observations. 20 18 17 22 18 At 90% confidence, you w

Margaret [11]

Answer:

The null hypothesis is

$H_o : \mu = 21$

The Alternative hypothesis is

$H_a : \mu< 21$

$\sigma_{\= x} = 0.8944$

$t = -2.236$

Yes the mean population is significantly less than 21.

Step-by-step explanation:

From the question we are given

a set of data

20 18 17 22 18

The confidence level is 90%

The sample size is n = 5

Generally the mean of the sample is mathematically evaluated as

$\= x = \frac{20 + 18 + 17 + 22 + 18}{5}$

$\= x = 19$

The standard deviation is evaluated as

$\sigma = \sqrt{ \frac{\sum (x_i - \= x)^2}{n} }$

$\sigma = \sqrt{ \frac{ ( 20- 19 )^2 + ( 18- 19 )^2 +( 17- 19 )^2 +( 22- 19 )^2 +( 18- 19 )^2 }{5} }$

$\sigma = 2$

Now the confidence level is given as 90 % hence the level of significance can be evaluated as

$\alpha = 100 - 90$

$\alpha = 10$ %

$\alpha =0.10$

Now the null hypothesis is

$H_o : \mu = 21$

the Alternative hypothesis is

$H_a : \mu< 21$

The standard error of mean is mathematically evaluated as

$\sigma_{\= x} = \frac{\sigma}{ \sqrt{n} }$

substituting values

$\sigma_{\= x} = \frac{2}{ \sqrt{5 } }$

$\sigma_{\= x} = 0.8944$

The test statistic is evaluated as

$t = \frac{\= x - \mu }{ \frac{\sigma }{\sqrt{n} } }$

substituting values

$t = \frac{ 19 - 21 }{ 0.8944 }$

$t = -2.236$

The critical value of the level of significance is obtained from the critical value table for z values as

$z_{0.10} = 1.28$

Looking at the obtained value we see that $z_{0.10}$ is greater than the test statistics value so the null hypothesis is rejected

6 0

3 years ago