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aleksklad [387]
3 years ago
13

For each set of three measures, determine if they can be angle measures of a triangle

Mathematics
1 answer:
Readme [11.4K]3 years ago
6 0

Answer:

Sum of all three angles =170

∘-180

Step-by-step explanation:

Solution

Correct option is

C

65

∘

,85

∘

,30

∘

Given set (A) 70

∘

,90

∘

,25

∘

Sum of all three angles =185

∘

>180

∘

This set cannot form a triangle

(B) 65

∘

,85

∘

,40

∘

Sum of all three angles =190

∘

>180

∘

This set cannot form a triangle

(C) 65,85

∘

,30

∘

Sum of all three angles =180

∘

This set can form a triangle

(D) 45

∘

,45

∘

,80

∘

Sum of all three angles =170

∘

<180

∘

This set cannot form a triangle

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Answer:

100

Step-by-step explanation:

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4 0
3 years ago
N = 20
lesantik [10]

Answer:

A. 30.144

Step-by-step explanation:

This is a Chi-Squared test

Given:

  • n=20
  • s=18
  • \textsf{H}_0:\sigma^2=384
  • \textsf{H}_1:\sigma^2 > 384
  • The significance level is 5%, so \alpha= 0.05

To use the Chi-Square distribution table, you need to know two values:

  • Degrees of freedom = (n - 1)
  • Significance level

Degrees of Freedom = n - 1 = 20 - 1 = 19

The significance level is 5%, so \alpha= 0.05

This is a one-tailed test since \textsf{H}_1:\sigma^2 > 384 so Upper tail area = 0.05

Reading from the table (attached), the critical value is:  30.144

(This means that we reject \textsf{H}_0 if the test statistic is greater than 30.144)

7 0
2 years ago
Verify that the conclusion of Clairaut’s Theorem holds, that is, uxy = uyx, u=tan(2x+3y)
choli [55]

Answer: Hello mate!

Clairaut’s Theorem says that if you have a function f(x,y) that have defined and continuous second partial derivates in (ai, bj) ∈ A

for all the elements in A, the, for all the elements on A you get:

\frac{d^{2}f }{dxdy}(ai,bj) = \frac{d^{2}f }{dydx}(ai,bj)

This says that is the same taking first a partial derivate with respect to x and then a partial derivate with respect to y, that taking first the partial derivate with respect to y and after that the one with respect to x.

Now our function is u(x,y) = tan (2x + 3y), and want to verify the theorem for this, so lets see the partial derivates of u. For the derivates you could use tables, for example, using that:

\frac{d(tan(x))}{dx} = 1/cos(x)^{2} = sec(x)^{2}

\frac{du}{dx}  =  \frac{2}{cos^{2}(2x + 3y)} = 2sec(2x + 3y)^{2}

and now lets derivate this with respect to y.

using that \frac{d(sec(x))}{dx}= sec(x)*tan(x)

\frac{du}{dxdy} = \frac{d(2*sec(2x + 3y)^{2} )}{dy}  = 2*2sec(2x + 3y)*sec(2x + 3y)*tan(2x + 3y)*3 = 12sec(2x + 3y)^{2}tan(2x + 3y)

Now if we first derivate by y, we get:

\frac{du}{dy}  =  \frac{3}{cos^{2}(2x + 3y)} = 3sec(2x + 3y)^{2}

and now we derivate by x:

\frac{du}{dydx} = \frac{d(3*sec(2x + 3y)^{2} )}{dy}  = 3*2sec(2x + 3y)*sec(2x + 3y)*tan(2x + 3y)*2 = 12sec(2x + 3y)^{2}tan(2x + 3y)

the mixed partial derivates are equal :)

7 0
3 years ago
There are 48 pounds of apples in a bushel. If apple pickers pick 320 pounds of apples, how many bushels are there and how many p
34kurt
There are 6 bushels and 6 pounds of apples.. 
5 0
3 years ago
Read 2 more answers
Please explain how to do this. 7th grade math. Worth 20 points.
Marrrta [24]

\large\mathfrak{{\pmb{\underline{\blue{To\:find}}{\blue{:}}}}}

The value of x.

\large\mathfrak{{\pmb{\underline{\orange{Solution}}{\orange{:}}}}}

\longrightarrow{\green{x=12° }} 

\large\mathfrak{{\pmb{\underline{\red{Step-by-step\:explanation}}{\red{:}}}}}

We know that,

\sf\pink{Sum\:of\:angles\:of\:a\:triangle\:=\:180°}

➪ 95° + (3x +4)° + (4x - 3)° = 180°

➪ 3x + 4x + 95° + 4° -3° = 180°

➪ 7x + 96° = 180°

➪ 7x = 180° - 96°

➪ 7x = 84°

➪ x = \frac{84°}{7}

➪ x = 12°

Therefore, the value of x is 12°.

Now, the three angles of the triangle are 95°, 40° and 45° respectively.

\large\mathfrak{{\pmb{\underline{\purple{To\:verify}}{\purple{:}}}}}

✒ 95° + 40° + 45° = 180°

✒ 180° = 180°

✒ L. H. S. = R. H. S.

\boxed{Hence\:verified.}

8 0
3 years ago
Read 2 more answers
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