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Agata [3.3K]
4 years ago
13

two angles of a triangle measure 58 degrees and 104 degrees what is the measure of the third angle? A) 18 degrees B) 28 degrees

C) 152 degrees D) 162 degrees
Mathematics
2 answers:
kifflom [539]4 years ago
8 0

Answer:

a) 18

Step-by-step explanation:

you first add the 58 and 104 degrees,

=162

then you subtract,

=180-162

=18

hope this helps!

Lelechka [254]4 years ago
4 0

Answer:

18 is the third angle

Step-by-step explanation:

The sum of the angles of a triangle add to 180 degrees

58+104 +x  = 180

Combine like terms

162+x = 180

Subtract 162

x = 180-162

x =18

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Answer: The greatest common factor, or GCF, is the greatest factor that divides two numbers. To find the GCF of two numbers: List the prime factors of each number. Multiply those factors both numbers have in common.

Step-by-step explanation:

4 0
3 years ago
An area is approximated to be 14 in 2 using a left-endpoint rectangle approximation method. A right- endpoint approximation of t
USPshnik [31]
The trapezoidal approximation will be the average of the left- and right-endpoint approximations.

Let's consider a simple example of estimating the value of a general definite integral,

\displaystyle\int_a^bf(x)\,\mathrm dx

Split up the interval [a,b] into n equal subintervals,

[x_0,x_1]\cup[x_1,x_2]\cup\cdots\cup[x_{n-2},x_{n-1}]\cup[x_{n-1},x_n]

where a=x_0 and b=x_n. Each subinterval has measure (width) \dfrac{a-b}n.

Now denote the left- and right-endpoint approximations by L and R, respectively. The left-endpoint approximation consists of rectangles whose heights are determined by the left-endpoints of each subinterval. These are \{x_0,x_1,\cdots,x_{n-1}\}. Meanwhile, the right-endpoint approximation involves rectangles with heights determined by the right endpoints, \{x_1,x_2,\cdots,x_n\}.

So, you have

L=\dfrac{b-a}n\left(f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1})\right)
R=\dfrac{b-a}n\left(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n)\right)

Now let T denote the trapezoidal approximation. The area of each trapezoidal subdivision is given by the product of each subinterval's width and the average of the heights given by the endpoints of each subinterval. That is,

T=\dfrac{b-a}n\left(\dfrac{f(x_0)+f(x_1)}2+\dfrac{f(x_1)+f(x_2)}2+\cdots+\dfrac{f(x_{n-2})+f(x_{n-1})}2+\dfrac{f(x_{n-1})+f(x_n)}2\right)

Factoring out \dfrac12 and regrouping the terms, you have

T=\dfrac{b-a}{2n}\left((f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1}))+(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n))\right)

which is equivalent to

T=\dfrac12\left(L+R)

and is the average of L and R.

So the trapezoidal approximation for your problem should be \dfrac{14+21}2=\dfrac{35}2=17.5\text{ in}^2
4 0
3 years ago
6x – 3y = 8
Nat2105 [25]

Answer:

Step-by-step explanation:

<em>The linear equation where:</em>

\large \boldsymbol {}  \sf y=\underbrace{m}_{slope }x \ +\underbrace{b} _{y -intersept}

Solution :

\displaystyle \sf \#13. \\\\   6x-3y=8 \\\\\ -3y=8-6x \\\\y=-\frac{8-6x}{3}  \\\\ \boxed{\sf y=2x-2\frac{1}{3} }} \\\\ slope =2 \\\\ y-intersept  =-2\frac{1}{3} \\\\-------------      

 \sf \#14 . \\\\\\ 7x=5y+2 \\\\5y=7x-2 \\\\y =\dfrac{7x-2}{5}  \\\\ \boxed{\sf y=1,4x-0,4} \\\\slope = 1,4 \\\\y-intersept  =-0,4 \\\\---------------  

    \dispalystye \sf \#15. \\\\ -6y+4x=8  \  |\div2 \\\\-3y+2x=4 \\\\ y=-\dfrac{4-2x}{3}  \\\\ y=\boxed{\sf \frac{2x-4}{3} } \\\\slope = \dfrac{2}{3}  \\\\ y-intersept  = -\dfrac{4}{3 }\\\\ -----------------

\sf \#16   .\\\\ x+y=2x+3 \\\\y=2x-x+3 \\\\\boxed{\sf y=1\cdot x+3}  \\\\slope =1 \\\\y-intersept =3

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Mariulka [41]
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4 years ago
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kupik [55]

Comparing the numbers, it is found that the tenths value first determines which of the numbers 6.399 or 6.400 is the larger number.

The decimal numbers are 6.399 and 6.400.

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The tenths digit is the <u>first in which there is a difference</u>, hence, it determines which of the numbers 6.399 or 6.400 is the larger number.

A similar problem is given at brainly.com/question/17248958

8 0
2 years ago
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