All categories

3 years ago

How many triangles can be constructed with angles measuring 120°, 40°, and 10°? one more than one none

2 answers:

4 0

I might be wrong but i believe you can make more than one with those 3 angles. the 120 and 40 can make one and if angled right can make 2 and the 10 deg on one of the other angles lines would give you a 3rd.

Margaret [11]3 years ago

3 0

The interior angles of a triangle must sum to 180 degrees.

$\rm 120 + 40 + 10\ne180$

These angles do not sum to the proper amount. No triangle can be constructed using all three angles.

You might be interested in

In order to answer the question correctly, please use the following image below:

Maru [420]

Answer:

see below

Step-by-step explanation:

The angle where chords cross is the average of the intercepted arcs. Here, that is ...

(37° +46°)/2 = (83°)/2 = 41.5°

Angle SUT is 41.5°.

_____

Comment on the error

The measure of an arc cannot be arbitrarily said to be the same as the angle where the chords cross. It will be the same if (a) the chords cross at the circle center, or (b) the opposite intercepted arc has the same measure. Neither of these conditions hold here.

8 0

3 years ago

Read 2 more answers

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

The table shows the battery life of four different mobile phones.

DanielleElmas [232]

10 * 0.775 = 7.75 So C

5 0

2 years ago

Erin is buying cupcakes for her birthday party. Each cupcake costs $1.50. How many guests can she invite if her budget is $80 an

GuDViN [60]

Answer:

About 21 People Can Go To Erin’s Party.

Step-by-step explanation:

Equation:

(80-16)= 1.5x*2

How To Solve It:

step 1: (80-16)= 1.5x*2

step 2: 64 =1.5x*2

step 3: 64=3x

step 4: 64/3=3x/3

Answer: About 21 People Can Go To Erin’s Party.

3 0

3 years ago

The 2nd, 6th, 8th terms of an A.P. form a G.P. , find the common ratio and the general term of the G.P.

melisa1 [442]

The terms of an arithmetic progression, can form consecutive terms of a geometric progression.

The common ratio is: $\mathbf{r = \frac{a + 5d}{a + d}}$
The general term of the GP is: $\mathbf{a_n = (a + d) \times (\frac{a + 5d}{a + d})^{n-1}}$