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3 years ago

Find the measures of the angles of an isosceles obtuse triangle with one angle that is 32 degrees.

Mathematics

1 answer:

SSSSS [86.1K]3 years ago

4 0

32 degrees, 32 degrees and 116 degrees

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12 x 1) +(6x1000) it has to be decimal form

Firdavs [7]

Answer:

I got 6.012 times 10^3

Step-by-step explanation:

I'm not sure how to make it into decimal form but maybe that number can help you. Hopefully it's correct

6 0

3 years ago

How to solve this problem?

valentinak56 [21]

Very nice handwriting but the math and English are confusing.

Let's assume we're told

$\displaystyle 45 = \sum_{i=1}^9 (x_i - 10)^2$

The subscript is important.

I think we're told the similar sum with 11 gives the smallest possible value for the sum. This is a rather cagey way of telling us 11 is the mean of the nine points. The mean is the number which minimizes the sum of squared deviations.

$\displaystyle 45 = \sum_{i=1}^9 (x_i - 10)^2 = \sum x_i^2 - 20 \sum x_i + 9(100)$

$\displaystyle \sum x_i^2= 20 \sum x_i - 900$

If 11 is the mean, the sum of the points is 9(11)=99.

$\displaystyle \sum x_i^2= 20 (99) - 900 = 1080$

Answer: 1080

6 0

4 years ago

You inherit $5000 from your long lost Uncle Harold. The bad news is that the money must sit in a bank account for the next ten y

erastovalidia [21]

Answer:

5360

Step-by-step explanation:

5000*1.072=5360

6 0

2 years ago

Is this a linear pair? Does it share a common side?

stiv31 [10]

Answer:

Buddy Actually even I am unablr to get it

Step-by-step explanation:

Sorry

4 0

3 years ago

Find the expression for the electric field, E [infinity] , of the ring as the point P becomes very far from the ring ( x ≫ R ) i

Anon25 [30]

Answer:

The expression of the field E as the point P becomes very far from the ring is:

$\vec{E}(x)=\displaystyle\frac{q}{4\pi\epsilon_0} \frac{sgn(x)}{x^2}\vec{x} \\\left \{ {{\vec{E}(x)=\frac{q}{4\pi\epsilon_0} \frac{1}{x^2}\vec{x} \mapsto x>0} \atop {\vec{E}(x)=\frac{q}{4\pi\epsilon_0} \frac{-1}{x^2}\vec{x} \mapsto x< 0 }} \right.$

Step-by-step explanation:

The Electric field expression is:

$\vec{E}(x)=\displaystyle\frac{q}{4\pi\epsilon_0} \frac{x}{(R^2+x^2)^{\frac{3}{2}}}\vec{x}$

To determine the asked expression we use limits. If we consider that x≫R, this is the same as considering the radius insignificant respect the x distance. Therefore we can considerate than from this distance X, the radius R tends to zero:

$\displaystyle\lim_{R \to{}0}{\vec{E}(x)}=\lim_{R \to{}0}{\frac{q}{4\pi\epsilon_0} \frac{x}{(R^2+x^2)^{\frac{3}{2}}}\vec{x}}\rightarrow\frac{q}{4\pi\epsilon_0} \frac{x}{(0^2+x^2)^{\frac{3}{2}}}\vec{x}=\frac{q}{4\pi\epsilon_0} \frac{x}{(x^2)^{\frac{3}{2}}}\vec{x}=\frac{q}{4\pi\epsilon_0} \frac{\cancel{x}}{|x|^{\cancel{3}}}\vec{x}=\displaystyle\frac{q}{4\pi\epsilon_0} \frac{sgn(x)}{x^2}\vec{x}$

5 0

4 years ago