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3 years ago

Consider the triangle.

Mathematics

2 answers:

mr Goodwill [35]3 years ago

8 0

For this case we observe that the three angles of the triangle are equal. Also, we have by definition:

A triangle is equilateral, when the three sides have the same length (in addition, the three internal angles measure 60 degrees or $\frac {\pi} {3}$ .

Answer:

The three sides have the same length

Aleonysh [2.5K]3 years ago

5 0

ANSWER

The three sides have the same length

EXPLANATION

The given triangle has all angles equal to 60°

A triangle that has all angles equal is called an equilateral triangle.

Another property about an equilateral triangle is that, it has all sides equal.

Therefore,the three sides have the same length.

The third option is correct.

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Your teacher is giving a test worth 200 points. There are a total of 32 four-point and

adelina 88 [10]

A+b=32
4a+7b=200

a = 15
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There are 15 four pointers and 20 seven pointers. You can plug in the numbers to check.

I believe this is correct. If not i am extremely sorry.
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3 years ago

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At which root does the graph of f(x)=(x+4)^6(x+7)^5 cross the x axis

jarptica [38.1K]

Answer:

-7.

Step-by-step explanation:

At the point -4 the graph just touches the x axis ( as its an even power) but at x = -7 it crosses the x axis ( because of the odd power 5).

The graph rises from the left and pass through the x axis at (-7,0).

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3 years ago

Julissa is running a 10-kilometer race at a constant pace. After running for 18 minutes, she completes 2 kilometers. After runni

vodomira [7]

First, we need her speed as it will serve as the slope of the equation.
Distance = 2 km
time = 18 min
Speed = 1/9 km/min
k = (1/9)t
Where k is the distance in km and t is the time in minutes.

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3 years ago

(-2/3, sqrt5/3) is a point on a unit circle. Find the cosine, cosecant, and sine of the angle.

Nana76 [90]

Look at the picture.

$\csc\theta=\dfrac{1}{\sin\theta}=\dfrac{1}{\frac{y}{r}}=\dfrac{r}{y}$

We have the right triangle x, y and r. From the Pythagorean theorem we have:

$r^2=x^2+y^2\to r=\sqrt{x^2+y^2}$

We have the point

$\left(-\dfrac{2}{3};\ \dfrac{\sqrt5}{3}\right)$

Substitute:

$r=\sqrt{\left(-\dfrac{2}{3}\right)^2+\left(\dfrac{\sqrt5}{3}\right)^2}\\\\r=\sqrt{\dfrac{4}{9}+\dfrac{5}{9}}\\\\r=\sqrt{\dfrac{9}{9}}\\\\r=1$

$\csc\theta=\dfrac{1}{\frac{\sqrt5}{3}}=\dfrac{3}{\sqrt5}=\dfrac{3\cdot\sqrt5}{\sqrt5\cdot\sqrt5}=\boxed{\dfrac{3\sqrt5}{5}}$

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3 years ago

Given: KLMN is a trapezoid m∠K = 90°, m∠N = 45° LK = LM = 10 Find: KN, Area of KLMN

Ilya [14]

Let KLMN be a trapezoid (see added picture). From the point M put down the trapezoid height MP, then quadrilateral KLMP is square and KP=MP=10.
A triangle MPN is right and <span>isosceles, because
</span>m∠N=45^{0}, m∠P=90^{0}, so m∠M=180^{0}-45^{0}-90^{0}=45^{0}.Then PN=MP=10.
The ttapezoid side KN consists of two parts KP and PN, each of them is equal to 10, then KN=20 units.
Area of KLMN is egual to $A= \frac{LM+KN}{2} *MP= \frac{10+20}{2} *10=150$ sq. units.

5 0

3 years ago