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2 years ago

If triangle LMN has an obtuse angle at vertex M, which statements could be true? Check all that apply.

Mathematics

2 answers:

nirvana33 [79]2 years ago

5 0

Answer: Triangle LMN is an obtuse triangle.

The angle at vertex L is acute.

The angle at vertex N is acute.

Step-by-step explanation:

Here, triangle LMN has an obtuse angle at vertex M,

Thus, by the definition of obtuse angle triangle LMN is an obtuse triangle,

Now, Angle M is obtuse,

⇒ 90° < m∠ M < 180°

Since, by the property of a triangle,

m∠ L + m∠ M + m∠ N = 180°

⇒ m∠ M = 180° - ( m∠ L+ m∠ N )

⇒ 90° < 180° - ( m∠ L+ m∠ N ) < 180°

⇒ 90° - 180° < - ( m∠ L+ m∠ N ) < 0

⇒ -90° < - ( m∠ L+ m∠ N ) < 0

⇒ 90° > ( m∠ L+ m∠ N ) > 0 ( Since, a < b ⇒ - a > - b)

⇒ 90° > m∠ L and 90° > m∠ N

⇒ Both angle L and angle N are acute.

musickatia [10]2 years ago

4 0

Triangle LMN is an obtuse triangle
The angle at vertex L is acute
The angle at vertex N is acute

1,2,5

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How many weeks of flea treatment would Jim’s dog get from one package if each treatment only lasted 3 weeks In a normal year whe

4vir4ik [10]

Solution

Number of weeks in a year = 52 weeks

If in a normal year when each package of flea treatment lasts for 4 weeks, then in a year there Jim's dog will have to be treated for

$\frac{52}{4} =13 times$

Where as, when the fleas are bad in a year, the treatment lasts for only 3 weeks.

Then in a year Jim's dog would get

$\frac{52}{3} =17 times$

So Jim's dog will get 17- 13 =4 treatments more.

4 treatments that are made in 3 weeks each will be 4×3 =12 weeks more treatment

4 0

2 years ago

A village was founded four hundred years ago by a group of 20 people. In this village, the population triples every one hundred

Paha777 [63]

Answer:

Step-by-step explanation:

Treat this like compound interest: Use A = P(1 + r)^t.

Here, P is the initial population and A is 3 times that, or 3P. Since P = 20 people, 3P = 60 people,

and this population is reached after 100 years.

We need to determine r, substitute its value into the formula A = P(1 + r)^t, and then determine the population of the village after 400 years.

60 = 20(1 + r)^100

Simplifying, 3 = (1 + r)^100.

Taking the natural log of both sides,

ln 3 = 100 ln (1 + r), or

ln 3

ln (1 + r) = ---------------

100

= 1.0986 / 100 = 0.01986

We must solve this for r. Raising e to the power ln (1 + r), on the left side of an equation, and raising e to the power 0. 01986 on the right side, we get:

1 + r = 3, so r must = 2.

Now find the pop of the village today. Use the same equation: A = P (1+r)^t.

A = 20(1 +2)^4 (hundreds),

A = 20(3)^4, or

A = 81

The population after 400 years is 81.

8 0

2 years ago

Write an expression that evaluates to true if the int associated with number_of_prizes is divisible (with no remainder) by the i

hodyreva [135]

We need to find the expression for " number_of_prizes is divisible number_of_participants". Also there should not remain any remainder left. On in order words, we can say the reaminder we get after division is 0.

Let us assume number of Prizes are = p and

Number of participants = n.

If we divide number of Prizes by number of participants and there will be not remainder then there would be some quotient remaining and that quotent would be a whole number.

Let us assume that quotent is taken by q.

So, we can setup an expression now.

Let us rephrase the statement .

" Number of Prizes ÷ Number of participants = quotient".

p ÷ n = q.

In fraction form we can write

p/n =q ; n ≠ 0.

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2 years ago

The square of a number decreased by 3 times the number 28 find all possible values for the number

stealth61 [152]

Question:

The square of a number decreased by 3 times the number is 28 find all possible values for the number

Answer:

The possible values of number are 7 and -4

Solution:

Given that the square of a number decreased by 3 times the number is 28

To find: all possible values of number

Let "a" be the unknown number

From given information,

square of a number decreased by 3 times the number = 28

$a^2 - 3a = 28$

$a^2 - 3a - 28 = 0$

Let us solve the above quadratic equation

$\text {For a quadratic equation } a x^{2}+b x+c=0, \text { where } a \neq 0$

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Using the above formula,

$\text { For } a^{2}-3 a-28=0 \text { we have } a=1, b=-3, c=-28$

$\begin{aligned}&a=\frac{-(-3) \pm \sqrt{(-3)^{2}-4(1)(-28)}}{2 \times 1}\\\\&a=\frac{3 \pm \sqrt{9+112}}{2}\\\\&a=\frac{3 \pm \sqrt{121}}{2}=\frac{3 \pm 11}{2}\\\\&a=\frac{3+11}{2} \text { or } a=\frac{3-11}{2}\\\\&a=7 \text { or } a=-4\end{aligned}$

Thus the possible values of number are 7 and -4

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2 years ago

Josiah kept track of how many songs of each genre were played in an hour from his MP3 player. The counts are displayed in the ta

Evgen [1.6K]

Answer:

1) He should have found the average of the number of rock songs by averaging 4 and 6 to get 5.

2) He did not multiply the numerator and denominator by the correct number to equal 1,500.

5) He should have multiplied the numerator and denominator by 75, not 30, because 20 x 75 = 1,500.

Step-by-step explanation:

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2 years ago

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