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3 years ago

Triangle LMN is a right triangle. true or false

Mathematics

2 answers:

Vlad1618 [11]3 years ago

8 0

Where is the triangle? You're missing a diagram

nlexa [21]3 years ago

8 0

The answer is that it is FALSE because a right angle forms 90 degrees. It not a right triangle
:) :) :)

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At his last track meet, Eli ran the 400-meter dash in 54 seconds. His friend Kimberly bet him that he couldn't run a whole kilom

Trava [24]

Answer:

Eli has to run for 2.25 minutes at this pace to win the bet.

Step-by-step explanation:

At his last track meet, Eli ran the 400-meter dash in 54 seconds.

So, the speed of Eli's run is $\frac{400}{54} \times 60 = 444.44$ meters per minute.

Now, if Eli is determined to run a whole kilometer with the same speed, then it will take Eli to run 1 kilometer within $\frac{1000}{444.44} = 2.25$ minutes.

Hence, Eli has to run for 2.25 minutes at this pace to win the bet. (Answer)

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2 years ago

What do you get when you combine the following numbers? -4-4-4-4

sladkih [1.3K]

Answer:

-16

Step-by-step explanation:

you just add all of them together

4 0

1 year ago

What is the answer to number 18?

dmitriy555 [2]

Answer: 3n ×9

Step-by-step explanation:

The product of 3 times a N number equals 3n or 3 ×N but you should write 3n then times it by 9

7 0

3 years ago

What is the coefficient of x2y3 in the expansion of (2x + y)5?

Zigmanuir [339]

Option C:

The coefficient of $x^{2} y^{3}$ is 40.

Solution:

Given expression:

$(2 x+y)^{5}$

Using binomial theorem:

$(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}$

Here $a=2 x, b=y$

Substitute in the binomial formula, we get

$(2x+y)^5=\sum_{i=0}^{5}\left(\begin{array}{l}5 \\i\end{array}\right)(2 x)^{(5-i)} y^{i}$

Now to expand the summation, substitute i = 0, 1, 2, 3, 4 and 5.

$$=\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}+\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1}+\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}+\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}$

$$+\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}+\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}$

Let us solve the term one by one.

$$\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}=32 x^{5}$

$$\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1} = 80 x^{4} y$

$$\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}= 80 x^{3} y^{2}$

$$\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}= 40 x^{2} y^{3}$

$$\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}= 10 x y^{4}$

$$\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}=y^{5}$

Substitute these into the above expansion.

$(2x+y)^5=32 x^{5}+80 x^{4} y+80 x^{3} y^{2}+40 x^{2} y^{3}+10 x y^{4}+y^{5}$

The coefficient of $x^{2} y^{3}$ is 40.

Option C is the correct answer.

5 0

2 years ago

Find the y -intercept of the line with equation y=4x+6

solmaris [256]

Answer: The y-intercept of this equation is 6.

Explanation: When doing slope intercept form, the formula is y = mx + b. "M" is slope, and "B" is y-intercept. In this equation, 4 is the slope and 6 is the y-intercept.

8 0

2 years ago