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2 years ago

7

Answer this please, I am really struggling, I know I have been toxic in the past - answering questions just for the points, but

now I have changed and I want you to answer this question sincerely, with no hate.
#BLM

1 answer:

dmitriy555 [2]2 years ago

8 0

Answer:

4x+4x+42+2x-12=180 Degrees (Angle Sum of a Triangle)

10x+30=180

10x=150

x=15

Step-by-step explanation:

Hope this helps! You will always be forgiven if you change. :)

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What is the perimeter of this equilateral triangle? <br> Choices <br> 4<br> 20<br> 40<br> 60

aliina [53]

Answer:

60

Step-by-step explanation:

If it is an equilateral triangle the sides are equal

24-x = 5x

Add x to each side

24-x+x = 5x+x

24 =6x

Divide by 6

24/6 =6x/6

4 =x

Each side is 5x or 5*4 = 20

Add the 3 equal sides

20+20+20 = 60

The perimeter is 60

8 0

3 years ago

Read 2 more answers

A rectangle has dimensions of 3 feet by 1 foot. What is the length of the diagonal of the

Andreyy89

Answer:

Step-by-step explanation:

3^2 + 1^2 = c^2 (diagonal squared)

9 + 1 = c^2

10 = c*2

$\sqrt{10}$ is the length

It is irrational because the decimal repeats forever

6 0

2 years ago

One the first day of school, each student is given $0.50 to attend. On day two, each student earns $1.00, day 3, $2.00, etc. How

aalyn [17]

Answer:

They earn $256 on the ninth day of school.

Step-by-step explanation:

A geometric sequence(in which each term is the previous term multiplied by the ratio) has the following format:

$a_{n+1} = r \times a_{n}$

In which r is the common ratio.

We can also find the nth term as a function of the first term, by the following equation:

$a_{n} = a_{1} \times r^{n}$

In which

$r = \frac{a_{n+1}}{a_{n}}$

In this question:

On the first day, $0.50. So $a_{1} = 0.5$ .

Day two $1.00, day three $2.00.

So

$r = \frac{1}{0.5} = \frac{2}{1} = 2$

Then

$a_{n} = 0.5 \times 2^{n}$

How much do students earn on the ninth day of school?

$a_{9} = 0.5 \times 2^{9} = 256$

They earn $256 on the ninth day of school.

3 0

3 years ago

Find the vectors T, N, and B at the given point. r(t) = < t^2, 2/3t^3, t >, (1, 2/3 ,1)

maxonik [38]

Answer with Step-by-step explanation:

We are given that

$r(t)=< t^2,\frac{2}{3}t^3,t >$

We have to find T,N and B at the given point t > (1, $2/3$ ,1)

$r'(t)=$

$\mid r'(t) \mid=\sqrt{(2t)^2+(2t^2)^2+1}=\sqrt{(2t^2+1)^2}=2t^2+1$

$T(t)=\frac{r'(t)}{\mid r'(t)\mid}=\frac{}{2t^2+1}$

Now, substitute t=1

$T(1)=\frac{}{2+1}=\frac{1}{3}$

$T'(t)=\frac{-4t}{(2t^2+1)^2} +\frac{1}{2t^2+1}$

$T'(1)=-\frac{4}{9}+\frac{1}{3}$

$T'(1)=\frac{1}{9}=$

$\mid T'(1)\mid=\sqrt{(\frac{-2}{9})^2+(\frac{4}{9})^2+(\frac{-4}{9})^2}=\sqrt{\frac{36}{81}}=\frac{2}{3}$

$N(1)=\frac{T'(1)}{\mid T'(1)\mid}$

$N(1)=\frac{}{\frac{2}{3}}=$

$N(1)=$

$B(1)=T(1)\times N(1)$

$B(1)=\begin{vmatrix}i&j&k\\\frac{2}{3}&\frac{2}{3}&\frac{1}{3}\\\frac{-1}{3}&\frac{2}{3}&\frac{-2}{3}\end{vmatrix}$

$B(1)=i(\frac{-4}{9}-\frac{2}{9})-j(\frac{-4}{9}+\frac{1}{3})+k(\frac{4}{9}+\frac{2}{9})$

$B(1)=-\frac{2}{3}i+\frac{1}{3}j+\frac{2}{3}k$

$B(1)=\frac{1}{3}$

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3 years ago

For 3x + 12x2 + 12x3, which of the following statements is true?

maksim [4K]

Answer:

A:the polynomial is a trinomial

8 0

3 years ago