Please help, I don’t get it

In the figure below, find the value of cose:

MariettaO [177]

Answer:

D

Step-by-step explanation:

cosΘ = $\frac{adjacent}{hypotenuse}$ = $\frac{x}{r}$

4 0

2 years ago

Please Help Me! It’s Math! Please!

nadezda [96]

Answer:

D. About 800 years

Step-by-step explanation:

Use the half-life equation:

A = A₀ (½)ⁿ

where A is the final amount,

A₀ is the initial amount,

and n is the number of half-lives.

0.90A₀ = A₀ (½)ⁿ

0.90 = (½)ⁿ

To solve for n, take log of both sides:

log 0.9 = n log 0.5

n = (log 0.9) / (log 0.5)

n = 0.152

It takes 0.152 half-lives. The half-life of carbon-14 is 5730 years.

0.152 × 5730 years = 871 years

The closest answer is D.

8 0

3 years ago

Line segment ST is dilated to create line segment S'T' using the dilation rule DQ,2. 25. Point Q is the center of dilation. Line

abruzzese [7]

The distance between points S' and S is x= 2.5 units.

<h2>Given that</h2>

Line segment ST is dilated to create line segment S'T' using the dilation rule DQ 2. 25.

Point Q is the center of dilation.

Line segment ST is dilated to create line segment S prime T prime.

The length of QT is 1. 2 and the length of QS is 2.

The length of SS prime is x and the length of TT prime is 1. 5.

<h3>We have to determine</h3>

What is x, the distance between points S' and S?

<h3>According to the question</h3>

Line segment ST is dilated to create line segment S'T' using the dilation rule DQ, 2.25.

Also, SQ = 2 units, TQ = 1.2 units, TT'=1.5, SS' = x.

Since the line ST is dilated to S'T' with the center of dilation Q, the triangles STQ and S'T'Q must be similar.

We know that the corresponding sides of two similar triangles are proportional.

So, from ΔSTQ and ΔS'T'Q.

$\dfrac{SQ}{S'Q} = \dfrac{TQ}{T'Q}\\
\\
\dfrac{SQ}{SQ+SS} = \dfrac{TQ}{TQ+TT'}\\
\\ 
\dfrac{2}{2+x} = \dfrac{1.2}{1.2+1.5}\\\rm 
\\
\dfrac{2}{2+x} = \dfrac{1.2}{2.7}\\\\ \dfrac{2}{2+x} = \dfrac{12}{27}\\\\2(27) = (2+x) 12\\\\ 54 = 24 + 12x\\
\\
12x = 54-24\\
\\
12x=30\\
\\
x = \dfrac{30}{12}\\
\\
x = 2.5$