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3 years ago

The polygons below are similar. Find the value of x.

Mathematics

2 answers:

satela [25.4K]3 years ago

6 0

Answer:

Step-by-step explanation:

Inessa05 [86]3 years ago

5 0

Answer:

x = 16

Step-by-step explanation:

We can use ratios to solve

8 x

-------- = ---------------

6 12

Using cross products

8 * 12 = 6x

Divide each side by 6

96 = 6x

Divide by 6

96/6 = 6x/6

16 = x

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Eight thousand six and thirty-four thousandths<br> * Write as a decimal

valentinak56 [21]

Answer:

8,006.034

Step-by-step explanation:

eight thousand six = 8,006

and = .

thirty-four thousandths = .034

5 0

3 years ago

Complete the equation. Round to the nearest hundredth where necessary.

Alexandra [31]

Answer:

Step-by-step explanation:

There are 1.61 km in 1 mi.

10 mi × (1.61 km/mi) = 16.1 km

7 0

3 years ago

Please help I'll mark brainliest

Anon25 [30]

Answer:

Step-by-step explanation:

Add up all the numbers and then divide by how many numbers there are.

3 0

2 years ago

Which graph shows the solution set of the compound inequality or 1.5x-1 > 6.5 or 7x+3 < -25 ?

Dmitry [639]

1.
<span>1.5x-1 > 6.5
1.5x>1+6.5
</span>1.5x>7.5, divide by 1.5

x>5, is represented by the region to the right of the vertical line x=5

2.

<span>7x+3 < -25
7x<-25-3
7x<-28, divide by 4:

x<-4

</span>x<-4, is represented by the region to the left of the vertical line x=-4

Answer: check the picture

6 0

2 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=prove%20that%5C%20%20%5Ctextless%20%5C%20br%20%2F%5C%20%20%5Ctextgreater%20%5C%20%5Cfrac%20%7B

inysia [295]

$\large \bigstar \frak{ } \large\underline{\sf{Solution-}}$

Consider, LHS

$\begin{gathered}\rm \: \dfrac { \tan \theta + \sec \theta - 1 } { \tan \theta - \sec \theta + 1 } \\ \end{gathered}$

We know,

$\begin{gathered}\boxed{\sf{ \:\rm \: {sec}^{2}x - {tan}^{2}x = 1 \: \: }} \\ \end{gathered} \\ \\ \text{So, using this identity, we get} \\ \\ \begin{gathered}\rm \: = \:\dfrac { \tan \theta + \sec \theta - ( {sec}^{2}\theta - {tan}^{2}\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered}$

We know,

$\begin{gathered}\boxed{\sf{ \:\rm \: {x}^{2} - {y}^{2} = (x + y)(x - y) \: \: }} \\ \end{gathered} \\$

So, using this identity, we get

$\begin{gathered}\rm \: = \:\dfrac { \tan \theta + \sec \theta - (sec\theta + tan\theta )(sec\theta - tan\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered}$

can be rewritten as

$\begin{gathered}\rm\:=\:\dfrac {(\sec \theta + tan\theta ) - (sec\theta + tan\theta )(sec\theta -tan\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:\dfrac {(\sec \theta + tan\theta ) \: \cancel{(1 - sec\theta + tan\theta )}} { \cancel{ \tan \theta - \sec \theta + 1} } \\ \end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:sec\theta + tan\theta \\\end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:\dfrac{1}{cos\theta } + \dfrac{sin\theta }{cos\theta } \\ \end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:\dfrac{1 + sin\theta }{cos\theta } \\ \end{gathered}$

<h2>Hence,</h2>

$\begin{gathered} \\ \rm\implies \:\boxed{\sf{ \:\rm \: \dfrac { \tan \theta + \sec \theta - 1 } { \tan \theta - \sec \theta + 1 } = \:\dfrac{1 + sin\theta }{cos\theta } \: \: }} \\ \\ \end{gathered}$

$\rule{190pt}{2pt}$

5 0

2 years ago