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2 years ago

Construct equation

Mathematics

1 answer:

ElenaW [278]2 years ago

6 0

Step-by-step explanation:

(x × 5) +6=41

5x + 6=41

5x =41-6

5x=35

x=(35/5)

x=7

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Please help answer the following question

Elena L [17]

Answer:

Step-by-step explanation:

The unmarked angle of the triangle is also 40o. The triangle is marked with two equal sides (isosceles).

<2 + 40 + 40 = 180 Combine the left

<2 + 80 = 180 Subtract 80 from both sides

<2 = 180 - 80

<2 = 100 degrees.

That's not the answer. We have to solve for x

<2 = x + 106

100 = x + 106 Subtract 106 from both sides

100 - 106 = x

x = - 6

8 0

2 years ago

Subtract 4 from n then multiply by 6

geniusboy [140]

Answer:

multiply 4 and 6 and the answer should be n-24

Step-by-step explanation:

n-4*6

n-24

3 0

2 years ago

A group of 5 painters takes 21 hours to paint a house. How long would it take a group of 7 painters to paint the same house?

VladimirAG [237]

Answer:

29.4

Step-by-step explanation:

if 5 painters takes 21 hours that makes it 4.5 per person so multiply that by 7 and there you go!

5 0

2 years ago

Read 2 more answers

Can segments with lengths of 15, 20, and 36 form a triangle?

strojnjashka [21]

Ok, so remember

the legnth of the longest side must be LESS THAN the sum of the measures of the other 2 sides
if no longest side (becasue it has 2 longest sides or 3 equal sides), then it can form a triangle

so the longest side is 36

36 must be less than the sum of 15 and 20
36<15+20
36<35
false

therfor it cannot form a triangle
try it yourself, you cannot connect them that way

5 0

2 years ago

<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