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ad-work [718]
3 years ago
11

Please help me correct answer if you said it correct I will give you brainlist please!​

Mathematics
1 answer:
oee [108]3 years ago
5 0
The answer would be 5 like they said
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What is the length of the hypotenuse for a right triangle with legs of length 6 and 8?
Vlad [161]

Answer:

hyp = sqRoot(6^2 + 8^2) = sqRoot(100) = 10

Step-by-step explanation:

6 0
3 years ago
Find the sum of the geometric series 40 + 40(1.005) + 40(1.005)^2 + ⋯ + 40(1.005)^11.
KiRa [710]

Answer:

The sum is 493.4

Step-by-step explanation:

In order to find the value of the sum, you have to apply the geometric series formula, which is:

\sum_{i=1}^{n} ar^{i-1} = \frac{a(1-r^{n})}{1-r}

where i is the starting point, n is the number of terms, a is the first term and r is the common ratio.

The finite geometric series converges to the expression in the right side of the equation. Therefore, you don't need to calculate all the terms. You can use the expression directly.

In this case:

a=40

b= 1.005

n=12 (because the first term is 40 and the last term is 40(1.005)^11 )

Replacing in the formula:

\frac{a(1-r^{n})}{1-r} = \frac{40(1-1.005^{12})}{1-1.005}

Solving it:

The sum is 493.4

4 0
3 years ago
Find all possible values of α+
const2013 [10]

Answer:

\rm\displaystyle  0,\pm\pi

Step-by-step explanation:

please note that to find but α+β+γ in other words the sum of α,β and γ not α,β and γ individually so it's not an equation

===========================

we want to find all possible values of α+β+γ when <u>tanα+tanβ+tanγ = tanαtanβtanγ</u><u> </u>to do so we can use algebra and trigonometric skills first

cancel tanγ from both sides which yields:

\rm\displaystyle  \tan( \alpha )  +  \tan( \beta ) =  \tan( \alpha )  \tan( \beta )  \tan( \gamma )  -  \tan( \gamma )

factor out tanγ:

\rm\displaystyle  \tan( \alpha )  +  \tan( \beta ) =   \tan( \gamma ) (\tan( \alpha )  \tan( \beta ) -  1)

divide both sides by tanαtanβ-1 and that yields:

\rm\displaystyle   \tan( \gamma ) =  \frac{ \tan( \alpha )  +  \tan( \beta ) }{ \tan( \alpha )  \tan( \beta )    - 1}

multiply both numerator and denominator by-1 which yields:

\rm\displaystyle   \tan( \gamma ) =   -  \bigg(\frac{ \tan( \alpha )  +  \tan( \beta ) }{ 1 - \tan( \alpha )  \tan( \beta )   } \bigg)

recall angle sum indentity of tan:

\rm\displaystyle   \tan( \gamma ) =   -  \tan( \alpha  +  \beta )

let α+β be t and transform:

\rm\displaystyle   \tan( \gamma ) =   -  \tan( t)

remember that tan(t)=tan(t±kπ) so

\rm\displaystyle   \tan( \gamma ) =    -\tan(   \alpha   +\beta\pm k\pi )

therefore <u>when</u><u> </u><u>k </u><u>is </u><u>1</u> we obtain:

\rm\displaystyle   \tan( \gamma ) =    -\tan(   \alpha   +\beta\pm \pi )

remember Opposite Angle identity of tan function i.e -tan(x)=tan(-x) thus

\rm\displaystyle   \tan( \gamma ) =    \tan(   -\alpha  -\beta\pm \pi )

recall that if we have common trigonometric function in both sides then the angle must equal which yields:

\rm\displaystyle  \gamma  =      -   \alpha   -  \beta \pm \pi

isolate -α-β to left hand side and change its sign:

\rm\displaystyle \alpha  +  \beta  +   \gamma  =  \boxed{ \pm \pi  }

<u>when</u><u> </u><u>i</u><u>s</u><u> </u><u>0</u>:

\rm\displaystyle   \tan( \gamma ) =    -\tan(   \alpha   +\beta \pm 0 )

likewise by Opposite Angle Identity we obtain:

\rm\displaystyle   \tan( \gamma ) =    \tan(   -\alpha   -\beta\pm 0 )

recall that if we have common trigonometric function in both sides then the angle must equal therefore:

\rm\displaystyle  \gamma  =      -   \alpha   -  \beta \pm 0

isolate -α-β to left hand side and change its sign:

\rm\displaystyle \alpha  +  \beta  +   \gamma  =  \boxed{ 0  }

and we're done!

8 0
3 years ago
Read 2 more answers
What is 35 divided by 97. 3?
Whitepunk [10]
0.35971223021 is the product

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3 0
3 years ago
I don’t understand the equation
ankoles [38]

you need to find to find the value of x and y.

2x>(-15-5y)

x>(-15-5y)/2

now put this in to other equation and solve for y and then put the y value in and find x.

6 0
3 years ago
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