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goblinko [34]
3 years ago
11

I WILL MARK BRAINLIEST!

Mathematics
2 answers:
rusak2 [61]3 years ago
7 0
Hey looks like u need help I'll help you,


So you'll need your cal for this. But I'll help anyway


A= x(4x-2) then distribute the x to get A=4x²-2x

Then plug in the 8 and get A=4(8)²-2(8)
A=256-16
A=240
Answer is 240


ipn [44]3 years ago
3 0
The answer is 4x2-5x;Area= 114 in<span>2 i think</span>
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Find the distance between the following points<br> R(-2, 3), S(3, 15)
aleksley [76]
Your answer will be: 13
5 0
3 years ago
Plss help!!Need this ASAP!!Daniel began filling an empty swimming pool
beks73 [17]

Answer:

15 hours

Step-by-step explanation:

To find the gallons per hour: 230/5=46, this means that it take 46 gallons per hour, you add 46 to 230, and you do that 15 times and you'll get 920. The pool will be filled at 11:00 pm.

5 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
The first, third and thirteenth terms of an arithmetic sequence are the first 3 terms of a geometric sequence. If the first term
Salsk061 [2.6K]

Answer:

The first three terms of the geometry sequence would be 1, 5, and 25.

The sum of the first seven terms of the geometric sequence would be 127.

Step-by-step explanation:

<h3>1.</h3>

Let d denote the common difference of the arithmetic sequence.

Let a_1 denote the first term of the arithmetic sequence. The expression for the nth term of this sequence (where n\! is a positive whole number) would be (a_1 + (n - 1)\, d).

The question states that the first term of this arithmetic sequence is a_1 = 1. Hence:

  • The third term of this arithmetic sequence would be a_1 + (3 - 1)\, d = 1 + 2\, d.
  • The thirteenth term of would be a_1 + (13 - 1)\, d = 1 + 12\, d.

The common ratio of a geometric sequence is ratio between consecutive terms of that sequence. Let r denote the ratio of the geometric sequence in this question.

Ratio between the second term and the first term of the geometric sequence:

\displaystyle r = \frac{1 + 2\, d}{1} = 1 + 2\, d.

Ratio between the third term and the second term of the geometric sequence:

\displaystyle r = \frac{1 + 12\, d}{1 + 2\, d}.

Both (1 + 2\, d) and \left(\displaystyle \frac{1 + 12\, d}{1 + 2\, d}\right) are expressions for r, the common ratio of this geometric sequence. Hence, equate these two expressions and solve for d, the common difference of this arithmetic sequence.

\displaystyle 1 + 2\, d = \frac{1 + 12\, d}{1 + 2\, d}.

(1 + 2\, d)^{2} = 1 + 12\, d.

d = 2.

Hence, the first term, the third term, and the thirteenth term of the arithmetic sequence would be 1, (1 + (3 - 1) \times 2) = 5, and (1 + (13 - 1) \times 2) = 25, respectively.

These three terms (1, 5, and 25, respectively) would correspond to the first three terms of the geometric sequence. Hence, the common ratio of this geometric sequence would be r = 25 /5 = 5.

<h3>2.</h3>

Let a_1 and r denote the first term and the common ratio of a geometric sequence. The sum of the first n terms would be:

\displaystyle \frac{a_1 \, \left(1 - r^{n}\right)}{1 - r}.

For the geometric sequence in this question, a_1 = 1 and r = 25 / 5 = 5.

Hence, the sum of the first n = 7 terms of this geometric sequence would be:

\begin{aligned} & \frac{a_1 \, \left(1 - r^{n}\right)}{1 - r}\\ &= \frac{1 \times \left(1 - 2^{7}\right)}{1 - 2} \\ &= \frac{(1 - 128)}{(-1)} = 127 \end{aligned}.

7 0
2 years ago
A right rectangular prism is packed with cubes of side length fraction 1 over 5 inch. If the prism is packed with 15 cubes along
Ann [662]

Given:

A right rectangular prism is packed with cubes of side length \dfrac{1}{5} inch.

Prism is packed with 15 cubes along the length, 6 cubes along the width, and 5 cubes along the height.

To find:

The volume of the prism

Solution:

Side length of each cube =\dfrac{1}{5} inch.

15 cubes along the length, so length of prism is

l=15\times \dfrac{1}{5}=3\text{ inches}

6 cubes along the width, so width of prism is

b=6\times \dfrac{1}{5}=\dfrac{6}{5}\text{ inches}

5 cubes along the height, so height of prism is

h=5\times \dfrac{1}{5}=1\text{ inch}

Now,

Volume of rectangular prism is

V=l\times b\times h

V=3\times \dfrac{6}{5}\times 1

V=\dfrac{18}{5}

V=\dfrac{3\times 5+3}{5}

V=3\dfrac{3}{5}\text{ cubic inches}

Therefore, the correct option is B.

7 0
3 years ago
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