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Allushta [10]
3 years ago
7

What is the value of x? (giving brainliest and thanks to all!)

Mathematics
1 answer:
astra-53 [7]3 years ago
4 0

Answer:

x° = 64°

Explaination:

This is a right angled triangle so one of the angle is 90°.

So, x° + 90° = 154° (Exterior angle property)

x° = 154° - 90° = 64°

Therefore x° = 64°

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Solve for W in A = 2(L + W)
Roman55 [17]

\Large\begin{aligned}\\\\&A = 2(L + W)\\&A = 2L+2W\\&2W=A-2L\\&W=\dfrac{A-2L}{2}\end

4 0
2 years ago
A fruit company sells two types of boxes large and small. A delivery of two large boxes and three small boxes has a total weight
lara31 [8.8K]

Answer:

the large box weight is 6.50 kg and the small box weight is 13.75 kg

Step-by-step explanation:

The computation of each type of box weight is as follows:

Let us assume the large box be x

And, the small box be y

So,

2x + 3y = 47.......(i)

6x + 5y = 115........(ii)

Multiply by 3 in equation 1

6x + 9y = 141

6x + 5y = 115

Now subtract the last equation from the above one

4y = 26

y = 6.50

For x, it would be

2x+ 3(6.50) = 47

2x + 19.5 = 47

2x = 47 - 19.50

2x = 27.50

x = 13.75

Hence, the large box weight is 6.50 kg and the small box weight is 13.75 kg

4 0
3 years ago
Write out the first four terms of the series to show how the series starts. Then find the sum of the series or show that it dive
Nostrana [21]

Answer:

The first four terms of the series are

(9+3),(\frac97+\frac35),(\frac9{7^2}+\frac3{5^2}),(\frac9{7^3}+\frac3{5^3})

\sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n} = 14.25

Step-by-step explanation:

We know that

Sum of convergent series is also a convergent series.

We know that,

\sum_{k=0}^\infty a(r)^k

If the common ratio of a sequence |r| <1 then it is a convergent series.

The sum of the series is \sum_{k=0}^\infty a(r)^k=\frac{a}{1-r}

Given series,

\sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n}

=(9+3)+(\frac97+\frac35)+(\frac9{7^2}+\frac3{5^2})+(\frac9{7^3}+\frac3{5^3})+.......

The first four terms of the series are

(9+3),(\frac97+\frac35),(\frac9{7^2}+\frac3{5^2}),(\frac9{7^3}+\frac3{5^3})

Let

S_n=\sum_{n=0}^\infty \frac{9}{7^n}    and     t_n=\sum_{n=0}^\infty \frac{3}{5^n}

Now for S_n,

S_n=9+\frac97+\frac{9}{7^2}+\frac9{7^3}+.......

    =\sum_{n=0}^\infty9(\frac 17)^n

It is a geometric series.

The common ratio of S_n is \frac17

The sum of the series

S_n=\sum_{n=0}^\infty \frac{9}{7^n}

    =\frac{9}{1-\frac17}

    =\frac{9}{\frac67}

    =\frac{9\times 7}{6}

    =10.5

Now for t_n

t_n= 3+\frac35+\frac{3}{5^2}+\frac3{5^3}+.......

    =\sum_{n=0}^\infty3(\frac 15)^n

It is a geometric series.

The common ratio of t_n is \frac15

The sum of the series

t_n=\sum_{n=0}^\infty \frac{3}{5^n}

    =\frac{3}{1-\frac15}

    =\frac{3}{\frac45}

    =\frac{3\times 5}{4}

    =3.75

The sum of the series is \sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n}

                                        = S_n+t_n

                                       =10.5+3.75

                                       =14.25

4 0
3 years ago
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