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3 years ago

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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3 years ago

Solve for W in A = 2(L + W)

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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