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2 years ago

11

Can someone answer these questions I really need help! I’ll give brainliest if I can!

1 answer:

Vedmedyk [2.9K]2 years ago

5 0

Answer:

ill help u but

Step-by-step explanation:

<h2>MARK ME BRAINLIEST PLZZZZZZZZZZZzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzMARK ME BRAINLIEST PLZZZZZZZZZZZzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz</h2>

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If 5k = -25, then 5k - 1 = -25 - 1<br> Segment proof

Pavlova-9 [17]

Yeah cuz u could +1 to each side of the second equation to get rid of both the -1’s and that is equivalent to the first equation :)

6 0

2 years ago

Geometry question on finding x.

Dmitrij [34]

Answer:

$x =63^o$

Step-by-step explanation:

<u>Given </u><u>:</u><u>-</u><u> </u>

$\text{ $\overline{MN} $ and $\overline{ML}$ are tangents .}$

We know that the tangents are perpendicular to the radius at the point of contact . And here OLMN is a quadrilateral. Also we know that , the angle sum property a triangle is 360° .

<u>According</u><u> to</u><u> question</u><u> </u><u>:</u><u>-</u><u> </u>

$\implies x + 117^o + 90^o +90^o = 360^o \\\\\implies x + 297^o = 360^o \\\\\implies x = 360^o - 297^o \\\\\implies x = 63^o$

<u>Hence</u><u> the</u><u> </u><u>value</u><u> of</u><u> x</u><u> </u><u>is </u><u>6</u><u>3</u><u>°</u><u> </u><u>.</u>

6 0

2 years ago

Solve the equation. Round to the nearest tenth if necessary.

ASHA 777 [7]

The answer will be 30

5 0

2 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

Need help with all questions please. :)

vovangra [49]

3. a) p∝1/m
P=k/m
48=k/9
48×9=k
k=432

the equation is: p=432/m

b) p∝1/m
P=k/m
P=432/12
P=36

5 0

3 years ago