You have botles of water how many can you buy with a 20$ and a bottle is 1.25

The points P and Q in the following figure define the line segment (PQ⎯⎯⎯⎯⎯) P Q . The steps involved in using a compass to cons

kenny6666 [7]

You haven't provided the steps.

mathisfun.com/geometry/construct-linebisect.html

Here is a useful link to the correct steps. The instructions may not be exactly the same but I think you can do it.

3 0

3 years ago

Please find the exact length of the midsegment of trapezoid JKLM with vertices J(6, 10), K(10, 6), L(8, 2), and M(2, 2). Thank y

I am Lyosha [343]

Answer:

the exact length of the midsegment of trapezoid JKLM = $\mathbf{ = 3 \sqrt{5} }$ i.e 6.708 units on the graph

Step-by-step explanation:

From the diagram attached below; we can see a graphical representation showing the mid-segment of the trapezoid JKLM. The mid-segment is located at the line parallel to the sides of the trapezoid. However; these mid-segments are X and Y found on the line JK and LM respectively from the graph.

Using the expression for midpoints between two points to determine the exact length of the mid-segment ; we have:

$\mathbf{ YX = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} }$

$\mathbf{ YX = \sqrt{(8-5)^2+(8-2)^2} }$

$\mathbf{ YX = \sqrt{(3)^2+(6)^2} }$

$\mathbf{ YX = \sqrt{9+36} }$

$\mathbf{ YX = \sqrt{45} }$

$\mathbf{ YX = \sqrt{9*5} }$

$\mathbf{ YX = 3 \sqrt{5} }$

Thus; the exact length of the midsegment of trapezoid JKLM = $\mathbf{ = 3 \sqrt{5} }$ i.e 6.708 units on the graph

8 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

Bananas cost $.65 per pound. Write an equation that can be used to find C, the total cost of P pounds of bananas.

aksik [14]

The equation for this is

$.65p=c

In order to find the total cost we must

have a number in which is multiplied by

$.65.

7 0

2 years ago

EASY MATH ILL GIVE BRIANLIST PLEASE HELP, IT HAS TO BE CORRECT

Ipatiy [6.2K]

Here is the answer and steps. Answer:-3/5

3 0

2 years ago

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You have botles of water how many can you buy with a 20$ and a bottle is 1.25​

You have botles of water how many can you buy with a 20$ and a bottle is 1.25