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3 years ago

What decimal is equivalent to -23/20

Mathematics

1 answer:

worty [1.4K]3 years ago

8 0

Answer:

shksjajajakakakskaakaakkaja

-23/20-1 3/20 anamjananananamanamamasnsjjjjjjjjjjjhjgtolbhj

Step-by-step explanation:

jsksjshelejelejekwjwkwl

snmsmsmammaamamms

w.wmmamanama

sssnsnsns hen kwjgwkwgwjwjwnqnqnq

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Nvm I got it now lol

Usimov [2.4K]

Answer:

nice bro good for you :)

Step-by-step explanation:

3 0

2 years ago

Sally is selling strawberries at the farmer’s market. She sells 5 boxes in the morning, and then sells half of the remaining box

Vilka [71]

Answer:

23 boxes

Step-by-step explanation:

x=5+9*2

5 0

3 years ago

Read 2 more answers

Solve the rational equation 8-6/x=5+12/x

Firdavs [7]

Answer:

X = 6

Step-by-step explanation:

4 0

3 years ago

Please Answer Quick 70 Points !

Kipish [7]

Answer:

The ordered pair (7, 19) is a solution to the first equation because it makes the first equation true

The ordered pair (7, 19) is not a solution to the system because it makes at least one of the equations false

Step-by-step explanation:

we have

$2x-y=-5$ -----> First equation

$x+3y=22$ -----> Second equation

Solve the system of equations by graphing

Remember that the solution of the system is the intersection point both graphs

The solution is the point (1,7)

The ordered pair (7,19) lie on the first line, so is a solution of the first equation

see the attached figure

<u><em>Verify each statement</em></u>

case 1) The ordered pair (7, 19) is a solution to the first equation because it makes the first equation true

The statement is true

case 2) The ordered pair (7, 19) is a solution to the second equation because it makes the second equation true

The statement is false, because the ordered pair (7,19) not lie on the second line

case 3) The ordered pair (7, 19) is not a solution to the system because it makes at least one of the equations false

The statement is true

case 4) The ordered pair (7, 19) is a solution to the system because it makes both equations true

The statement is false,because it makes only the first equation true

6 0

3 years ago

Prove by mathematical induction that

postnew [5]

For $n=1$ , on the left we have $\cos\theta$ , and on the right,

$\dfrac{\sin2\theta}{2\sin\theta}=\dfrac{2\sin\theta\cos\theta}{2\sin\theta}=\cos\theta$

(where we use the double angle identity: $\sin2\theta=2\sin\theta\cos\theta$ )

Suppose the relation holds for $n=k$ :

$\displaystyle\sum_{n=1}^k\cos(2n-1)\theta=\dfrac{\sin2k\theta}{2\sin\theta}$

Then for $n=k+1$ , the left side is

$\displaystyle\sum_{n=1}^{k+1}\cos(2n-1)\theta=\sum_{n=1}^k\cos(2n-1)\theta+\cos(2k+1)\theta=\dfrac{\sin2k\theta}{2\sin\theta}+\cos(2k+1)\theta$

So we want to show that

$\dfrac{\sin2k\theta}{2\sin\theta}+\cos(2k+1)\theta=\dfrac{\sin(2k+2)\theta}{2\sin\theta}$

On the left side, we can combine the fractions:

$\dfrac{\sin2k\theta+2\sin\theta\cos(2k+1)\theta}{2\sin\theta}$

Recall that

$\cos(x+y)=\cos x\cos y-\sin x\sin y$

so that we can write

$\dfrac{\sin2k\theta+2\sin\theta(\cos2k\theta\cos\theta-\sin2k\theta\sin\theta)}{2\sin\theta}$

$=\dfrac{\sin2k\theta+\sin2\theta\cos2k\theta-2\sin2k\theta\sin^2\theta}{2\sin\theta}$

$=\dfrac{\sin2k\theta(1-2\sin^2\theta)+\sin2\theta\cos2k\theta}{2\sin\theta}$

$=\dfrac{\sin2k\theta\cos2\theta+\sin2\theta\cos2k\theta}{2\sin\theta}$

(another double angle identity: $\cos2\theta=\cos^2\theta-\sin^2\theta=1-2\sin^2\theta$ )

Then recall that

$\sin(x+y)=\sin x\cos y+\sin y\cos x$

which lets us consolidate the numerator to get what we wanted:

$=\dfrac{\sin(2k+2)\theta}{2\sin\theta}$

and the identity is established.

8 0

3 years ago