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REY [17]
3 years ago
5

Help ASAP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics
1 answer:
natali 33 [55]3 years ago
8 0

Answer:

A Swedish court said Friday that rapper ASAP Rocky will be detained for another week as prosecutors decide whether to formally charge him following a fight last month in downtown Stockholm.

The rapper, born Rakim Mayers, has been in Swedish custody, along with two of his associates, since early July after a June 30 altercation that Rocky’s supporters say he tried to avoid. Under the court’s decision Friday, the men could be detained until at least July 25.

Concerns have mounted for the men amid reports that they are being held in “inhumane conditions,” including long daily periods of solitary confinement. A growing number of lawmakers and celebrities, including Kim Kardashian and Kanye West, has called for their release.

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What is the solution of 6(a-5)-4a=4​
Romashka [77]

Answer:

Hi! The answer to your question is a = 17

Step-by-step explanation:

☆*: .。..。.:*☆☆*: .。..。.:*☆☆*: .。..。.:*☆☆*: .。..。.:*☆

☆Brainliest is greatly appreciated!☆

Hope this helps!!

- Brooklynn Deka

6 0
3 years ago
the hypotenuse of a right triangle is 24ft long. The length of one leg is 20ft more than the other. Find the length of the legs.
Amiraneli [1.4K]
By the Pythagorean Theorem, the hypotenuse squared is equal to the sum of the sides squared...

h^2=x^2+y^2  

We are told that y=x+20 and h=24 so

x^2+(x+20)^2=24^2

x^2+x^2+40x+400=576

2x^2+40x+400=576

2x^2+40x=176

x^2+20x=88

x^2+20x+100=188

(x+10)^2=188

x+10=±√188

x=-10±√188, x>0 so

x=-10+√188 ft

y=10+√188 ft

If you want approximations...

x≈3.71 ft

y≈23.71 ft
4 0
4 years ago
What’s the Missing Side? <br> PLEASEE HELPP
frez [133]

Answer:

missing side =19 ft

Step-by-step explanation:

area of a triangle =base *height/2

57=base*6/2

57*2=base*6

114=base*6

114/6=base

19 ft=base

6 0
3 years ago
James uses 5/6 meter if butcher paper to make 1 sign how many meters of paper will he need to make 3 signs
bekas [8.4K]

Answer:

2.5 meters of butcher paper.

Step-by-step explanation:

5/6(per 1 sign) times 3(signs) = 2.5

7 0
2 years ago
Read 2 more answers
Solve sinX+1=cos2x for interval of more or equal to 0 and less than 2pi
Igoryamba

Answer:

Question 1: \sin(x)+1=\cos^2(x)

Answer to Question 1: x=0, \pi \frac{3\pi}{2}

Question 2: \sin(x)+1=\cos(2x)

Answer to Question 2: 0,\pi,\frac{7\pi}{6}, \frac{11\pi}{6}

Question:

I will answer the following two questions.

Condition: 0\le x

Question 1: \sin(x)+1=\cos^2(x)

Question 2: \sin(x)+1=\cos(2x)

Step-by-step explanation:

Question 1: \sin(x)+1=\cos^2(x)

Question 2: \sin(x)+1=\cos(2x)

Question 1:

\sin(x)+1=\cos^2(x)

I will use a Pythagorean Identity so that the equation is in terms of just one trig function, \sin(x).

Recall \sin^2(x)+\cos^2(x)=1.

This implies that \cos^2(x)=1-\sin^2(x). To get this equation from the one above I just subtracted \sin^2(x) on both sides.

So the equation we are starting with is:

\sin(x)+1=\cos^2(x)

I'm going to rewrite this with the Pythagorean Identity I just mentioned above:

\sin(x)+1=1-\sin^2(x)

This looks like a quadratic equation in terms of the variable: \sin(x).

I'm going to get everything to one side so one side is 0.

Subtracting 1 on both sides gives:

\sin(x)+1-1=1-\sin^2(x)-1

\sin(x)+0=1-1-\sin^2(x)

\sin(x)=0-\sin^2(x)

\sin(x)=-\sin^2(x)

Add \sin^2(x) on both sides:

\sin(x)+\sin^2(x)=-\sin^2(x)+\sin^2(x)

\sin(x)+\sin^2(x)=0

Now the left hand side contains terms that have a common factor of \sin(x) so I'm going to factor that out giving me:

\sin(x)[1+\sin(x)]=0

Now this equations implies the following:

\sin(x)=0 or 1+\sin(x)=0

\sin(x)=0 when the y-coordinate on the unit circle is 0. This happens at 0, \pi, or also at 2\pi. We do not want to include 2\pi because of the given restriction 0\le x.

We must also solve 1+\sin(x)=0.

Subtract 1 on both sides:

\sin(x)=-1

We are looking for when the y-coordinate is -1.

This happens at \frac{3\pi}{2} on the unit circle.

So the solutions to question 1 are 0,\pi,\frac{3\pi}{2}.

Question 2:

\sin(x)+1=\cos(2x)

So the objective at the beginning is pretty much the same. We want the same trig function.

\cos(2x)=\cos^2(x)-\sin^2(x) by double able identity for cosine.

\cos(2x)=(1-\sin^2(x))-\sin^2(x) by Pythagorean Identity.

\cos(2x)=1-2\sin^2(x) (simplifying the previous equation).

So let's again write in terms of the variable \sin(x).

\sin(x)+1=\cos(2x)

\sin(x)+1=1-2\sin^2(x)

Subtract 1 on both sides:

\sin(x)+1-1=1-2\sin^2(x)-1

\sin(x)+0=1-1-2\sin^2(x)

\sin(x)=0-2\sin^2(x)

\sin(x)=-2\sin^2(x)

Add 2\sin^2(x) on both sides:

\sin(x)+2\sin^2(x)=-2\sin^2(x)+2\sin^2(x)

\sin(x)+2\sin^2(x)=0

Now on the left hand side there are two terms with a common factor of \sin(x) so let's factor that out:

\sin(x)[1+2\sin(x)]=0

This implies \sin(x)=0 or 1+2\sin(x)=0.

The first equation was already solved in question 1. It was just at x=0.

Let's look at the other equation: 1+2\sin(x)=0.

Subtract 1 on both sides:

2\sin(x)=-1

Divide both sides by 2:

\sin(x)=\frac{-1}{2}

We are looking for when the y-coordinate on the unit circle is \frac{-1}{2}.

This happens at \frac{7\pi}{6} or also at \frac{11\pi}{6}.

So the solutions for this question 2 is 0,\pi,\frac{7\pi}{6}, \frac{11\pi}{6}.

8 0
3 years ago
Read 2 more answers
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