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Naya [18.7K]
2 years ago
14

PLS HELP WITH THIS (this is algebra 1 btw I don't know why it says college) Question in picture

Mathematics
1 answer:
postnew [5]2 years ago
7 0

Answer:

b

Step-by-step explanation:

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Simplify the following number by using the imaginary number i square root of -19
lakkis [162]

Answer:

Exact form:   -√19

Decimal form:  −4.35889894...

3 0
3 years ago
Can any one please help me in number 4 please I really need help please
Reil [10]

Answer:

24

Step-by-step explanation:

4 0
3 years ago
Solve sin 0 + 1 = cos20 on the interval 0 ≤ 0 < 2pi. Show work please!
yan [13]

Answer:

\theta=\frac{\pi}{2},\frac{3\pi}{2}\frac{2\pi}{3}\frac{4\pi}{3}

Step-by-step explanation:

You need 2 things in order to solve this equation:  a trig identity sheet and a unit circle.

You will find when you look on your trig identity sheet that

cos(2\theta)=1-2sin^2(\theta)

so we will make that replacement, getting everything in terms of sin:

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

Now we will get everything on one side of the equals sign, set it equal to 0, and solve it:

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

We can factor out the sin(theta), since it's common in both terms:

sin(\theta)(2sin(\theta)+1)=0

Because of the Zero Product Property, either

sin(\theta)=0 or

2sin(\theta)+1=0

Look at the unit circle and find which values of theta have a sin ratio of 0 in the interval from 0 to 2pi.  They are:

\theta=\frac{\pi}{2},\frac{3\pi}{2}

The next equation needs to first be solved for sin(theta):

2sin(\theta)+1=0 so

2sin(\theta)=-1 and

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

Go back to your unit circle and find the values of theta where the sin is -1/2 in the interval.  They are:

\theta=\frac{2\pi}{3},\frac{4\pi}{3}

7 0
3 years ago
I need help on 1/2/3/4
vaieri [72.5K]

You may need to sit down with your parents or with your teacher and
go over how to add and subtract fractions.

1).  "Perimeter" means the distance all the way around the square.
With a square, all 4 sides are the same length.  With <u>this</u> square,
every side is 1-1/4 inches long.

Perimeter = length of all 4 sides= (1-1/4) + (1-1/4) + (1-1/4) + (1-1/4) =

                                               (1 + 1 + 1 + 1) + (1/4 + 1/4 + 1/4 + 1/4) =

                                                         4           +             4/4  =  <em>5 inches</em> .

2).  (2-3/8) + (1-7/8) = (2 + 1) + (3/8 + 7/8) =

                                        (3)   +    (10/8) =

                                         3    +      1-1/4 =        <em>4-1/4 .</em>

3).  The difference is  (1-1/6) minus (5/6) .

Before you start to do the subtraction, write the (1-1/6)  as  (7/6) .

Then the subtraction is    (7/6) - (5/6)  =  2/6  =  <em>1/3</em> .

4).  This one is almost the same kind of problem as #3. 
It's another subtraction.

If you need (2-1/4) all together, and you already have (1-3/8),
then the amount you still have to find, or borrow, or buy, is the
difference between those two numbers.

           (2-1/4) minus (1-3/8) .

The trick is to write the (2-1/4) in some form that you'll be able to
subtract (1-3/8) from it.  When I learned how to do that, it was called
'borrowing', but I think now it's called 'regrouping'.

We need to work on (2-1/4):

-- take 1 from the 2, and change it into fourths.
 
                 2-1/4  =  1 and 4/4 and 1/4  =  1 and 5/4

-- Now, take those 5/4, and turn them into eighths.                
   Each fourth makes 2 eighths.  So  5/4  =  10/8.

Now, the (2-1/4) has turned into  1-10/8 .
We did NOT change the value.  It's still the same amount
as 2-1/4 , but it's just written in a different way.
And now the subtraction is easy:

         (2-1/4) minus (1-3/8)  = 

       (1-10/8) minus (1-3/8)  =  (zero and 7/8).

You need  <em>7/8 inch</em>  more string than you already have.

         

5 0
3 years ago
--8W + 5 = -43<br> what’s this?
nataly862011 [7]

Answer:

it is the equation which you have to solve.

3 0
3 years ago
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