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KatRina [158]
4 years ago
12

What is the value of the expression below? 3/8+(-4/5)+(-3/8)+5/4

Mathematics
2 answers:
Tamiku [17]4 years ago
7 0
Before I put in the actual expression, let's figure out which sign is which. A positive sign and a negative sign make a negative sign. This applies for the first two pairs of signs. That would make it :

\frac{3}{8} - \frac{4}{5} - \frac{3}{8} + \frac{5}{4}

First thing is first : let's subtract the first two fractions. \frac{3}{8} - \frac{4}{5}. We would have to find the common multiple between 8 and 5, and that would be 40. So we change them both to 40, and we do to the numerator what we do to the denominator. \frac{15}{40} - \frac{32}{40}. That makes \frac{-17}{40}. (Now we subtract this and the next fraction. Again, we find the common multiple. The common multiple is 40.)

\frac{-17}{40} - \frac{3}{8} turns into \frac{-17}{40} - \frac{15}{40}. (-17 - 15 is -32.)

\frac{-32}{40} + \frac{5}{4} : (use common multiple, it changes) \frac{-32}{40} + \frac{50}{40}.

(-32 + 50 is 18.) \frac{18}{40}.

Simplified : \frac{9}{20}

ANS. IN FRACTION FORM : \frac{9}{20}
ANS. IN DECIMAL FORM : 0.45
Crazy boy [7]4 years ago
4 0
It will help you!

=-17/40+-3/8+5/4

= -4/5+5/4

= 9/20 

And the answer is 9/20 or decimals (0.45)

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mrs_skeptik [129]

Let's do

\\ \rm\dashrightarrow y=\dfrac{3}{x-h}+k

Release k for some while

If

  • we take h=0

So

  • y=3/x

So vertical asymptote is at origin now

It mentioned that it's at x=-5 so we need to change x

  • put -5 in place of h

\\ \rm\dashrightarrow y=\dfrac{3}{x-(-5)}

\\ \rm\dashrightarrow y=\dfrac{3}{x+5}

  • Vertical asymptote at x=-5

Now

  • for k=0 horizontal asymptote at origin

But it's given

  • y is at 12

Same put y=12 in place of k

\\ \rm\dashrightarrow y=\dfrac{3}{x+5}+12

  • h=-5
  • k=12

Graph attached for verification

3 0
2 years ago
The denarius was a unit of currency in ancient rome. Suppose it costs the roman government 101010 denarii per day to support 444
givi [52]

<u>the correct question is</u>

The denarius was a unit of currency in ancient rome. Suppose it costs the roman government 10 denarii per day to support 4 legionaries and 4 archers. It only costs 5 denarii per day to support 2 legionaries and 2 archers. Use a system of linear equations in two variables. Can we solve for a unique cost for each soldier?

Let

x-------> the cost to support a legionary per day

y-------> the cost to support an archer per day

we know that

4x+4y=10 ---------> equation 1

2x+2y=5 ---------> equation 2

If you multiply equation 1 by 2

2*(2x+2y)=2*5-----------> 4x+4y=10

so

equation 1 and equation 2 are the same

The system has infinite solutions-------> Is a consistent dependent system

therefore

<u>the answer is</u>

We cannot solve for a unique cost for each soldier, because there are infinite solutions.

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4 0
3 years ago
Derive these identities using the addition or subtraction formulas for sine or cosine: sinacosb=(sin(a+b)+sin(a-b))/2
Sergeu [11.5K]

Answer:

The work is in the explanation.

Step-by-step explanation:

The sine addition identity is:

\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b).

The sine difference identity is:

\sin(a-b)=\sin(a)\cos(b)-\cos(a)\sin(a).

The cosine addition identity is:

\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b).

The cosine difference identity is:

\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b).

We need to find a way to put some or all of these together to get:

\sin(a)\cos(b)=\frac{\sin(a+b)+\sin(a-b)}{2}.

So I do notice on the right hand side the \sin(a+b) and the \sin(a-b).

Let's start there then.

There is a plus sign in between them so let's add those together:

\sin(a+b)+\sin(a-b)

=[\sin(a+b)]+[\sin(a-b)]

=[\sin(a)\cos(b)+\cos(a)\sin(b)]+[\sin(a)\cos(b)-\cos(a)\sin(b)]

There are two pairs of like terms. I will gather them together so you can see it more clearly:

=[\sin(a)\cos(b)+\sin(a)\cos(b)]+[\cos(a)\sin(b)-\cos(a)\sin(b)]

=2\sin(a)\cos(b)+0

=2\sin(a)\cos(b)

So this implies:

\sin(a+b)+\sin(a-b)=2\sin(a)\cos(b)

Divide both sides by 2:

\frac{\sin(a+b)+\sin(a-b)}{2}=\sin(a)\cos(b)

By the symmetric property we can write:

\sin(a)\cos(b)=\frac{\sin(a+b)+\sin(a-b)}{2}

3 0
3 years ago
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