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Anastaziya [24]
3 years ago
11

Pls help

Mathematics
1 answer:
denpristay [2]3 years ago
5 0
The cost of the bond was 90/100 times $5,000, that is, $4,500 total. 

<span>The annual interest is 5% of $5,000, that is, $250. </span>

<span>The current yield is 5% divided by (90/100), that is 5.555%; round to 5.6% as instructed. </span>

<span>The yield that real bond buyers would be more interested in is the yield to maturity, but this cannot be calculated without knowing the term (number of years). If it's a short term bond that will pay you back $5,000 in just a few years, that would add several percent to the yield, but if it's a 15-year bond the growth of the $4,500 to $5,000 adds only a fraction of 1% to the yield.</span>
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Simplify <br><br><img src="https://tex.z-dn.net/?f=%28x%5E%7B2%7D%20%20%2B%203x%20-%204%29%20%2B%20%284%5E%7B2%7D%20%20-%205%29%
xxTIMURxx [149]

X^2-4x+4 is the answer, to save you points next time, use an app called M8thw8y

3 0
3 years ago
Read 2 more answers
Which description does NOT guarantee that a quadrilateral is a square?
Ivahew [28]
Let's go through the choices one by one

------------------------------------------
Choice A

If all sides are congruent, then this figure is a rhombus (by definition). If all angles are congruent, then we have a rectangle. Combine the properties of a rhombus with the properties of a rectangle and we have a square.

In terms of "algebra", you can think
rhombus+rectangle = square

Or you can draw out a venn diagram. One circle represents the set of all rhombuses; another circle represents the set of all rectangles. The overlapping region is the set of all squares. The overlapping region is inside both circles at the same time.

So we can rule out choice A. This guarantees we have a square when we want something that isn't a guarantee.

------------------------------------------
Choice B

If we had a parallelogram with perpendicular diagonals, then we can prove that we have a rhombus (all four sides congruent). However, we don't know anything about the four angles of this parallelogram. Are they congruent? We don't know. So we can't prove this figure is a rectangle. The best we can say is that it's a rhombus. It may or may not be a rectangle. There isn't enough info about the rectangle & square part.

This is why choice B is the answer. We have some info, but not enough to be guaranteed everytime.

------------------------------------------
Choice C

This is a repeat of choice A. Having "all right angles" is the same as saying "all angles congruent". This is because "right angle" is the same as saying "90 degrees". So we can rule out choice C for identical reasons as we did with choice A.

------------------------------------------
Choice D

As mentioned before in choice A, if we know that a quadrilateral is a rectangle and a rhombus at the same time, then the figure is also a square. This is always true, so we are guaranteed to have a square. We can cross choice D off the list.

------------------------------------------

Once again, the final answer is choice B


3 0
3 years ago
I NEED THIS ANSWERED AS SOON AS POSSIBLE! How much material is needed to construct a triangular tent that is 6 feet wide, 4 feet
Elena-2011 [213]

To figure out this problem, you need to find the surface area of a triangular prism.

a=5ft

b=5ft

c=6ft

h=8ft

A= (a*h) + (b*h) + (c*h) + 1/2√-a⁴ + 2(a*b)² + 2(a*c) - b⁴ + 2(b*c) - c⁴

=152 square feet

5 0
3 years ago
Alexander Litvinenko was poisoned with 10 micrograms of the radioactive substance Polonium-210. Since radioactive decay follows
koban [17]

Answer:

The amount of Polonium-210 left in his body after 72 days is 6.937 μg.

Step-by-step explanation:

The decay rate of Polonium-210 is the following:

N(t) = N_{0}e^{-\lambda t}     (1)

Where:

N(t) is the quantity of Po-210 at time t =?

N₀ is the initial quantity of Po-210 = 10 μg

λ is the decay constant  

t is the time = 72 d  

The decay rate is 0.502%, hence the quantity that still remains in Alexander is 99.498%.    

First, we need to find the decay constant:

\lambda = \frac{ln(2)}{t_{1/2}}    (2)

Where t(1/2) is the half-life of Po-210 = 138.376 days

By entering equation (2) into (1) we have:

N(t) = N_{0}e^{-\frac{ln(2)}{t_{1/2}}*t}} = 10* \frac{99.498}{100}*e^{-\frac{ln(2)}{138.376}*72} = 6.937 \mu g    

Therefore, the amount of Polonium-210 left in his body after 72 days is 6.937 μg.  

I hope it helps you!  

8 0
2 years ago
How do you write the number for 3,152,308
Sergeu [11.5K]

If you mean in word form, then it should be as follows;

Three million one hundred and fifty-two thousand three hundred and eight.

If you don't mean word form, you will have to be a little more specific.

8 0
3 years ago
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