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Amanda [17]
3 years ago
15

Last year, Susan opened an investment account with $6800 . At the end of the year, the amount in the account had decreased by 6.

5% . How much is this decrease in dollars? How much money was in her account at the end of last year?
Mathematics
1 answer:
Vadim26 [7]3 years ago
7 0
442 is the decrease
6358 money in her account at the end of last year
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Answer: -80 and -3

Step-by-step explanation: By signed number, do you mean a negative or positive number? If so that means that number one is -80 because she lost 80 dollars which mean that she LOST the money. And the second number is -3 because the number DECREASED

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Rewrite the expression using gcf and distributive property 56x7
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(cf)times g

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How are area and volume similar? both are 2-dimensional both take up an amount of space both are 3-dimensional both have length,
VARVARA [1.3K]

Answer:

128

Step-by-step explanation:

The area of the larger rectangle is \begin{align*}8(16)=128 \ \mathrm{units}^2\end{align*} and the area of the smaller rectangle is \begin{align*}5(10)=50 \ \mathrm{units}^2.\end{align*} If we compare the areas in a ratio, it is \begin{align*}50:128 \ = \ 25:64 \ = \ 5^2:8^2.\end{align*}

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Can anyone help me here asapp,, I am in this question for nearly an hour
zhannawk [14.2K]

Answer:

See below

Step-by-step explanation:

Let side AB equal x. Since triangle ABC is equilateral, sides AB, BC, and Ac are all the same length, x. In any isosceles triangle(equilateral is a type of isosceles triangle) the median is the same as the altitude and angle bisector. This means we can say that AD is also a median. A median splits a side into two equal sections, so we can say BD = DC = x / 2. We are given that DC = CE, so we can also say CE = DC = x / 2. Now, we can use the pythagorean theorem to find the length of AD. So we get the equation:

AB^2 - BD^2 = AD^2

We have the values of AB and BD, so we can substitute them and solve for AD:

x^2 - (x/2)^2 = AD^2

x^2 - x^2 / 4 = AD^2

AD^2 = 3x^2 / 4

AD = x√3 / 2

DE is equal to the sum of DC and CE because of segment addition postulate, so we can say DE = DC + CE = x / 2 + x/ 2 = x. We can again use the pythagorean theorem to find the length of AE:

AD^2 + DE^2 = AE^2

(x√3 / 2)^2 + x^2 = AE^2

3x^2 / 4 + x^2 = AE^2

AE^2 = 7x^2 / 4

AE = x√7 / 2

Now, we know(from before) that AE squared is 7x^2 / 4. We can say EC squared is x^2 / 4 because EC is x / 2 and x / 2 squared is x^2 / 4. We can also notice that AE squared is 7 times EC squared because 7x^2 / 4 = 7 * x^2 / 4

Therefore, we can come to the conclusion AE^2 = 7 EC^2

7 0
3 years ago
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