Is depositing $85 in a bank account a positive or a negative number

1 answer:

3 0

To "deposit" into a bank account means you're adding money into it.
If the accounts money contained was "0" before you added the 85$ then the amount in the account is 85$.
It is positive.

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Looking the area in the triangle

TEA [102]

There’s nothing there..?

7 0

3 years ago

#if a sum become rs 6480 in 3 years and rs 7776 in 4 years interest being compounded annually, find the sum and rate of interest

Andrei [34K]

Answer:

The rate of interest is 20% and the sum is $3,750

Step-by-step explanation:

In order to calculate the sum and rate of interest we would have to make the following calculation:

rate of interest= (sum in 4 years-sum in 3 years)*100/sum in 3 years*1

According to the given data we have the following:

sum in 4 years=$7,776

sum in 3 years=$6,480

Therefore, sum in 4 years-sum in 3 years=$7,776-$6,480=$1,296

Therefore, rate of interest=$1,296*100/$6,480*1

rate of interest=20%

To calculate the sum we would have to make the following calculation:

FV=PV(1+20%)∧3

$6,480=PV(1,20)∧3

PV=$3,750

Sum is $3,750

8 0

3 years ago

Quadrilateral ABCD is similar to Quadrilateral EFGH. Diagonal AC has length 7 and diagonal EG has length 13. What is the scale f

olasank [31]

Answer:

$\large \boxed{\text{A. }\dfrac{13}{7}\text{; B. }\dfrac{17}{14} \text{; C. 507 in}^{2}}$

Step-by-step explanation:

A. Scale factor

When you dilate an object by a scale factor, you multiply its line lengths by the same number.

If EF/AB = 13/7, the scale factor is 13/7.

B. Length of EF

$\begin{array}{rcl}\dfrac{EF}{AB} & = & \dfrac{13}{7}\\\\\dfrac{EF}{\frac{17}{26}} & = & \dfrac{13}{7}\\\\EF & = & \dfrac{13}{7}\times\dfrac{17}{26}\\\\ & = &\dfrac{1}{7}\times\dfrac{17}{2}\\\\ & = & \mathbf{\dfrac{17}{14}}\\\end{array}\\\text{The length of EF is $\large \boxed{\mathbf{ \dfrac{17}{14}}}$}$

C. Area of EFGH

If the lengths in a shape are all multiplied by a scale factor, then the areas will be multiplied by the scale factor squared.

ABCD is dilated by a scale factor of 13/7, so its area is dilated by a scale factor of

$\left(\dfrac{13}{7} \right)^{2} = \dfrac{169}{49}$

The area of its dilated image EFGH is

$\text{Area of EFGH} = \text{147 in}^{2} \times \dfrac{\text{169}}{\text{49}} = 3 \times 169\text{ in}^{2} = 507 \text{ in}^{2}\\\\\text{The area of EFGH is $\large \boxed{\textbf{507 in}^{\mathbf{2}}}$}$