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Slav-nsk [51]
3 years ago
8

Rhombus LMNO is shown with its diagonals. Rhombus L M N O is shown. Diagonals are drawn from point M to point O and from point L

to point N and intersect at point P. All sides are congruent. The length of LN is 28 centimeters. What is the length of LP?
Mathematics
1 answer:
LUCKY_DIMON [66]3 years ago
3 0

In a Rhombus the diagonals intersect and divides it in two equal halves.

Therefore LP = 28/2

LP = 14cm

Answered by GauthMath if you like please click thanks and comment thanks

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Austin was budgeted $ 825 to spend on chairs for his upcoming event. If each chair costs $ 15 , how many chairs can he purchase?
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Answer:

55 chairs

Step-by-step explanation:

825/15=55

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Find the median of the following set of data. Round to the nearest tenth if necessary. 26.1, 8.4, 11.4, 44.1, 32.3, 46, 41, 18.5
timofeeve [1]

Answer:25

Step-by-step explanation:

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Perry made $20,000 per year.He soon he received a promotion(raise) with a 20% increase in pay.What is his new salary?How much is
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Hello! The salary goes up 20% due to the promotion. His original salary was $20,000. To find out his new salary, you have to do is multiply the original price by the rate. 20,000 * 20% (0.2) is 4,000. His promotion is $4,000. Now, we add. 20,000 + 4,000 is 24,000. There. Perry's new salary is $24,000.
5 0
4 years ago
Clark and Lana take a 30-year home mortgage of $128,000 at 7.8%, compounded monthly. They make their regular monthly payments fo
svp [43]

Answer:

Step-by-step explanation:

From the given information:

The present value of the house = 128000

interest rate compounded monthly r = 7.8% = 0.078

number of months in a year n= 12

duration of time t = 30 years

To find their regular monthly payment, we have:

PV = P \begin {bmatrix}  \dfrac{1 - (1 + \dfrac{r}{n})^{-nt}}{\dfrac{r}{n}}    \end {bmatrix}

128000 = P \begin {bmatrix}  \dfrac{1 - (1 + \dfrac{0.078}{12})^{- 12*30}}{\dfrac{0.078}{12}}    \end {bmatrix}

128000 = 138.914 P

P = 128000/138.914

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∴ Their regular monthly payment P = $921.433

To find the unpaid balance when they begin paying the $1400.

when they begin the payment ,

t = 30 year - 5years

t= 25 years

PV= 921.433 \begin {bmatrix}  \dfrac{1 - (1 - \dfrac{0.078}{12})^{25*30}}{\dfrac{0.078}{12}}    \end {bmatrix}

PV = $121718.2714

C) In order to estimate how many payments of $1400  it will take to pay off the loan, we have:

121718.2714 =  \begin {bmatrix}  \dfrac{1300  (1 - \dfrac{12.078}{12}))^{-nt}}{\dfrac{0.078}{12}}    \end {bmatrix}

121718.2714 = 200000  \begin {bmatrix}  (1 - \dfrac{12.078}{12}))^{-nt}   \end {bmatrix}

\dfrac{121718.2714}{200000 } =  \begin {bmatrix}  (1 - \dfrac{12.078}{12}))^{-nt}   \end {bmatrix}

0.60859 =  \begin {bmatrix}  (1 - \dfrac{12}{12.078}))^{nt}   \end {bmatrix}

0.60859 = (0.006458)^{nt}

nt = \dfrac{0.60859}{0.006458}

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How much interest will they save by paying the loan using the number of payments from part (c)?

The total amount of interest payed on $921.433 = 921.433 × 30(12) years

= 331715.88

The total amount paid using 921.433 and 1300 = (921.433 × 60 )+( 1300 + 94.238)

= 177795.38

The amount of interest saved = 331715.88  - 177795.38

The amount of interest saved = $153920.5

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3 years ago
<img src="https://tex.z-dn.net/?f=1%20%2B%20%20log_%7B2%7D%28x%20-%202%29%20%20%3D%20%20%20log_%7B2%7Dx" id="TexFormula1" title=
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Move all the logarithms on the left hand side, and all the constants on the other:

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Use the rule of logarithms

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To rewrite the equation as

\log_2\left(\dfrac{x-2}{x}\right) = -1

Evaluate 2 to the power of each side:

\dfrac{x-2}{x} = 2^{-1} = \dfrac{1}{2}

Multiply both sides by 2x:

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