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yuradex [85]
2 years ago
14

I NEED HELP PLEASE directions:Work and Solution

Mathematics
1 answer:
Volgvan2 years ago
4 0

Answer:

2(x + 4) / 6(x² - 3x - 28)

Step-by-step explanation:

Area of a rectangle = length × width

Length = 2/(x² - 3x - 28)

Width = x² - 16/6x - 24

= (x + 4)(x - 4) / 6(x - 4)

= (x + 4) / 6

Area of a rectangle = length × width

= 2/(x² - 3x - 28) × (x + 4) / 6

= 2(x + 4) / (x² - 3x - 28)6

= 2(x + 4) / 6x² - 18x - 168

= 2(x + 4) / 6(x² - 3x - 28)

Area of a rectangle =

2(x + 4) / 6(x² - 3x - 28)

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A car is purchase at $20000.The value of the car depreciates at 12.5% a year. What will the value of the car be to the nearest c
Rudik [331]

Answer: $4603.82

Step-by-step explanation:

Given

The initial cost of a car is P=\$20,000

The rate of depreciation is i=12.5\%

Depreciated value after n years is given by

\Rightarrow P(1-\frac{i}{100})^n\quad \text{where n=no. of years}

Value after 11 years

\Rightarrow 20,000(1-\frac{12.5}{100})^{11}\\\\\Rightarrow 20000(0.875)^{11}=\$4603.82

6 0
3 years ago
Please help me with this problem
inn [45]

Answer:  Option 1

<u>Step-by-step explanation:</u>

\begin {array}{l|c|c||l}&\underline{\quad Length\quad}&\underline{\quad Width\quad}&\underline{\qquad Area\qquad}\\ Current&3x&x&3x\cdot x=3x^2\\Option\ 1\ (triple)&3(3x)=9x&3(x)=3x&9x\cdot 3x=27x^2\\Option\ 2\ (3\ closets)&3x&x&3(3x\cdot x)=9x^2\\\end{array}

Option 1 has a greater area than option 2.

4 0
3 years ago
Read 2 more answers
Divide the number in the thousands place by itself and then multiply the answer by 0. Write your answer in the tenths place
Ivanshal [37]
Number: 1111
1111/1111=1
1*0= 0
Now put 0 in the tenths place
01

Hope this Helped :)
5 0
3 years ago
(64-8)÷4 answer ASAP please​
Ann [662]

Answer:

14

Step-by-step explanation:

64-8=56

56/4=14

4 0
3 years ago
Verify the trigonometric identity. convert the left side of the relationship to look like the right side.
Tomtit [17]

Answer:

Verified

Step-by-step explanation:

Use \tan{x}=\frac{\sin{x}}{\cos{x}}  , \sin^2{x}+\cos^2{x}=1 and \sec{x}=\frac{1}{\cos{x}}

LHS=\tan^2{x}+1=\frac{\sin^2{x}}{\cos^2{x}} + 1=\frac{\sin^2{x}+\cos^2{x}}{\cos^2{x}}=\frac{1}{\cos^2{x}}=\sec^2{x}=RHS

5 0
2 years ago
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