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timama [110]
3 years ago
11

Write an equation in slope intercept form for the line that passes through (1,8) and (2,11)

Mathematics
2 answers:
svet-max [94.6K]3 years ago
8 0

Answer:

both b and d cross those points

Step-by-step explanation:

solved them

SVETLANKA909090 [29]3 years ago
3 0

im not really sure but im tginking b

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Describe the transformation required to obtain the graph of the given function from the basic trigonometric graph.
mamaluj [8]

Answer:

Option c. Reflection across the x-axis and vertical stretch by a factor of 7

Step-by-step explanation:

If the graph of the function y = cf(x) represents the transformations made to the graph of y = f(x) then, by definition:

If 0 then the graph is compressed vertically by a factor a.

If |c| > 1 then the graph is stretched vertically by a factor a.

If c then the graph is reflected on the x axis.

In this problem we have the function y = -7secx and our paretn function is y = secx

therefore it is true that

c = -7\\\\|-7|> 1\\\\-7.

Therefore the graph of y = secx is stretched vertically by a factor of 7 and is reflected on the x-axis

Finally the answer is Option c

6 0
3 years ago
Could anyone please help?​
natima [27]

Answer:

b+3a

Step-by-step explanation:

MN=AN+AM

AN=0.5*AB

AB=OA+OB=4a+2b

AN=2a+b

AM=OA-OM=4a-3a=a

MN=2a+b+a=3a+b

3 0
3 years ago
Read 2 more answers
The current year’s depreciation expense using the straight-line method and assuming only one depreciable property, is P20,000. W
denpristay [2]

Answer:

Machine's useful number of years = 9 years

Step-by-step explanation:

Using the straight line method, depreciation is calculated as the difference between the cost of the equipment minus the salvage value, all divided by the number of useful years.

Yearly Depreciation

= (Cost - Salvage value) ÷ (Number of useful years)

Yearly depreciation = P20,000

Cost = P200,000

Salvage Value = P20,000

Number of useful years = n

20000 = (200000 - 20000) ÷ n

20000 = (180000/n)

n = (180000/20000) = 9 years

Hope this Helps!!!

8 0
3 years ago
Orlando invests $1000 at 6% annual interest compounded daily and Bernadette invests $1000 at 7%
Montano1993 [528]

Answer:

6 Years

Step-by-step explanation:

Orlando invests $1000 at 6% annual interest compounded daily.

Orlando's investment = A=1000(1+\frac{0.06}{365})^{(365\times t)}

Bernadette invests $1000 at 7% simple interest.

Bernadette's investment = A = 1000(1+0.07×t)

By trail and error method we will use t = 5

Bernadette's investment will be after 5 years

1000(1 + 0.07 × 5)

= 1000(1 + 0.35)

= 1000 × 1.35

= $1350

Orlando's investment after 5 years

A=1000(1+\frac{0.06}{365})^{(365\times 5)}

   = 1000(1+0.000164)^{1825}

  = 1000(1.000164)^{1825}

  = 1000(1.349826)

  = 1349.825527 ≈ $1349.83

After 5 years Orlando's investment will not be more than Bernadette's.

Therefore, when we use t = 6

After 6 years Orlando's investment will be = $1433.29

and Bernadette's investment will be = $1420

So, after 6 whole years Orlando's investment will be worth more than Bernadette's investment.

8 0
3 years ago
A survey is taken of 104 people who vacationed at a
ExtremeBDS [4]

Answer:

What's the question?

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
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