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worty [1.4K]
2 years ago
5

M= -2 b=15 write the equation of the line​

Mathematics
1 answer:
NeTakaya2 years ago
8 0

Answer:

y=-2x+15

Step-by-step explanation:

Hi there!

We are given that m=-2, b=15, and we want to write the equation of the line, given these values

We can write the line in slope-intercept form, which is y=mx+b, where m is the slope and b is the y intercept, since we are given the values of both m and b.

Since we know the values of m and b (-2 and 15 respectively), we can substitute those numbers in as the variables they equal to.

Substitute -2 as m:

y=-2x+b

Substitute 15 as b:

y=-2x+15

Hope this helps!

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Josephine purchased a used vehicle that depreciates under a straight-line
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Answer:

D. $4000

Step-by-step explanation:

Initial value of the car is given $5000

Salvage value given $500

Useful life of car given is 5 years

Formula

Depreciation for n years is given as:

d_{n}= \textrm{Salvage value}\times \textrm{Number of years used}

d_{n}=\textrm{Salvage Value}\times n

Plug in 2 for n. This gives,

d_{2}=500\times 2=1000

∴ Depreciation after 2 years =$ 1000

Value of car after 2 years = Initial value - Depreciation after 2 years

Value of car after 2 years = 5000- 1000=4000

Therefore, value of car after 2 years = $ 4000

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2 years ago
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B. closer to 4 than to five
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n △ABC, point P∈ AB is so that AP:BP=1:3 and point M is the midpoint of segment CP. Find the area of △ABC if the area of △BMP is
monitta

Answer: The area of ABC is 56 m².

Explanation:

It is given that in △ABC, point P∈ AB is so that AP:BP=1:3 and point M is the midpoint of segment CP.

Since point P divides the line AB in 1:3, therefore the area of triangle APC and BPC is also in ratio 1:3. To prove this draw a perpendicular h on AB from C.

\frac{\text{Area of } \triangle BCP}{\text{Area of } \triangle ABC} =\frac{\frac{1}{2}\times BP\times CH}{\frac{1}{2}\times AB\times CH} =\frac{BP}{AB}= \frac{3}{4}

Since the area of BPC is \frac{3}{4}th part of total area, therefore area of APC is  \frac{1}{4}th part of total area.

The point M is the midpoint of CP, therefore the area of BMP and BMC is equal by midpoint theorem.

\text{Area of } \triangle BMP=\text{Area of } \triangle BMC

21=\text{Area of } \triangle BMC

Area of BPC is,

\text{Area of } \triangle BPC=\text{Area of } \triangle BMP+\text{Area of } \triangle BMC

\text{Area of } \triangle BPC=21+21

\text{Area of } \triangle BPC=42

Area of APC is,

\text{Area of } \triangle APC=\frac{1}{3}\times \text{Area of } \triangle BPC

\text{Area of } \triangle APC=\frac{1}{3}\times 42

\text{Area of } \triangle APC=14

Area of ABC is,

\text{Area of } \triangle ABC=\text{Area of } \triangle APC+\text{Area of } \triangle BPC

\text{Area of } \triangle ABC=14+42=56

Therefore, the area of ABC is 56 m².

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