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2 years ago

Find the equation of the line shown.

Mathematics

1 answer:

ExtremeBDS [4]2 years ago

3 0

Answer: y = -1/4x + 2
explanation: line passes through y axis on 2, so 2 is your y-intercept. rise/run is ur slope, which is -1/4 .

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Given the ellipse represented by the following equation, where are the foci?<br><br> VIEW IMAGE

kicyunya [14]

Answer:

Step-by-step explanation:

81 - 16 = 65

formula c^2 = a^2 - b^2

c^2 = 81 - 16

√c^2 = √ 65

c = √ 65

7 0

2 years ago

F(x) = 3x <br>reflected over y-axis

mylen [45]

Answer:

g(x)=-3x

Step-by-step explanation:

Reflecting any equation over the y-axis will make it negative, hence 3x->-3x

8 0

3 years ago

What is the answer to c-(5c-3)

oee [108]

Add like terms to get -4c = 3. divide by ,-4 each side so c = -3/4

4 0

3 years ago

V=4/3 3.14 18*18*18 what is the volume of the sphere

podryga [215]

Answer:

24416.64

Step-by-step explanation:

Based on the standard formula,4/3×π×r^3

The π has been given already and r=18

4/3×3.14×18^3

V=24416.64

8 0

3 years ago

The value V of a certain automobile that is t years old can be modeled by V(t) = 14,651(0.81). According to the model, when will

lakkis [162]

Answer:

a) The car will be worth $8000 after 2.9 years.

b) The car will be worth $6000 after 4.2 years.

c) The car will be worth $1000 after 12.7 years.

Step-by-step explanation:

The value of the car after t years is given by:

$V(t) = 14651(0.81)^{t}$

According to the model, when will the car be worth V(t)?

We have to find t for the given value of V(t). So

$V(t) = 14651(0.81)^{t}$

$(0.81)^t = \frac{V(t)}{14651}$

$\log{(0.81)^{t}} = \log{(\frac{V(t)}{14651})}$

$t\log{(0.81)} = \log{(\frac{V(t)}{14651})}$

$t = \frac{\log{(\frac{V(t)}{14651})}}{\log{0.81}}$

(a) $8000

V(t) = 8000

$t = \frac{\log{(\frac{8000}{14651})}}{\log{0.81}} = 2.9$

The car will be worth $8000 after 2.9 years.

(b) $6000

V(t) = 6000

$t = \frac{\log{(\frac{6000}{14651})}}{\log{0.81}} = 4.2$

The car will be worth $6000 after 4.2 years.

(c) $1000

V(t) = 1000

$t = \frac{\log{(\frac{1000}{14651})}}{\log{0.81}} = 12.7$

The car will be worth $1000 after 12.7 years.

5 0

2 years ago