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bearhunter [10]
3 years ago
11

Point A is located at (0,4) and point B is located at (-2,-3). Find the x value for the point that is 1/4 the distance from poin

t A to point B
Mathematics
1 answer:
fiasKO [112]3 years ago
8 0
To solve these questions, use this formula: X3= F(X1-X2) Y3= F(Y1-Y2) Where X3 is the X value in the coordinate your solving for, F is the fraction, X1 is the first X and X2 is the second X Plug in your numbers to fit this: X3= 1/4 x (0--2) Y3= 1/4 x (4- -3) By solving you get: (-.5, 1.75)
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Evaluate 3x+y when x=1/3 and y=−7/4. Write your answer as a fraction in simplest form.
SOVA2 [1]

Answer:

3x+y

=3*1/3-7/4

=1-7/4

=(4-7)/4

=-3/4

6 0
2 years ago
Please answer it now in two minutes
zimovet [89]

Answer:

59.0

Step-by-step explanation:

Given a right angled triangle, ∆XYZ, to know which trigonometric ratio formula to apply in finding the measure of angle X, note the following:

Opposite side to angle X = 6

Hypotenuse = 7

Therefore, we would apply the following trigonometric ratio formula to solve for m<X:

sin X = \frac{6}{7}

sin X = 0.8571

X = sin^{-1}(0.8571)

X = 58.99

m < X = 59.0 (rounded to nearest tenth)

4 0
3 years ago
Graph the function. y=2-x
oee [108]
Its the 3rd from the left good luck

4 0
3 years ago
Best Buy is having a sale on computers. Katherine picks out a computer that was originally $1099.99
borishaifa [10]

Answer:

$901.9918

Step-by-step explanation:

$901.9918

7 0
3 years ago
Animal populations are not capable of unrestricted growth because of limited habitat and food supplies. Under such conditions th
trasher [3.6K]

Answer:

(a) 100 fishes

(b) t = 10: 483 fishes

    t = 20: 999 fishes

    t = 30: 1168 fishes

(c)

P(\infty) = 1200

Step-by-step explanation:

Given

P(t) =\frac{d}{1+ke^-{ct}}

d = 1200\\k = 11\\c=0.2

Solving (a): Fishes at t = 0

This gives:

P(0) =\frac{1200}{1+11*e^-{0.2*0}}

P(0) =\frac{1200}{1+11*e^-{0}}

P(0) =\frac{1200}{1+11*1}

P(0) =\frac{1200}{1+11}

P(0) =\frac{1200}{12}

P(0) = 100

Solving (a): Fishes at t = 10, 20, 30

t = 10

P(10) =\frac{1200}{1+11*e^-{0.2*10}} =\frac{1200}{1+11*e^-{2}}\\\\P(10) =\frac{1200}{1+11*0.135}=\frac{1200}{2.485}\\\\P(10) =483

t = 20

P(20) =\frac{1200}{1+11*e^-{0.2*20}} =\frac{1200}{1+11*e^-{4}}\\\\P(20) =\frac{1200}{1+11*0.0183}=\frac{1200}{1.2013}\\\\P(20) =999

t = 30

P(30) =\frac{1200}{1+11*e^-{0.2*30}} =\frac{1200}{1+11*e^-{6}}\\\\P(30) =\frac{1200}{1+11*0.00247}=\frac{1200}{1.0273}\\\\P(30) =1168

Solving (c): \lim_{t \to \infty} P(t)

In (b) above.

Notice that as t increases from 10 to 20 to 30, the values of e^{-ct} decreases

This implies that:

{t \to \infty} = {e^{-ct} \to 0}

So:

The value of P(t) for large values is:

P(\infty) = \frac{1200}{1 + 11 * 0}

P(\infty) = \frac{1200}{1 + 0}

P(\infty) = \frac{1200}{1}

P(\infty) = 1200

5 0
3 years ago
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