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2 years ago

What is the scale factor of 5.8 and 1.45

Mathematics

2 answers:

Marianna [84]2 years ago

4 0

No clue at all, Soorieee

wlad13 [49]2 years ago

3 0

Answer:the scale factor is 4.

Step-by-step explanation:hope this helps!

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If a tank holds 5000 gallons of water, which drains from the bottom of the tank in 40 minutes, then Torricelli's Law gives the v

pochemuha

Answer:

V'(t) = $-250(1 - \frac{1}{40}t)$

If we know the time, we can plug in the value for "t" in the above derivative and find how much water drained for the given point of t.

Step-by-step explanation:

Given:

V = $5000(1 - \frac{1}{40}t )^2$ , where 0≤t≤40.

Here we have to find the derivative with respect to "t"

We have to use the chain rule to find the derivative.

V'(t) = $2(5000)(1 - \frac{1}{40} t)d/dt (1 - \frac{1}{40}t )$

V'(t) = $2(5000)(1 - \frac{1}{40} t)(-\frac{1}{40} )$

When we simplify the above, we get

V'(t) = $-250(1 - \frac{1}{40}t)$

If we know the time, we can plug in the value for "t" and find how much water drained for the given point of t.

4 0

3 years ago

mikah bought a dozen cupcakes for her friends birthday. if she paid $3.48 for the cupcakes what was the price per cupcake?

maria [59]

Answer: 0.29 or 29 cents each :)

anymore questions ask me ;)

6 0

3 years ago

What is 2/r=12/18? Chcjcjxjxkxkx

Alexeev081 [22]

Answer: r = 3

Step-by-step explanation:

Step 1: Cross-multiply.

2r = 12/18

(2)*(18)=12*r

36=12r

Step 2: Flip the equation.

12r=36

Step 3: Divide both sides by 12.

12r/12 = 36/12

r = 3

8 0

2 years ago

This equation represents an exponential function that passes through the point (2, 80).

miss Akunina [59]

we know that

If the point belongs to the graph, then the point must satisfy the equation

we will proceed to verify each case

case A.) $f(x)=4(x^{5})$

The point is $(2,80)$

Verify if the point satisfy the equation

For $x=2$ find the value of y in the equation and compare with the y-coordinate of the point

$f(2)=4(2^{5})$

$f(2)=128$

$128\neq 80$

therefore

the equation $f(x)=4(x^{5})$ not passes through the point $(2,80)$

case B.) $f(x)=5(x^{4})$

The point is $(2,80)$

Verify if the point satisfy the equation

For $x=2$ find the value of y in the equation and compare with the y-coordinate of the point

$f(2)=5(2^{4})$

$f(2)=80$

$80=80$

therefore

the equation $f(x)=5(x^{4})$ passes through the point $(2,80)$

case C.) $f(x)=4(5^{x})$

The point is $(2,80)$

Verify if the point satisfy the equation

For $x=2$ find the value of y in the equation and compare with the y-coordinate of the point

$f(2)=4(5^{2})$

$f(2)=100$

$100\neq 80$

therefore

the equation $f(x)=4(5^{x})$ not passes through the point $(2,80)$

case D.) $f(x)=5(4^{x})$

The point is $(2,80)$

Verify if the point satisfy the equation

For $x=2$ find the value of y in the equation and compare with the y-coordinate of the point

$f(2)=5(4^{2})$

$f(2)=80$

$80=80$

therefore

the equation $f(x)=5(4^{x})$ passes through the point $(2,80)$

therefore

the answer is

$f(x)=5(x^{4})$

$f(x)=5(4^{x})$

6 0

2 years ago

Read 2 more answers

In general, the probability that a blood donor has Type A blood is 0.40.Consider 8 randomly chosen blood donors, what is the pro

postnew [5]

Answer:

The probability that more than half of them have Type A blood in the sample of 8 randomly chosen donors is P(X>4)=0.1738.

Step-by-step explanation:

This can be modeled as a binomial random variable with n=8 and p=0.4.

The probability that k individuals in the sample have Type A blood can be calculated as:

$P(x=k) = \dbinom{n}{k} p^{k}(1-p)^{n-k}\\\\\\P(x=k) = \dbinom{8}{k} 0.4^{k} 0.6^{8-k}\\\\\\$

Then, we can calculate the probability that more than 8/2=4 have Type A blood as:

$P(X>4)=P(X=5)+P(X=6)+P(X=7)+P(X=8)\\\\\\P(x=5) = \dbinom{8}{5} p^{5}(1-p)^{3}=56*0.0102*0.216=0.1239\\\\\\P(x=6) = \dbinom{8}{6} p^{6}(1-p)^{2}=28*0.0041*0.36=0.0413\\\\\\P(x=7) = \dbinom{8}{7} p^{7}(1-p)^{1}=8*0.0016*0.6=0.0079\\\\\\P(x=8) = \dbinom{8}{8} p^{8}(1-p)^{0}=1*0.0007*1=0.0007\\\\\\\\P(X>4)=0.1239+0.0413+0.0079+0.0007=0.1738$

3 0

3 years ago