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zimovet [89]
3 years ago
9

There is a relay race with 4 runners. The total distance run in the relay race is m

Mathematics
1 answer:
almond37 [142]3 years ago
4 0
4m=how far they run in the relay race
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Someone help ASAP<br> serious answers only or i will report
nadezda [96]
T represents hours, so if c(1.5)=c(t) as it mentioned in the problem, then 1.5 equals hours, and c represents cost, so if cost + time equals nine then I think it's a
3 0
4 years ago
Find the distance d(A,B) between points A and B.
Misha Larkins [42]

Answer:

A(4,-5) ; B(4,3)

distance d(A,B) = \sqrt{( 4 - 4)^{2} + (3 -(-5))^{2} } = 8

7 0
3 years ago
Assuming that the equations define x and y implicitly as differentiable functions x=f(t),y=g(t) find the slope of the curve x=f(
Mumz [18]
The given equations are
x(t+1)-4t \sqrt{x} =9            (1)
2y+4y^{3/2}=t^{3}+t           (2)

When t=0, obtain
x=9 \\ 2y+4y^{3/2}=0 \,\,=\ \textgreater \ \, y(1+2 \sqrt{y} )=0 \,=\ \textgreater \ \,y=0

Obtain derivatives of (1) and find x'(0).
x' (t+1) + x - 4√x - 4t*[(1/2)*1/√x = 0
x' (t+1) + x - 4√x -27/√x = 0
When t=0, obtain
x'(0) + x(0) - 4√x(0) = 0
x'(0) + 9 - 4*3 = 0
x'(0) = 3
Here, x' means \frac{dx}{dt}.

Obtain the derivative of (2) and find y'(0).
2y' + 4*(3/2)*(√y)*(y') = 3t² + 1
When t=0, obtain
2y'(0) +6√y(0) * y'(0) = 1
2y'(0) = 1 
y'(0) = 1/2.
Here, y' means \frac{dy}{dt}.

Because \frac{dy}{dx} = \frac{dy}{dt} / \frac{dx}{dt}, obtain
\frac{dy}{dx} |_{t=0}\, =  \frac{1/2}{3}= \frac{1}{6}

Answer:
The slope of the curve at t=0 is 1/6.



3 0
3 years ago
A new health drink has 130% of the recommended daily allowance (RDA) for a certain vitamin. The RDA for this vitamin is 30 mg. H
dsp73

Answer:

39 mg

Step-by-step explanation:

To get the number of milligrams, all we have to do is to find 130% of the RDA

mathematically, we have this as;

130% of 30

= 130/100 * 30 = 39 mg

7 0
3 years ago
NEED ANSWERED ASAP
fomenos
The answer is 10 2/3, 5

Proof:

Solve the following system:
{x/2 + y/3 = 7 | (equation 1)
{x/4 + (2 y)/3 = 6 | (equation 2)

Subtract 1/2 × (equation 1) from equation 2:
{x/2 + y/3 = 7 | (equation 1)
{0 x+y/2 = 5/2 | (equation 2)

Multiply equation 1 by 6:
{3 x + 2 y = 42 | (equation 1)
{0 x+y/2 = 5/2 | (equation 2)

Multiply equation 2 by 2:
{3 x + 2 y = 42 | (equation 1)
{0 x+y = 5 | (equation 2)

Subtract 2 × (equation 2) from equation 1:
{3 x+0 y = 32 | (equation 1)
{0 x+y = 5 | (equation 2)

Divide equation 1 by 3:
{x+0 y = 32/3 | (equation 1)
{0 x+y = 5 | (equation 2)
Collect results:
Answer: {x = 10 2/3, y = 5
8 0
4 years ago
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