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3 years ago

The temperature in degrees Celsius can be expressed by the function C(f)=59(f−32)C(f)=59(f−32) where f is the temperature in deg

rees Fahrenheit. Find the temperature in degrees Celsius (to the nearest degree) if it is 55 °f.

Mathematics

1 answer:

Fiesta28 [93]3 years ago

4 0

$\dfrac{5}{9}\cdot (55-32)=\dfrac{5}{9}\cdot 23=\dfrac{115}{9}\approx 13^\circ C$

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Determine the equation of the tangent to x− 3x +1 at the point (3,1)

wariber [46]

9514 1404 393

Answer:

y = 3x -8

Step-by-step explanation:

We assume you want the tangent to the parabola y = x² -3x +1 at the given point. The slope is found using the derivative of the function at that point.

y' = 2x -3

At x=3, the slope is ...

y' = 2(3) -3 = 3

The equation of the line through point (3, 1) with a slope of 3 is ...

y -1 = 3(x -3) . . . . use the point-slope form of the equation for a line

y = 3x -9 +1 . . . . . eliminate parentheses, add 1

y = 3x -8

4 0

3 years ago

Y''+y'+y=0, y(0)=1, y'(0)=0

mars1129 [50]

Answer:

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Step-by-step explanation:

A second order linear , homogeneous ordinary differential equation has form $ay''+by'+cy=0$ .

Given: $y''+y'+y=0$

Let $y=e^{rt}$ be it's solution.

We get,

$\left ( r^2+r+1 \right )e^{rt}=0$

Since $e^{rt}\neq 0$ , $r^2+r+1=0$

{ we know that for equation $ax^2+bx+c=0$ , roots are of form $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ }

We get,

$y=\frac{-1\pm \sqrt{1^2-4}}{2}=\frac{-1\pm \sqrt{3}i}{2}$

For two complex roots $r_1=\alpha +i\beta \,,\,r_2=\alpha -i\beta$ , the general solution is of form $y=e^{\alpha t}\left ( c_1\cos \beta t+c_2\sin \beta t \right )$

i.e $y=e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Applying conditions y(0)=1 on $e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$ , $c_1=1$

So, equation becomes $y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

On differentiating with respect to t, we get

$y'=\frac{-1}{2}e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )+e^{\frac{-t}{2}}\left ( \frac{-\sqrt{3}}{2} \sin \left ( \frac{\sqrt{3}t}{2} \right )+c_2\frac{\sqrt{3}}{2}\cos\left ( \frac{\sqrt{3}t}{2} \right )\right )$

Applying condition: y'(0)=0, we get $0=\frac{-1}{2}+\frac{\sqrt{3}}{2}c_2\Rightarrow c_2=\frac{1}{\sqrt{3}}$

Therefore,

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

3 0

3 years ago

Which pair of numbers is relatively prime?

tamaranim1 [39]

Not A.(17 is a common factor)
B. only common factor is 1
not C (7 is a common factor)
D. only common factor is 1

Both B and D are relatively prime.

3 0

3 years ago

Round 409514.963259 to the nearest tenth.

olga_2 [115]

Answer:

Step-by-step explanation:

409515 Is the answer because you round up because the hundredths place number is greater than 5 so you round up, BUT you gotta roung 9 up but then you gotta add 1 to the ones place

6 0

3 years ago

Read 2 more answers

Answer this equation

Rudik [331]

Answer:

The solution to the box is

a = 2.1

b = 5.9

c = 0.9

d = 10

Step-by-step explanation:

To answer the equation, we simply name the boxes a,b,c and d.

Such that

a + b = 8 ---- (1)

b - c = 5 ------ (2)

d * c = 9 ------ (3)

a * d = 21 ------- (4)

Make d the subject of formula in (3)

d * c = 9 ---- Divide both sides by c

d * c/c = 9/c

d = 9/c

Substitute 9/c for d in (4)

a * d = 21

a * 9/c = 21

Multiply both sides by c

a * 9/c * c = 21 * c

a * 9 = 21 * c

9a = 21c ------ (5)

Make b the subject of formula in (1)

a + b = 8

b = 8 - a

Substitute 8 - a for b in (2)

b - c = 5

8 - a - c = 5

Collect like terms

-a - c = 5 - 8

-a - c = -3

Multiply both sides by -1

-1(-a - c) = -1 * -3

a + c = 3

Make a the subject of formula

a = 3 - c

Substitute 3 - c for a in (5)

9a = 21c becomes

9(3 - c) = 21c

Open bracket

27 - 9c = 21c

Collect like terms

27 = 21c + 9c

27 = 30c

Divide both sides by 30

27/30 = 30c/30

27/30 = c

0.9 = c

c = 0.9

Recall that a = 3 - c

So, a = 3 - 0.9

a = 2.1

From (1)

a + b = 8

2.1 + b = 8

b = 8 - 2.1

b = 5.9

From (3)

d * c = 9

Substitute 0.9 for c

d * 0.9 = 9

Divide both sides by 0.9

d * 0.9/0.9 = 9/0.9

d = 9/0.9

d = 10.

Hence, the solution to the box is

a = 2.1

b = 5.9

c = 0.9

d = 10

3 0

3 years ago