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podryga [215]
3 years ago
8

The temperature on Monday was -23° F. The temperature on Tuesday was 18° higher. What was the temperature on Tuesday?

Mathematics
1 answer:
Sedaia [141]3 years ago
6 0

Answer: C is correct

Step-by-step explanation:

-23+18= -5

You might be interested in
What is the value of -2xy if x = -1 and y = 6?<br><br> 8<br> 12<br> 26<br> -12
Elena L [17]
12 would be your answer
-2*-1= 2
2*6=12
so 12
3 0
3 years ago
A sled is being held at rest on a slope that makes an angle theta with the horizontal. After the sled is released, it slides a d
Alenkasestr [34]

Answer:

μ =  Sin θ * d₁ / (d₂ - Cos θ*d₁)

d₂ = (d₁*Sin θ) / μ

Step-by-step explanation:

a) We apply The work-energy theorem

W = ΔE

W = - Ff*d

Ff = μ*N = μ*m*g

<em>Distance 1:</em>

- Ff*d₁ = Ef - Ei

⇒  - (μ*m*g*Cos θ)*d₁ = (Kf+Uf) - (Ki+Ui) = (Kf+0) - (0+Ui) = Kf - Ui

Kf = 0.5*m*vf² = 0.5*m*v²

Ui = m*g*h = m*g*d₁*Sin θ

then

- (μ*m*g*Cos θ)*d₁ = 0.5*m*v² - m*g*d₁*Sin θ  

⇒   - μ*g*Cos θ*d₁ = 0.5*v² - g*d₁*Sin θ   <em>(I)</em>

 

<em>Distance 2:</em>

<em />

- Ff*d₂ = Ef - Ei

⇒  - (μ*m*g)*d₂ = (0+0) - (Ki+0) = - Ki

Ki = 0.5*m*vi² = 0.5*m*v²

then

- (μ*m*g)*d₂ = - 0.5*m*v²

⇒   μ*g*d₂ = 0.5*v²     <em>(II)</em>

<em />

<em>If we apply (I) + (II)</em>

- μ*g*Cos θ*d₁ = 0.5*v² - g*d₁*Sin θ

μ*g*d₂ = 0.5*v²

 ⇒ μ*g (d₂ - Cos θ*d₁) = v² - g*d₁*Sin θ   <em>  (III)</em>

Applying the equation (for the distance 1) we get v:

vf² = vi² + 2*a*d = 0² + 2*(g*Sin θ)*d₁   ⇒   vf² = 2*g*Sin θ*d₁ = v²

then (from the equation <em>III</em>) we get

μ*g (d₂ - Cos θ*d₁) = 2*g*Sin θ*d₁ - g*d₁*Sin θ

⇒  μ (d₂ - Cos θ*d₁) = Sin θ * d₁

⇒   μ =  Sin θ * d₁ / (d₂ - Cos θ*d₁)

b)

If μ is a known value

d₂ = ?

We apply The work-energy theorem again

W = ΔK   ⇒   - Ff*d₂ = Kf - Ki

Ff = μ*m*g

Kf = 0

Ki = 0.5*m*v² = 0.5*m*2*g*Sin θ*d₁ = m*g*Sin θ*d₁

Finally

- μ*m*g*d₂ = 0 - m*g*Sin θ*d₁   ⇒   d₂ = Sin θ*d₁ / μ

3 0
3 years ago
Find the circumference of a circle whose.<br> A. Diameter is 14cm<br> B. Radius is 3.5cm
Nikitich [7]

Hello!

Circumference of a Circle

C=πd

or

C=2πr

Plug in the values

C=14π

C≈44 cm

C=2π×3.5

C≈22 cm

Hope everything is clear.

Let me know if you have any questions!
#KeepLearning

;-)

5 0
2 years ago
Find the maximum and minimum values of the function f(x,y)=2x2+3y2−4x−5 on the domain x2+y2≤100. The maximum value of f(x,y) is:
ryzh [129]

First find the critical points of <em>f</em> :

f(x,y)=2x^2+3y^2-4x-5=2(x-1)^2+3y^2-7

\dfrac{\partial f}{\partial x}=2(x-1)=0\implies x=1

\dfrac{\partial f}{\partial y}=6y=0\implies y=0

so the point (1, 0) is the only critical point, at which we have

f(1,0)=-7

Next check for critical points along the boundary, which can be found by converting to polar coordinates:

f(x,y)=f(10\cos t,10\sin t)=g(t)=295-40\cos t-100\cos^2t

Find the critical points of <em>g</em> :

\dfrac{\mathrm dg}{\mathrm dt}=40\sin t+200\sin t\cos t=40\sin t(1+5\cos t)=0

\implies\sin t=0\text{ OR }1+5\cos t=0

\implies t=n\pi\text{ OR }t=\cos^{-1}\left(-\dfrac15\right)+2n\pi\text{ OR }t=-\cos^{-1}\left(-\dfrac15\right)+2n\pi

where <em>n</em> is any integer. We get 4 critical points in the interval [0, 2π) at

t=0\implies f(10,0)=155

t=\cos^{-1}\left(-\dfrac15\right)\implies f(-2,4\sqrt6)=299

t=\pi\implies f(-10,0)=235

t=2\pi-\cos^{-1}\left(-\dfrac15\right)\implies f(-2,-4\sqrt6)=299

So <em>f</em> has a minimum of -7 and a maximum of 299.

4 0
3 years ago
2x + 12 = 1/3 (7/3x+4/3)
gladu [14]

Answer:

Step-by-step explanation:

2x + 12 = 1/3 (7/3x+4/3)

2x+12=\frac{1}{3}*\frac{7x}{3}+\frac{1}{3}*\frac{4}{3}\\\\2x+12=\frac{7x}{9}+\frac{4}{9}\\\\2x-\frac{7x}{9}=\frac{4}{9}-12\\\\\frac{2x*9}{1*9}-\frac{7x}{9}=\frac{4}{9}-\frac{12*9}{1*9}\\\\\frac{18x}{9}-\frac{7x}{9}=\frac{4}{9}-\frac{108}{9}\\\\\frac{18x-7x}{9}=\frac{4-108}{9}\\\\\frac{11x}{9}=\frac{-104}{9}\\\\\\x=\frac{-104*9}{11*9}=\frac{-104}{11}=-9\frac{5}{11}

8 0
3 years ago
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