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vladimir1956 [14]
3 years ago
5

En una estación de esquí la temperatura más alta ha sido de -2°C, y la

Mathematics
1 answer:
Andreas93 [3]3 years ago
7 0
21 degrees C ????????????
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Kadeem rode his bike at a constant speed. He rode 1 mile in 5 minutes. Which equation represents the amount of time in minutes,
sashaice [31]

Answer:

A

Step-by-step explanation:

Speed is the rate of change of distance over time.

The equation represents the amount of time t is \mathbf{t = 5 d}t=5d

Speed is calculated using:

\mathbf{Speed = \frac{distance}{time}}Speed=timedistance

When distance = 1 mile, and time = 5 minutes.

We have:

\mathbf{Speed = \frac{1}{5}}Speed=51

When distance = d, and time = t

We have:

\mathbf{Speed = \frac{d}{t}}Speed=td

Substitute \mathbf{Speed = \frac{1}{5}}Speed=51

\mathbf{\frac{d}{t} = \frac 15}td=51

Cross multiply

\mathbf{t = 5 \times d}t=5×d

\mathbf{t = 5 d}t=5d

Hence, the equation represents the amount of time t is \mathbf{t = 5 d}t=5d

5 0
2 years ago
Read 2 more answers
Verify the identity. Show your work. 1 + sec^2xsin^2x = sec^2x
allsm [11]
\bf \textit{Pythagorean Identities}
\\\\
1+tan^2(\theta)=sec^2(\theta)\\\\
-------------------------------\\\\
1+sec^2(x)sin^2(x)=sec^2(x)
\\\\\\
1+\cfrac{1}{cos^2(x)}\cdot sin^2(x)\implies 1+\cfrac{sin^2(x)}{cos^2(x)}\\\\\\ 1+tan^2(x)\implies sec^2(x)
4 0
3 years ago
2 3/7 + (-1 3/4) i need help please
julsineya [31]

Answer:

1/28

Step-by-step explanation:

7 0
3 years ago
6 1/2 - 3 2/5 please help really need it
tatuchka [14]
Answer is 3 1/10 


6 1/2 - 3 2/5 =
13/2 - 17/5 =
65/10 - 34/10 = 
31/10 = 3 1/10
4 0
3 years ago
Which of the following is equivalent to 2 ln e^ln 5x = 2 ln 15?
neonofarm [45]

Answer:

x  = 3

Step-by-step explanation:

Given in the question an equation,

2lne^{ln5x}=2ln15

Step 1

e^{lnx}=x

so,

2ln(5x)=2ln15

Step 2

cancel 2 on both sides of the equation

ln{5x}=ln15

Step 3

ln{5x}-ln15=0

ln\frac{5x}{15}=0

Step 4

ln\frac{x}{3}=0

Step 5

e^{ln\frac{x}{3}}=e^{0}

Step 6

x/3 = 1

x = 3(1)

x = 3

8 0
3 years ago
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