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3 years ago

Which tempature is the warmest

Mathematics

2 answers:

Mademuasel [1]3 years ago

4 0

Answer:

D.) 21 degrees fahrenheit.

Step-by-step explanation:

From warmest to coldest:

d. 21 degrees

a. 16 degrees

b. -16 degrees

c. -21 degrees

Hope this helps you : )

Sphinxa [80]3 years ago

4 0

Answer:

D. 21 degrees Fahrenheit

Step-by-step explanation:

We can get rid of letters B and C because we now if the temperature is in the negatives it is cold. And since 21 is greater than 16 we know that 21 is warmest.

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What is the direct variation between x and y when x=10 and y=250

ioda

You can use equations to describe the relationship between the two numbers.

x+240=y

y-240=x

7 0

3 years ago

The map of a mall is overlaid with a numeric

Veseljchak [2.6K]

Answer:

$\sqrt{118}$ or 10.8627805

Step-by-step explanation:

Start by taking the difference of each value of both points

Find the difference in x values

78-2 = 76

Find the difference in y values

46-4 = 42

Add the two differences

76+42=118

Take the square root of this value

$\sqrt{118}$

3 0

3 years ago

Please help in the question it should be if t=-6 when r =-2 its an error

Natasha2012 [34]

Answer:

r = - $\frac{12}{7}$

Step-by-step explanation:

Given that r varies inversely as t , then the equation relating them is

r = $\frac{k}{t}$ ← k is the constant of variation

To find k use the condition t = - 6 when r = - 2, then

- 2 = $\frac{k}{-6}$ ( multiply both sides by - 6 )

12 = k, thus

r = $\frac{12}{t}$ ← equation of variation

when t = - 7, then

r = $\frac{12}{-7}$ = - $\frac{12}{7}$

7 0

3 years ago

Let f (x) = 3x − 1 and ε > 0. Find a δ > 0 such that 0 < ∣x − 5∣ < δ implies ∣f (x) − 14∣ < ε. (Find the largest

s344n2d4d5 [400]

Answer:

This proves that f is continous at x=5.

Step-by-step explanation:

Taking f(x) = 3x-1 and $\varepsilon>0$ , we want to find a $\delta$ such that $|f(x)-14|$

At first, we will assume that this delta exists and we will try to figure out its value.

Suppose that $|x-5|$ . Then

$|f(x)-14| = |3x-1-14| = |3x-15|=|3(x-5)| = 3|x-5|< 3\delta$ .

Then, if $|x-5|$ , then $|f(x)-14|$ . So, in this case, if $3\delta \leq \varepsilon$ we get that $|f(x)-14|$ . The maximum value of delta is $\frac{\varepsilon}{3}$ .