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SOVA2 [1]
2 years ago
5

Jim is on his way home in his car. He has driven 20 miles so far which is 1/2 of the way home. What is the total length of his D

rive​
Mathematics
1 answer:
kati45 [8]2 years ago
8 0

Answer:40 miles per hour

Step-by-step explanation:

This is because, If 20 miles is 1/2 of his drive home then 40 miles should be the rest. Multiply 20 by 2 since it is 1/2

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What is the solution to 5x-6y>42
Shalnov [3]

Answer:

y< 5/6 - 7

Step-by-step explanation:

First, you have to send 5x to the other side so youĺl -5x and the equation will be -6y>-5x+42

Then, you divide the equation on the left side by -6 because youll want y by itself. Remember, when you divide by a negative, you flip the greater than/equal to sign. itll now be y<-5/6 - 7

done

6 0
3 years ago
How many terms are In 5x-2 ​
vagabundo [1.1K]

✧・゚: *✧・゚:*    *:・゚✧*:・゚✧

                  Hello!

✧・゚: *✧・゚:*    *:・゚✧*:・゚✧

❖ There are 2 terms in 5x - 2. Each number (including the variable) counts as a term.

~ ʜᴏᴘᴇ ᴛʜɪꜱ ʜᴇʟᴘꜱ! :) ♡

~ ᴄʟᴏᴜᴛᴀɴꜱᴡᴇʀꜱ

4 0
2 years ago
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Pls help me I’m failing math
maw [93]

Answer:

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Step-by-step explanation:

6 0
2 years ago
We are standing on the top of a 320 foot tall building and launch a small object upward. The object's vertical altitude, measure
STALIN [3.7K]

Answer:

The highest altitude that the object reaches is 576 feet.

Step-by-step explanation:

The maximum altitude reached by the object can be found by using the first and second derivatives of the given function. (First and Second Derivative Tests). Let be h(t) = -16\cdot t^{2} + 128\cdot t + 320, the first and second derivatives are, respectively:

First Derivative

h'(t) = -32\cdot t +128

Second Derivative

h''(t) = -32

Then, the First and Second Derivative Test can be performed as follows. Let equalize the first derivative to zero and solve the resultant expression:

-32\cdot t +128 = 0

t = \frac{128}{32}\,s

t = 4\,s (Critical value)

The second derivative of the second-order polynomial presented above is a constant function and a negative number, which means that critical values leads to an absolute maximum, that is, the highest altitude reached by the object. Then, let is evaluate the function at the critical value:

h(4\,s) = -16\cdot (4\,s)^{2}+128\cdot (4\,s) +320

h(4\,s) = 576\,ft

The highest altitude that the object reaches is 576 feet.

6 0
3 years ago
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