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Pani-rosa [81]
3 years ago
15

Small roads with lower speed limits are known as:

Mathematics
1 answer:
Law Incorporation [45]3 years ago
6 0

Answer:

  B. secondary roads

Step-by-step explanation:

While all of streets, county roads, and secondary roads are generally designed to carry less traffic and/or have lower speed limits than interstate highways, the generic name for such roads is ...

  secondary roads.

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Solve the formula for the indicated variable.
kolbaska11 [484]

Answer: b = p - a -c  

Step-by-step explanation:

p= a + b + c  Solving for b means getting b on one side by itself.

p = a + b + c    To solve for b first subtract a from both sides to get,

-b         -b

p - a =  b + c   Now subtract  c from both sides   to get

-c              -c

p - a - c = b

b = p - a -c

3 0
3 years ago
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Graph the equation using the slope and y-intercept <br> y = 2x - 4
salantis [7]
Put a dot on your y-axis at -4 (the one going up and down), from there go up two, and over one. your second dog should be at (1,-2). connect your dots and that’s your graphed equation !
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3 years ago
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Find the fifth term in a sequence: 15,20,25...<br> A) 20<br> B) 45<br> C) 35 <br> D) 40
guajiro [1.7K]

Answer:

C. 35

Step-by-step explanation:

Use the equation: y= 10+5x

8 0
3 years ago
Tanya runs a catering business. Based on her records, her weekly profit can be approximated by the function, where x is the numb
uranmaximum [27]

answer is in image attached

7 0
3 years ago
Can you help me i don’t know the answers
Advocard [28]
1)

An irrational number is a number that a) can't be written as a fraction of two whole numbers AND b) is an infinite decimal without any sort of pattern.

For the first answer choice, clearly \frac{1}{3} does not pass the first criterion so we look at the second choice.

Let's come back to \sqrt{2} and \pi.

\frac{2}{9} doesn't meet our first criterion, and let's skip \sqrt{3} for now.

It is often easier to disprove an irrational number than to prove one. There are a few famous irrationals to know (although there is an infinite number of irrationals). The most common are \sqrt{2},  \pi, e,  \sqrt{3}. For now, it's just helpful to know these and recognize them.

So we can check off \sqrt{2},  \pi and \sqrt{3}.

2) 

For this next question, we know that \sqrt{64} = 8. Clearly this isn't irrational. Likewise, \frac{1}{2} isn't irrational. \frac{16}{4} =  \frac{4}{4} = 1, which is rational, leaving only \frac{ \sqrt{20}}{5} =  \frac{2 \sqrt{5} }{5}. By process of elimination, this is the correct answer. Indeed, \sqrt{5} is an irrational number.

3) This notation means that we have 0.3636363636... and so on, to an infinite number of digits. It is called a repeating decimal.

But it can be written as a fraction because its pattern repeats, unlike for an irrational number.

Let's say x=0.36363636.... Would you agree that 100x=36.36363636...? (We choose to multiply by 100 because there are two decimals that repeat. For 1, choose 10, for 3 choose 1,000, and so on.)

Now, let's subtract x from 100x and solve.

100x=36.36363636\\-x \ \ \ \ \ \ \ -0.36363636\\99x=36\\\\x= \dfrac{36}{99}= \dfrac{4}{11}

Voila!
4 0
3 years ago
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