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Nikolay [14]
3 years ago
14

Need help graphing these 8x + y = 4 x+y= -3

Mathematics
1 answer:
Tomtit [17]3 years ago
5 0

Answer:

y=-8x+4

y=-x-3

Step-by-step explanation:

for the first one start at 4 on the y-axis, then go down 8 and right 1

and for the second one start on negative 3 on the y-axis, then go down one and right one.

hope this helps

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Chandler Heisinger buys a monthly metro card, which entitles him to unlimited subway travel in New York City, for $81 per month.
ANTONII [103]
Let r be the number of rides Chandler takes in a month. Then the cost with the MetroCard is still $81, but the cost without the MetroCard is 2r. So we can set up an equation representing what we want: "The cost with a MetroCard of r rides in a month is less than the cost without a MetroCard." In equations,
81\ \textless \ 2r \\ r\ \textgreater \ 81/2\ \textgreater \ 40 \\
r \geq 41
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7 0
4 years ago
How many times bigger is 1.8x10^27 than 1x10^25
victus00 [196]
To find the number of times it is bigger

= \dfrac{1.8 \times 10^{27} }{1 \times 10^{25}}

= 1.8 \times 10^{27-25}

= 1.8 \times 10^{2}

= 180 \text { times}
3 0
3 years ago
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Each of 15 students will give a 1 1/2 minute speech in English class. How long will it take to get to speeches
erik [133]

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6 0
3 years ago
Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used.
Oduvanchick [21]

Answer:

  see attached

Step-by-step explanation:

The Pythagorean theorem can be used to find the hypotenuse associated with each pair of legs. That tells you ...

  c² = a² +b² . . . . . legs a, b; hypotenuse c

__

<h3>alternate form of Pythagorean theorem</h3>

For the purpose of this problem, it is convenient to consider a slightly different form of the equation.

For legs √a and √b, the hypotenuse √c is given by ...

  (√c)² = (√a)² +(√b)²

  c = a +b

That is ...

  legs √a, √b ⇒ hypotenuse √(a+b)

__

<h3>application to this problem</h3>

Since the legs are (mostly) given in terms of square roots, the value under the radical for the hypotenuse is simply the sum of those:

legs: √1, √2 ⇒ hypotenuse √(1+2) = √3

legs: √2, √3 ⇒ hypotenuse √(2+3) = √5

legs: √5, √3 ⇒ hypotenuse √(5+3) = √8

legs: √5, √1 ⇒ hypotenuse √(5+1) = √6

_____

<em>Additional comment</em>

You may not see the leg lengths given as square roots very often. This is a rather unusual set of problems for hypotenuse length.

6 0
2 years ago
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Please HELP I DONT KNOW WHTA IM DOING!!
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BD is 18

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