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3 years ago

6

Solve the system of equations below by graphing. Y=-5x and y=x-6

1 answer:

11Alexandr11 [23.1K]3 years ago

5 0

9514 1404 393

Answer:

(x, y) = (1, -5)

Step-by-step explanation:

Using a graphing calculator to solve equations such as these is quick and easy. It is one of my favorite methods. The attached graph shows the solution is ...

(x, y) = (1, -5)

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F(x)=14-2x<br> If f(x)=20, find x

mixer [17]

Answer:

x= -3

Step-by-step explanation:

20=14-2x

-14 -14

6 = -2x

÷-2 ÷-2

-3 =x

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3 years ago

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What are two numbers that add up to 43 and their difference is 11?

LiRa [457]

Answer:

27 and 16

Step-by-step explanation:

x + y = 43

x - y = 11

You can add up the two equations so the ys are cancelled out.

2x = 54

x = 27

Plug x in to find y.

27 - y = 11

y = 16

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3 years ago

During the spring and summer months, the Happy Home Motel sees a rise in travelers. The manager of the motel turned in this repo

Korvikt [17]

Answer:

342

Step-by-step explanation:

Let's see the difference is each successive values:

232 - 217 = 15

252 - 232 = 20

277 - 252 = 25

307 - 277 = 30

As we can see, each month the increase is 5 more than previous months' increase. <em>Clearly from the pattern we can surmise that the next month (August), it will increase by 35 from the value of July.</em>

Number of visitors in August = Number of Visitors in July + 35 = 307 + 35 = 342

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3 years ago

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2 divided by 1/3 in simpleast form

leva [86]

Answer:

6

Step-by-step explanation:

When dividing by a fraction flip it over and multiply

2 / (1/3) = 2 * 3/1

2 / (1/3) = 6/1

2 / (1/3) = 6

8 0

3 years ago

Let Q(x, y) be the predicate "If x < y then x2 < y2," with domain for both x and y being R, the set of all real numbers.

Levart [38]

Answer:

a) Q(-2,1) is false

b) Q(-5,2) is false

c)Q(3,8) is true

d)Q(9,10) is true

Step-by-step explanation:

Given data is $Q(x,y)$ is predicate that $x$ then $x^{2}$ . where $x,y$ are rational numbers.

a)

when $x=-2, y=1$

Here $-2$ that is $x$ satisfied. Then

$(-2)^{2}$

$4$ this is wrong. since $4>1$

That is $x^{2}$ $>y^{2}$ Thus $Q(x,y)$ $=Q(-2,1)$ is false.

b)

Assume $Q(x,y)=Q(-5,2)$ .

That is $x=-5, y=2$

Here $-5$ that is $x$ this condition is satisfied.

Then

$(-5)^{2}$

$25$ this is not true. since $25>4$ .

This is similar to the truth value of part (a).

Since in both $x$ satisfied and $x^{2} >y^{2}$ for both the points.

c)

if $Q(x,y)=Q(3,8)$ that is $x=3$ and $y=8$

Here $3$ this satisfies the condition $x$ .

Then $3^{2}$

$9$ This also satisfies the condition $x^{2}$ .

Hence $Q(3,8)$ exists and it is true.

d)

Assume $Q(x,y)=Q(9,10)$

Here $9$ satisfies the condition $x$

Then $9^{2}$

$81$ satisfies the condition $x^{2}$ .

Thus, $Q(9,10)$ point exists and it is true. This satisfies the same values as in part (c)

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3 years ago