All categories

3 years ago

Is the ordered pair a solution to the system of linear equations?

1 answer:

5 0

-7x + 2y = 0
6x + 6y = 0
ordered pair: (−2,2)

the way to figure out if that ordered pair is a solution is pretty simple, you just plug the ordered pair into both equations, and check if the outcome is true.

-7(-2) + 2(2) = 0
14 + 4 = 0
18 = 0 is NOT true, so this isn't a solution. still, check your second equation as well.

6x + 6y = 0
6(-2) + 6(2) = 0
-12 + 12 = 0

for this equation, it is a solution. however, because the solution must apply to the system as a whole, and not just one of the equations, the ordered pair still is not a solution.

You might be interested in

Will give brainliest answer

lidiya [134]

The square root of 10 because

sqrt(10) ^2 * 3.14 =

10*3.14 = 31.4

7 0

3 years ago

Read 2 more answers

I need help with this question please ASAP before tomorrow! Thank you :)

luda_lava [24]

23- right handed females

7 0

3 years ago

Read 2 more answers

The function f(x) = 2x2 + 2 represents the rate of a bus traveling in miles per hour. The function g(x) = x + 2 represents the t

AlekseyPX

Answer:

(f • g)(3) = 100, the distance in miles the bus traveled

Step-by-step explanation:

The product of (miles/hour) and (hours) is (miles), so only answers with units of distance are applicable. The numbers are ...

f(3) = 2·3² +2 = 20 . . . . miles per hour

g(3) = 3+2 = 5 . . . . hours

so (f·g)(3) = f(3)·g(3) = 20·5 = 100 . . . . miles

This means only the first answer choice is applicable.

6 0

3 years ago

Read 2 more answers

i f N(-7,-1) is a point on the terminal side of ∅ in standard form, find the exact values of the trigonometric functions of ∅.

Natali [406]

Answer:

$sin\ \varnothing = \frac{-1}{10}\sqrt 2$

$cos\ \varnothing = \frac{-7}{10}\sqrt 2$

$tan\ \varnothing = \frac{1}{7}$

$cot\ \varnothing = 7$

$sec\ \varnothing = \frac{-5}{7}\sqrt 2$

$csc\ \varnothing = -5\sqrt 2$

Step-by-step explanation:

Given

$N = (-7,-1)$ --- terminal side of $\varnothing$

Required

Determine the values of trigonometric functions of $\varnothing$ .

For $\varnothing$ , the trigonometry ratios are:

$sin\ \varnothing = \frac{y}{r}$ $cos\ \varnothing = \frac{x}{r}$ $tan\ \varnothing = \frac{y}{x}$

$cot\ \varnothing = \frac{x}{y}$ $sec\ \varnothing = \frac{r}{x}$ $csc\ \varnothing = \frac{r}{y}$

Where:

$r^2 = x^2 + y^2$

$r = \sqrt{x^2 + y^2$

In $N = (-7,-1)$

$x = -7$ and $y = -1$

So:

$r = \sqrt{(-7)^2 + (-1)^2$

$r = \sqrt{50$

$r = \sqrt{25 * 2$

$r = \sqrt{25} * \sqrt 2$

$r = 5 * \sqrt 2$

$r = 5 \sqrt 2$

Solving the trigonometry functions

$sin\ \varnothing = \frac{y}{r}$

$sin\ \varnothing = \frac{-1}{5\sqrt 2}$

Rationalize:

$sin\ \varnothing = \frac{-1}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$sin\ \varnothing = \frac{-\sqrt 2}{5*2}$

$sin\ \varnothing = \frac{-\sqrt 2}{10}$

$sin\ \varnothing = \frac{-1}{10}\sqrt 2$

$cos\ \varnothing = \frac{x}{r}$

$cos\ \varnothing = \frac{-7}{5\sqrt 2}$

Rationalize

$cos\ \varnothing = \frac{-7}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$cos\ \varnothing = \frac{-7*\sqrt 2}{5*2}$

$cos\ \varnothing = \frac{-7\sqrt 2}{10}$

$cos\ \varnothing = \frac{-7}{10}\sqrt 2$

$tan\ \varnothing = \frac{y}{x}$

$tan\ \varnothing = \frac{-1}{-7}$

$tan\ \varnothing = \frac{1}{7}$

$cot\ \varnothing = \frac{x}{y}$

$cot\ \varnothing = \frac{-7}{-1}$

$cot\ \varnothing = 7$

$sec\ \varnothing = \frac{r}{x}$

$sec\ \varnothing = \frac{5\sqrt 2}{-7}$

$sec\ \varnothing = \frac{-5}{7}\sqrt 2$

$csc\ \varnothing = \frac{r}{y}$

$csc\ \varnothing = \frac{5\sqrt 2}{-1}$

$csc\ \varnothing = -5\sqrt 2$

3 0

3 years ago

1/4 of the original price is marked down if the sales price is 128.00 what is the original price DONT not round your answer to t

mr_godi [17]

Answer:

Ipguifiyfohvlbouviyfiyfyivojviyc itcitditcyivuy if it go

7 0

3 years ago