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3 years ago

ONE QUESTION ONLY!!!

2 answers:

6 0

Let first number = X and 2nd number = Y

Equation 1: 5X + 8Y = -13

Equation 2: 5X + 10Y = -15

Multiply equation 1 by -1 to get: -5X + -8Y = 13

Now add the 2 equations together:
-5X + 5X = 0

-8Y + 10Y = 2Y

13 + -15 = -2

The total of the 2 equations is 2Y = -2

divide both sides by 2 to solve for Y
Y = -2 / 2
Y = -1
Now we know the 2nd number is -1, so replace Y in the 1st equation and solve for X

5X +8(-1) = -13

5x + -8 = -13

Add 8 to both sides:
5x = -5

Divide both sides by 5 to get x

X = -5 / 5
X = -1

Both numbers are -1

To check replace X and Y with -1 and solve:

1st equation: 5(-1) +8(-1) = -5 + -8 = -13 TRUE

2nd equation: 5(-1) +10(-1) = -5 + -10 = -15 TRUE

Both numbers are -1

Sedaia [141]3 years ago

3 0

1st number be x, 2nd number be y

Adding 5 times the first number and 8 times the second number gives a total of -13

so 5x+8y=-13

Adding 5 times the first number and 10 times the second number gives -15

so 5x+10y=-15

subtracting the above two eqns

(5x+8y)-(5x+10y) = -13-(-15)

-2y=2

y=-1

put back in first eqn 5x+8(-1)=-13

5x=-5

x=-1

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7.8 is what percent of 78?

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Answer:

10 %

Step-by-step explanation:

let the percent be x

according to the question

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6 0

3 years ago

Read 2 more answers

Maths trig functions <br> please help

Nikitich [7]

1. According to the plots, the curves intersect when $x=90^\circ$ and $x=360^\circ$ .

We can confirm this algebraically.

$\sin(x) + 2 = -\cos(x) + 3$

$\sin(x) + \cos(x) = 1$

$\sqrt2 \sin\left(x + 45^\circ\right) = 1$

$\sin\left(x + 45^\circ\right) = \dfrac1{\sqrt2}$

$x + 45^\circ = \sin^{-1}\left(\dfrac1{\sqrt2}\right) + 360^\circ n \text{ or } x + 45^\circ = 180^\circ - \sin^{-1}\left(\dfrac1{\sqrt2}\right) + 360^\circ n$

(where $n$ is an integer)

$x + 45^\circ = 45^\circ + 360^\circ n \text{ or } x + 45^\circ = 135^\circ + 360^\circ n$

$x = 360^\circ n \text{ or } x = 90^\circ + 360^\circ n$

We get the two solutions we found in the interval [0°, 360°] with $n=1$ in the first case, and $n=0$ in the second case.

2. We have $\sin(x)=0$ when $x \in \{0^\circ, \pm 180^\circ, \pm360^\circ, \ldots\}$ . For the given plot domain [0°, 360°], this happens when $180^\circ < x < 360^\circ$ .