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2 years ago

only number 4, and please include the working. also do it with the substitution method not elimination or any other.

Mathematics

1 answer:

sergiy2304 [10]2 years ago

6 0

Answer:

x = -2 and y = 1

Step-by-step explanation:

→ Rearrange x - y = -3 to make y the subject

-y = - x - 3 so y = x + 3

→ Substitute into 2x + y = -3

2x + x + 3 = -3

→ Simplify

3x + 3 = -3

→ Minus 3 from both sides

3x = -6

→ Divide both sides by 3

x = -2

→ Substitute x value into first equation

-4 + y = -3

→ Add -4 to both sides

y = 1

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Express the function y(t)= 4 sin 2πt + 15 cos 2πt in terms of (a) a sine term only. (b) Determine the amplitude, the period, the

Bond [772]

Answer:

See the step by step solution

Step-by-step explanation:

Consider a more general case of the form $a\sin(x)+b\cos(x)$ . We want to express it as a sine term only by using the following formula $\sin(\alpha+\beta) = \sin(\alpha)\cos(\beta)+\sin(\beta)\cos(\alpha)$

The trick consists on find a number $\beta \in [0,2\pi]$ such that $\cos(\beta) = a, \sin(\beta)=b$ . But, note that since $\cos^2(\beta)+\sen^2(\beta)=1$ it is most likely that $a^2+b^2neq 1$ . Then, we will use the following equations:

$\cos(\beta) =\frac{a}{\sqrt[]{a^2+b^2}=A,\sin(\beta) =\frac{b}{\sqrt[]{a^2+b^2}=B$

Then, problem turns out to be

$a\sin(x)+b\cos(x)= \sqrt[]{a^2+b^2}(A\sin(x)+B\cos(x)) =\sqrt[]{a^2+b^2}(\sin(x)\cos(\beta)+\sin(\beta)\cos(x))=\sqrt[]{a^2+b^2}\sin(x+\beta)$ ,

where $\beta = \tan^{-1}(\frac{B}{A})= \tan^{-1}(\frac{b}{a}). So, note that the frequency is the same. Then, a) Taking a=4 and b= 15 we have that 4 sin 2πt + 15 cos 2πt= [tex]\sqrt[]{15^2+4^2}\sin(2\pi t + \tan^{-1}\frac{15}{4})$

b) The amplitude is $\sqrt[]{15^2+4^2} = \sqrt[]{241}$ . Since $2\pi rad/s = 1 Hz$ the frequency is 1 Hz. The period is given by \frac{2\pi}{2\pi} = 1.

c) The function's graph is attached

7 0

2 years ago

Read 2 more answers

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Bumek [7]

The step function is characteristic of the floor or ceiling functions, which map a real number to an integer. There's clearly a factor of 2, indicated by two units of y for every one unit of x.

The open and filled circles indicate, for example, f(1)=2, f(1.01)=4. That means we're rounding up, so we're using the ceiling function.

$f(x) = 2 \lceil x \rceil$