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2 years ago

X=3y2x+4y=10solve using substitution

2 answers:

6 0

Answer- y=1 x=3

Substitute the first equation into the second one
2(3y)+4y=10
6y+4y=10
10y=10
Y=1
Place the answer for Y into the first equation
X=3(1)
X=3

Blababa [14]2 years ago

3 0

Answer:

Answer:

x=3 and y=1

Step-by-step explanation:

Let's solve your system by substitution.

x=3y;2x+4y=10

Step: Solvex=3yfor x:

x=3y

Step: Substitute3yforxin2x+4y=10:

2x+4y=10

23y+4y=10

10y=10(Simplify both sides of the equation)

10y

(Divide both sides by 10)

y=1

Step: Substitute1foryinx=3y:

x=3y

x=(3)(1)

x=3(Simplify both sides of the equation)

Answer:

x=3 and y=1

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2 years ago

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If g(x) = 2x2 + bx + 5 and g(1) = 4, what is the value of g(-1)?

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2 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

How do you Use Trigonometry to find length

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Right Triangles and the Pythagorean Theorem

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2 years ago

Please solve 5 f (Trigonometric Equations) #salute u if u solved it

Zanzabum

Answer:

$\beta=45\degree\:\:or\:\:\beta=135\degree$

Step-by-step explanation:

We want to solve $\tan \beta \sec \beta=\sqrt{2}$ , where $0\le \beta \le360\degree$ .

We rewrite in terms of sine and cosine.

$\frac{\sin \beta}{\cos \beta} \cdot \frac{1}{\cos \beta} =\sqrt{2}$

$\frac{\sin \beta}{\cos^2\beta}=\sqrt{2}$

Use the Pythagorean identity: $\cos^2\beta=1-\sin^2\beta$ .

$\frac{\sin \beta}{1-\sin^2\beta}=\sqrt{2}$

$\implies \sin \beta=\sqrt{2}(1-\sin^2\beta)$

$\implies \sin \beta=\sqrt{2}-\sqrt{2}\sin^2\beta$

$\implies \sqrt{2}\sin^2\beta+\sin \beta- \sqrt{2}=0$

This is a quadratic equation in $\sin \beta$ .

By the quadratic formula, we have:

$\sin \beta=\frac{-1\pm \sqrt{1^2-4(\sqrt{2})(-\sqrt{2} ) } }{2\cdot \sqrt{2} }$

$\sin \beta=\frac{-1\pm \sqrt{1^2+4(2) } }{2\cdot \sqrt{2} }$

$\sin \beta=\frac{-1\pm \sqrt{9} }{2\cdot \sqrt{2} }$

$\sin \beta=\frac{-1\pm3}{2\cdot \sqrt{2} }$

$\sin \beta=\frac{2}{2\cdot \sqrt{2} }$ or $\sin \beta=\frac{-4}{2\cdot \sqrt{2} }$

$\sin \beta=\frac{1}{\sqrt{2} }$ or $\sin \beta=-\frac{2}{\sqrt{2} }$

$\sin \beta=\frac{\sqrt{2}}{2}$ or $\sin \beta=-\sqrt{2}$

When $\sin \beta=\frac{\sqrt{2}}{2}$ , $\beta=\sin ^{-1}(\frac{\sqrt{2} }{2} )$

$\implies \beta=45\degree\:\:or\:\:\beta=135\degree$ on the interval $0\le \beta \le360\degree$ .

When $\sin \beta=-\sqrt{2}$ , $\beta$ is not defined because $-1\le \sin \beta \le1$

4 0

3 years ago

X=3y2x+4y=10solve using substitution​

X=3y2x+4y=10solve using substitution