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1 year ago

Solve the following system of linear equations using elimination. 3r- y=5 x=y=1

Mathematics

1 answer:

Natali [406]1 year ago

7 0

3x - y = 5 Equation l

x - y = 1 Equation ll

By elimination

-Multiply equation l by -1

-3x + y = -5

x - y = 1

-2x = -4

Solve for x

x = -4/-2

x = 2

-Find y

2 - y = 1

-y = 1 - 2

-y = -1

y = -1/-1

y = 1

Solurtion: x = 2, y = 1

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Find the area. The figure is not drawn to scale

poizon [28]

Answer:

96 square inches.

Step-by-step explanation:

From given picture we see that

BC=8 in

AD=10 in

Given that

AD=AB

Then AB=10 in

Apply Pythagorean theorem in triangle ABC to find AC

$(hypotenuse)^2=(base)^2+(perpendicular)^2$

$(AB)^2=(BC)^2+(AC)^2$

$(10)^2=(8)^2+(AC)^2$

$100=64+(AC)^2$

$100-64=(AC)^2$

$36=(AC)^2$

take square root

$6=AC$

Then area of triangle ABC $=\frac{1}{2}\left(base\right)\left(altitude\right)$

$=\frac{1}{2}\left(8\right)\left(6\right)=24$

There are total 4 congruent triangles.

Then total area of the given figure = 4(24)=96 square inches.

7 0

3 years ago

2x² - 5x+1 has roots alpha and beta. Find alpha⁴+beta⁴ without solving the equation.

shepuryov [24]

Answer:

Step-by-step explanation:

$\alpha+\beta=\dfrac{5}{2} \\\\\alpha*\beta=\dfrac{1}{2} \\\\\alpha^2+\beta^2=\dfrac{21}{4} \ (see\ previous\ post)\\\\(\alpha+\beta)^4=\dfrac{625}{16} \\\\=\alpha^4+\beta^4+4*\alpha^3*\beta+6*\alpha^2*\beta^2+4*\alpha*\beta^3\\\\=\alpha^4+\beta^4+4*(\alpha*\beta)(\alpha^2+\beta^2)+6*\alpha^2*\beta^2\\\\\alpha^4+\beta^4=(\alpha+\beta)^4-4*(\alpha*\beta)(\alpha^2+\beta^2)-6*\alpha^2*\beta^2\\\\= \dfrac{625}{16} -4*\dfrac{1}{2} *\dfrac{21}{4} -6*(\dfrac{1}{2})^2 \\\\$