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3 years ago

7

What is the sum of the solutions of the 2 equations below?

1 answer:

vitfil [10]3 years ago

5 0

Answer:

D. -8

Step-by-step explanation:

-14x = 280

So x = -20

9 + 5y = 69

5y=60

So y=12

-20+12 = -8

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Compute 1240.64/6.4 round the answer appropriately <br><br>thank you

liubo4ka [24]

1240.64 / 6.4 = 193.85

193.85 rounded is 194

3 0

3 years ago

A square and an equilateral triangle have equal perimeters. the area of the triangle is

Ivenika [448]

a - length of side of a square

t - length of side of a triangle

The perimeter of a square: $P_S=4a$

The perimeter of a triangle: $P_T=3t$

We have the area of a triangle: $A_T=16\sqrt3\ cm^2$

The formula of an area of an equilateral trinagle: $A_T=\dfrac{t^2\sqrt3}{4}$

Substitute:

$\dfrac{t^2\sqrt3}{4}=16\sqrt3$ <em>multiply both sides by 4</em>

$t^2\sqrt3=64\sqrt3$ <em>divide both sides by $\sqrt3$ </em>

$t^2=64\to t=\sqrt{64}\to t=8\ cm$

The perimeter of a triangle: $P_T=3(8)=24\ cm$

Substitute to the formula of a perimeter of a square:

$4a=24$ <em>divide both sides by 4</em>

$a=6\ cm$

The formula of a diagonal of a square: $d=a\sqrt2$

Substitute:

$d=6\sqrt2\ cm$

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3 years ago

Which study would be best done in a labrotory

Amiraneli [1.4K]

Answer:

Science? :)

Step-by-step explanation:

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2 years ago

X squared+ y squared = 2 y = 2x squared – 3 Which of the following describes the system?

cluponka [151]

Answer:

$x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}}$ and $y=-1,\frac{1}{2}$

The ordered pair solutions are $(-\sqrt{\frac{7}{4}},0.5)$ , $(\sqrt{\frac{7}{4}},0.5)$ , $(-1,-1)$ , and $(1,-1)$ .

Step-by-step explanation:

I'm assuming the system is $\left \{ {x^2+y^2=2} \atop {y=2x^2-3}} \right.$ :

$x^2+y^2=2$

$x^2+(2x^2-3)^2=2$

$x^2+(4x^4-12x^2+9)=2$

$x^2+4x^4-12x^2+9=2$

$4x^4-11x^2+9=2$

$4x^4-11x^2+7=0$

$x^4-11x^2+28=0$

$(x^2-7)(x^2-4)=0$

$(4x^2-7)(x^2-1)=0$

$4x^2-7=0$

$4x^2=7$

$x^2=\frac{7}{4}$

$x=\pm\sqrt{\frac{7}{4}}$

$x^2-1=0$

$x^2=1$

$x=\pm1$

$y=2x^2-3$

$y=2(\pm\sqrt{\frac{7}{4}})^2-3$

$y=2({\frac{7}{4}})-3$

$y=\frac{7}{2}-3$

$y=\frac{1}{2}$

$y=2x^2-3$

$y=2(\pm1)^2-3$

$y=2(1)-3$

$y=2-3$

$y=-1$

Therefore, $x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}}$ and $y=-1,\frac{1}{2}$

The ordered pair solutions are $(-\sqrt{\frac{7}{4}},0.5)$ , $(\sqrt{\frac{7}{4}},0.5)$ , $(-1,-1)$ , and $(1,-1)$ .

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2 years ago

Please help me! i dont know what to do... And my weekend depends on you

ludmilkaskok [199]

2) 5.3
3) -6
4) -6
5) 6

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3 years ago