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2 years ago

8

Based on the construction which statement must be true

1 answer:

svetlana [45]2 years ago

7 0

Answer:

A (I think)

Step-by-step explanation:

The whole thing is divided into two parts, so 'CBD' is half, or 1/2

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3x-2y=6<br> 2x-y=5 <br><br> Solve the system by substitution

nalin [4]

The answer is

X =3

because they both equal the amounts given

7 0

3 years ago

How do I factor 8v2-8v-448=0?

Shtirlitz [24]

Factor it should be
8(v-8)(v+7)

3 0

3 years ago

I’m taking a college algebra test tomorrow and I am super scared. Any advice? Thank you!!

Naily [24]

Answer:

pray to god about it and go sit somewhere quiet and think

Step-by-step explanation:

7 0

2 years ago

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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 $(1,-1)$ .

4 0

2 years ago

Anna said that the product of 78⋅112=7278·112=72.

sergejj [24]

Because she randomly added a 72 before the 78

3 0

3 years ago

Read 2 more answers