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mihalych1998 [28]

3 years ago

5

2-1 i need answer for this asappppppppppppppppppppppppppppppppppppp

2 answers:

True [87]3 years ago

6 0

Answer:

2-1 = 1

Step-by-step explanation:

Hopefully help......

mote1985 [20]3 years ago

3 0

Answer:

1

Step-by-step explanation:

say you have two apples, if you eat one, it subtracts one from the whole (2) and leaves you with one apple!1!!

i—

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PLSSS HELP WILL GIVE BRAINLIEST FOR FIRST CORRECT ANSWER!!!!!!!!!

Kryger [21]

Answer:

A fair die has 6 sides.

There are 3 odd numbers on a fair die so the probability would be 3/6

There are 4 numbers greater than 2 on a fair die o the probability of that would be 4/6

There is only one number that will make it square and that is 4. So the probability would be 1/6

6 0

2 years ago

Please help me !!! what’s the answer ?

diamong [38]

Answer:

negtive 1

Step-by-step explanation:

4 0

2 years ago

Help me out please!!!!

DedPeter [7]

Answer:

Answer (e)

Step-by-step explanation:

Note that this graph has a zero at x = 1 and a vertical asymptote at x = -4. Only Answer (e) corresponds to that. Try letting x = 1. y is zero, correct? Try letting x = -4. y is undefined, correct?

7 0

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 $(1,-1)$ .

4 0

1 year ago

Actually help pls people have been typing random stuff

agasfer [191]

Answer:

reading, chores, and then homework

Step-by-step explanation:

just accept it

3 0

2 years ago

Read 2 more answers