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2 years ago

C. Mrs. Hampton used 1/3 of a gallon of milk to feed three kittens. If she fed an equal amount

Mathematics

2 answers:

romanna [79]2 years ago

4 0

Answer: 14.22 ounces each

Step-by-step explanation:

128 ounces in a gallon divided by 3 divided by 3 again

tatyana61 [14]2 years ago

3 0

1/9 I hope this helps you out!

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Translate each verbal expression into an algebraic expression.

Semmy [17]

Answer:

10. 7 + 4a

11. 7a + 4

12. 12a - 4

13. 14(a - 4)

14. 2a + b

15. 2(a + b)

16. 1/3x - 5

17. 3x/8

18. ab / a+b

Step-by-step explanation:

Let x be the number.

Can't explain, I need to do homework.

Hope this helps :)

6 0

1 year ago

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jenny won the ping pong championship eight more times than gerardo. they have won a combined total of 32 championship

Lena [83]

X + x +8 = 32
2x + 8 = 32
2x = 24
x = 12
So gerardo has won 12 and jenny has won 20

3 0

2 years ago

The point P on the circle x2 + y2 = r2 that is also on the terminal side of an angle θ in standard position is given. Find the i

Brut [27]

Okay so you do not really need the circle equation. if you make a triangle with it's x length as 3 and it's y length as 4, you will be able to find the third length. Do Pythagorean theorem to find the hypotenuse.
the hypotenuse will be five after you calculate it.

sin is opposite over hypotenuse or Y over R

so... the sin is 4/5

make sure you know what quadrant your triangle is in for the negatives

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2 years ago

Solve irrational equation pls

rusak2 [61]

$\hbox{Domain:}\\
x^2+x-2\geq0 \wedge x^2-4x+3\geq0 \wedge x^2-1\geq0\\
x^2-x+2x-2\geq0 \wedge x^2-x-3x+3\geq0 \wedge x^2\geq1\\
x(x-1)+2(x-1)\geq 0 \wedge x(x-1)-3(x-1)\geq0 \wedge (x\geq 1 \vee x\leq-1)\\
(x+2)(x-1)\geq0 \wedge (x-3)(x-1)\geq0\wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle1,\infty) \wedge x\in(-\infty,1\rangle \cup\langle3,\infty) \wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle3,\infty)$

$
\sqrt{x^2+x-2}+\sqrt{x^2-4x+3}=\sqrt{x^2-1}\\
x^2-1=x^2+x-2+2\sqrt{(x^2+x-2)(x^2-4x+3)}+x^2-4x+3\\
2\sqrt{(x^2+x-2)(x^2-4x+3)}=-x^2+3x-2\\
\sqrt{(x^2+x-2)(x^2-4x+3)}=\dfrac{-x^2+3x-2}{2}\\
(x^2+x-2)(x^2-4x+3)=\left(\dfrac{-x^2+3x-2}{2}\right)^2\\
(x+2)(x-1)(x-3)(x-1)=\left(\dfrac{-x^2+x+2x-2}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-x(x-1)+2(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-(x-2)(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\dfrac{(x-2)^2(x-1)^2}{4}\\
4(x+2)(x-3)(x-1)^2=(x-2)^2(x-1)^2\\$
$
4(x+2)(x-3)(x-1)^2-(x-2)^2(x-1)^2=0\\
(x-1)^2(4(x+2)(x-3)-(x-2)^2)=0\\
(x-1)^2(4(x^2-3x+2x-6)-(x^2-4x+4))=0\\
(x-1)^2(4x^2-4x-24-x^2+4x-4)=0\\
(x-1)^2(3x^2-28)=0\\
x-1=0 \vee 3x^2-28=0\\
x=1 \vee 3x^2=28\\
x=1 \vee x^2=\dfrac{28}{3}\\
x=1 \vee x=\sqrt{\dfrac{28}{3}} \vee x=-\sqrt{\dfrac{28}{3}}\\$

There's one more condition I forgot about
$-(x-2)(x-1)\geq0\\
x\in\langle1,2\rangle\\$

Finally
$x\in(-\infty,-2\rangle\cup\langle3,\infty) \wedge x\in\langle1,2\rangle \wedge x=\{1,\sqrt{\dfrac{28}{3}}, -\sqrt{\dfrac{28}{3}}\}\\
\boxed{\boxed{x=1}}$

3 0

2 years ago

Write 0.589 as a fraction

ohaa [14]

Answer:

590/1000

Step-by-step explanation:

6 0

2 years ago

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