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liubo4ka [24]
3 years ago
10

a pitcher contains 40 fluid ounces of iced tea. Shelby pours 3 cups of iced tea. how many pints of iced tea are left?

Mathematics
2 answers:
slava [35]3 years ago
8 0
8 ounce=1 cup
so 3 cups=8 times 3 ounces=24 ounces
40-24=16 counces left

8 ounces=1 cup 
16 ounces=2 times 8
16=2 cups
2 cups=1 pint
1 pint left


lord [1]3 years ago
4 0
Given;
40 fluid ounces of iced tea.

We know,
1 cup = 8 fluid ounce
1 pint = 2 cup

Now,
8 fluid ounce = 1 cup
1 fluid ounce = 1 / 8 cup
40 fluid ounce = (1 / 8) * 40 cup
So, 40 fluid ounce = 5 cup

Now,
We, used 3 cups of iced tea. So, we only have 2 cups of iced tea left.

We have,
2 cups = 1 pint

So, 1 pint of iced tea is left!


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Step-by-step explanation:

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kherson [118]

Answer:

\large\boxed{(x-2)^2+(y-1)^2=34}

Step-by-step explanation:

The equation of a circle in standard form:

(x-h)^2+(y-k)^2=r^2

(h, k) - center

r - radius

We have the endpoints of the diameter: (-1, 6) and (5, -4).

Midpoint of diameter is a center of a circle.

The formula of a midpoint:

\left(\dfrac{x_1+x_2}{2};\ \dfrac{y_1+y_2}{2}\right)

Substitute:

h=\dfrac{-1+5}{2}=\dfrac{4}{2}=2\\\\k=\dfrac{6+(-4)}{2}=\dfrac{2}{2}=1

The center is in (2, 1).

The radius length is equal to the distance between the center of the circle and the endpoint of the diameter.

The formula of a distance between two points:

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Substitute the coordinates of the points (2, 1) and (5, -4):

r=\sqrt{(5-2)^2+(-4-1)^2}=\sqrt{3^2+(-5)^2}=\sqrt{9+25}=\sqrt{34}

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(x-2)^2+(y-1)^2=(\sqrt{34})^2

3 0
3 years ago
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Stella [2.4K]

<u>Question:</u>

Find the number of real number solutions for the equation. x^2 + 5x + 7 = 0

<u>Answer:</u>

The number of real solutions for the equation x^{2}+5 x+7=0 is zero

<u>Solution:</u>

For a Quadratic Equation of form : a x^{2}+b x+c=0  ---- eqn 1

The solution is x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}  

Now , the given Quadratic Equation is x^{2}+5 x+7=0  ---- eqn 2

On comparing Equation (1) and Equation(2), we get

a = 1 , b = 5 and c = 7

In x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} , b^2 - 4ac is called the discriminant of the quadratic equation

Its value determines the nature of roots

Now, here are the rules with discriminants:

1) D > 0; there are 2 real solutions in the equation

2) D = 0; there is 1 real solution in the equation

3) D < 0; there are no real solutions in the equation

Now let solve for given equation

D= b^2 - 4ac\\\\D = 5^2 - 4(1)(7)\\\\D = 25 - 28 \\\\D = -3

Since -3 is less than 0, this means that there are 0 real solutions in this equation.

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