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bekas [8.4K]
3 years ago
13

Solve four and three fourths minus one and one third.

Mathematics
2 answers:
gtnhenbr [62]3 years ago
7 0

Answer:

Three and five twelfths; 3 5/12; 41/12

AnnZ [28]3 years ago
7 0

Answer:

41/12

Step-by-step explanation:

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Write the equation in slope intercept form y= 3 (x+ 1) + 5x
schepotkina [342]

Given:

y=3(x+1)+5x

Required:

To write the given equation in slope intercept form.

Explanation:

Consider

\begin{gathered} y=3(x+1)+5x \\  \\ y=3x+3+5x \\  \\ y=8x+3 \end{gathered}

Final Answer:

y=8x+3

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Answer:

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3 years ago
3(5x-2)=15x-6 is their a solution or no
fenix001 [56]

Answer: 0 = 0

Step-by-step explanation: there is no solution  

6 0
3 years ago
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NO LINKS!!! Please help me with these problems​
nadezda [96]
<h3>Answers:</h3>

7) Center= (-1,2) Radius= \boldsymbol{\sqrt{8}} Equation: (x+1)^2+(y-2)^2 = 8

8) Center= (3,13) Radius= 13 Equation: (x-3)^2+(y-13)^2 = 169

=========================================================

Explanation:

Problem 7

Let's find the distance from (-1,2) to (-3,4)

(x_1,y_1) = (-1,2) \text{ and } (x_2, y_2) = (-3,4)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(-1-(-3))^2 + (2-4)^2}\\\\d = \sqrt{(-1+3)^2 + (2-4)^2}\\\\d = \sqrt{(2)^2 + (-2)^2}\\\\d = \sqrt{4 + 4}\\\\d = \sqrt{8}\\\\

This is the radius because it stretches from the center to a point on the circle, so r = \sqrt{8}

Squaring both sides will get us r^2 = 8

One useful template for a circle is the equation (x-h)^2+(y-k)^2 = r^2\\\\

(h,k) is the center

r is the radius

Let's plug in the given center (h,k) = (-1,2) and the r^2 value we found earlier.

(x-h)^2+(y-k)^2 = r^2\\\\(x-(-1))^2+(y-2)^2 = 8\\\\(x+1)^2+(y-2)^2 = 8\\\\

You can confirm this by using a tool like Desmos. See below.

------------------------------------------------------------------------

Problem 8

The endpoints of the diameter are (-2,1) and (8,25)

The center is the midpoint of these endpoints.

The midpoint of the x coordinates is (-2+8)/2 = 3

The midpoint of the y coordinates is (1+25)/2 = 13

The center is (h,k) = (3,13)

Now find the distance from the center to one of the points on the circle, let's say to (8,25)

(x_1,y_1) = (3,13) \text{ and } (x_2, y_2) = (8,25)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(3-8)^2 + (13-25)^2}\\\\d = \sqrt{(-5)^2 + (-12)^2}\\\\d = \sqrt{25 + 144}\\\\d = \sqrt{169}\\\\d = 13\\\\

The radius is exactly 13 units.

So,

(x-h)^2+(y-k)^2 = r^2\\\\(x-3)^2+(y-13)^2 = 13^2\\\\(x-3)^2+(y-13)^2 = 169\\\\

is the equation of this particular circle.

Visual confirmation is shown below.

8 0
3 years ago
Read 2 more answers
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