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Marta_Voda [28]
2 years ago
15

Solve the following equation: 12 x three eighths PLS HELPPP THIS IS THE 2ND TIME ASAP

Mathematics
2 answers:
mr_godi [17]2 years ago
8 0

Answer:

12*3/8 we just turn the 12 into a improper fraction and multiply the top and the bottom with 3/8 top and bottom so

12/1*3/8

12*3=36

1*8=8

so 36/8 the answer is

D

Hope This Helps!!!

melamori03 [73]2 years ago
7 0

Answer:

Last option, 36

Step-by-step explanation:

12/1 × 3/8

=12×3/1×8

=36/8 (ans.)

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Find the slopes of the asymptotes of the hyperbola with the following equation.
frosja888 [35]
You are given the equation 36 = 9x² + 4y². You are asked to find the slopes of the asymptotes of the hyperbola. A hyperbola has the following general equation x²/a² + y²/b² = 1. the goal here is to find the slopes of the hyperbolic equation. So divide both sides by 36

36 = 9x² + 4y²
(1/36)[36 = 9x² + 4y²]
1 = x²/4 + y²/9
a² = 4
a = 2
and
b² = 9
b = 3

To find the slope, divide a/b and you will get 2/3.
3 0
3 years ago
One number is one-third a second number. if the sum of the numbers is 48, find the numbers.
Gnoma [55]

First number assume A

Second number assume B

A= 1/3B ==> B= 3A

A+B = 48 ==> A+ 3A= 48

4A= 48 ==> A= 12

B= 3A= 36

8 0
3 years ago
HELP!! Algebra help!! Will give stars thank u so much <333
Anna35 [415]

Answers:

  • Part a)  \bf{\sqrt{x^2+(x^2-3)^2}
  • Part b)  3
  • Part c)   2.24
  • Part d)  1.58

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

Work Shown:

Part (a)

The origin is the point (0,0) which we'll make the first point, so let (x1,y1) = (0,0)

The other point is of the form (x,y) where y = x^2-3. So the point can be stated as (x2,y2) = (x,y). We'll replace y with x^2-3

We apply the distance formula to say...

d = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\\\d = \sqrt{(0-x)^2+(0-y)^2}\\\\d = \sqrt{(0-x)^2+(-y)^2}\\\\d = \sqrt{x^2 + y^2}\\\\d = \sqrt{x^2 + (x^2-3)^2}\\\\

We could expand things out and combine like terms, but that's just extra unneeded work in my opinion.

Saying d = \sqrt{x^2 + (x^2-3)^2} is the same as writing d = sqrt(x^2-(x^2-3)^2)

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

Part (b)

Plug in x = 0 and you should find the following

d(x) = \sqrt{x^2 + (x^2-3)^2}\\\\d(0) = \sqrt{0^2 + (0^2-3)^2}\\\\d(0) = \sqrt{(-3)^2}\\\\d(0) = \sqrt{9}\\\\d(0) = 3\\\\

This says that the point (x,y) = (0,3) is 3 units away from the origin (0,0).

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

Part (c)

Repeat for x = 1

d(x) = \sqrt{x^2 + (x^2-3)^2}\\\\d(1) = \sqrt{1^2 + (1^2-3)^2}\\\\d(1) = \sqrt{1 + (1-3)^2}\\\\d(1) = \sqrt{1 + (-2)^2}\\\\d(1) = \sqrt{1 + 4}\\\\d(1) = \sqrt{5}\\\\d(1) \approx 2.23606797749979\\\\d(1) \approx 2.24\\\\

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

Part (d)

Graph the d(x) function found back in part (a)

Use the minimum function on your graphing calculator to find the lowest point such that x > 0.

See the diagram below. I used GeoGebra to make the graph. Desmos probably has a similar feature (but I'm not entirely sure). If you have a TI83 or TI84, then your calculator has the minimum function feature.

The red point of this diagram is what we're after. That point is approximately (1.58, 1.66)

This means the smallest d can get is d = 1.66 and it happens when x = 1.58 approximately.

6 0
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balu736 [363]

Answer:

y = -5/3 x + 40

Step-by-step explanation:

The general form is y = ax + b, where a is the slope and b is the y intercept.

So immediately you can see that b must be 40, since the line crosses the y axis at 40.

The slope can be calculated by finding two "nice" points on the line that are also grid lines.

How about (0,40) and (6, 30).

Then, calculate the slope by dividing the y difference by the x difference:

(30-40)/(6-0) = -10/6 = -5/3

So C must be the answer.

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2 years ago
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20 greater than or equal to X less than or equal to 60
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3 years ago
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