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3 years ago

How can you subtract two integers without number lines or counters

1 answer:

8 0

For example:

12 - 10 = ?

Start with the numbers to the far right; take your 2 from the 12 draw a curving line from that 2 to the 0 in 10; ask what is 2 minus zero? 2!

Draw a curving arrow from the 1 in 12 to the 1 in 10. And ask, what is one minus one ?
Zero.

Try that with other numbers

You might be interested in

Item 7

Mariulka [41]

Answer:

$A = 74.7^\circ$

$B = 42.5^\circ$

$C = 62.8^\circ$

Step-by-step explanation:

Given

$A = (-1,2) \to (x_1,y_1)$

$B = (2,8) \to (x_2,y_2)$

$C = (4,1) \to (x_3,y_3)$

Required

The measure of each angle

First, we calculate the length of the three sides of the triangle.

This is calculated using distance formula

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

For AB

$A = (-1,2) \to (x_1,y_1)$

$B = (2,8) \to (x_2,y_2)$

$d = \sqrt{(-1 - 2)^2 + (2 - 8)^2$

$d = \sqrt{(-3)^2 + (-6)^2$

$d = \sqrt{45$

So:

$AB = \sqrt{45$

For BC

$B = (2,8) \to (x_2,y_2)$

$C = (4,1) \to (x_3,y_3)$

$BC = \sqrt{(2 - 4)^2 + (8 - 1)^2$

$BC = \sqrt{(-2)^2 + (7)^2$

$BC = \sqrt{53$

For AC

$A = (-1,2) \to (x_1,y_1)$

$C = (4,1) \to (x_3,y_3)$

$AC = \sqrt{(-1 - 4)^2 + (2 - 1)^2$

$AC = \sqrt{(-5)^2 + (1)^2$

$AC = \sqrt{26$

So, we have:

$AB = \sqrt{45$

$BC = \sqrt{53$

$AC = \sqrt{26$

By representation

$AB \to c$

$BC \to a$

$AC \to b$

So, we have:

$a = \sqrt{53$

$b = \sqrt{26$

$c = \sqrt{45$

By cosine laws, the angles are calculated using:

$a^2 = b^2 + c^2 -2bc \cos A$

$b^2 = a^2 + c^2 -2ac \cos B$

$c^2 = a^2 + b^2 -2ab\ cos C$

$a^2 = b^2 + c^2 -2bc \cos A$

$(\sqrt{53})^2 = (\sqrt{26})^2 +(\sqrt{45})^2 - 2 * (\sqrt{26}) +(\sqrt{45}) * \cos A$

$53 = 26 +45 - 2 * 34.21 * \cos A$

$53 = 26 +45 - 68.42 * \cos A$

Collect like terms

$53 - 26 -45 = - 68.42 * \cos A$

$-18 = - 68.42 * \cos A$

Solve for $\cos A$

$\cos A =\frac{-18}{-68.42}$

$\cos A =0.2631$

Take arc cos of both sides

$A =\cos^{-1}(0.2631)$

$A = 74.7^\circ$

$b^2 = a^2 + c^2 -2ac \cos B$

$(\sqrt{26})^2 = (\sqrt{53})^2 +(\sqrt{45})^2 - 2 * (\sqrt{53}) +(\sqrt{45}) * \cos B$

$26 = 53 +45 -97.67 * \cos B$

Collect like terms

$26 - 53 -45= -97.67 * \cos B$

$-72= -97.67 * \cos B$

Solve for $\cos B$

$\cos B = \frac{-72}{-97.67}$

$\cos B = 0.7372$

Take arc cos of both sides

$B = \cos^{-1}(0.7372)$

$B = 42.5^\circ$

For the third angle, we use:

$A + B + C = 180$ --- angles in a triangle

Make C the subject

$C = 180 - A -B$

$C = 180 - 74.7 -42.5$

$C = 62.8^\circ$

8 0

2 years ago

A small box weighs 1 25 pounds. A large box weighs 3 1/3 time as much as the small box. how much does the large box weigh?

egoroff_w [7]

Answer:

416.7pounds

Step-by-step explanation:

Given parameters:

Weight of the small box = 125pounds

Unknown:

Weight of the larger box = ?

Solution:

The large box weighs $3\frac{1}{3}$ times as the small box:

So;

= 125pounds x $\frac{10}{3}$

= 416.7pounds

6 0

3 years ago

Solve 4 over x minus 4 equals the quantity of x over x minus 4, minus four thirds for x and determine if the solution is extrane

Vinil7 [7]

I'm thinking this is what the problem looks like: $\frac{4}{x-4}= \frac{x}{x-4}- \frac{4}{3}$ . The first thing to do is to move the $\frac{x}{x-4}$ over to the other side because it has a common denominator with the other side. Doing that and at the same time combining them over their common denominator looks like this: $\frac{4-x}{x-4}= -\frac{4}{3}$ . The best way to solve for x now is to cross-multiply to get 3(4-x)=-4(x-4). Distributing through the parenthesis is 12 - 3x = -4x + 16. Solving for x gives us x = 4. Of course when we sub a 4 back in for x we get real problems, don't we? Dividing by zero breaks every rule in math that there ever was! So, yes, the solution is extraneous.

6 0

3 years ago

Two number are in a ratio 4:3. Their sum is 70. Find the number

Semmy [17]

Answer:

30 AND 40

Step-by-step explanation:

SUM OF RATIO IS 4 + 3 = 7

1ST NUMBER : 4/7 × 70 = 40

2ND NUMBER : 3/7 × 70 = 30

3 0

3 years ago

The sides of a rhombus with angle of 60° are 6 inches. Find the area of the rhombus.

Vedmedyk [2.9K]

Sorry I'm not too sure but I know that you can probably find it using this formula:
Area = (1/2)(a)(b)(sin(C)) with C being the angle in the middle of both lines a and b in a triangle

Since a rhombus is pretty much two similar triangles....
Area = ((1/2)(6)(6)(sin6)) x 2
should give you the exact area

Sorry that I couldn't give an exact answer, I'm not too sure what in is because we were probably not taught the same things. Does it mean inch or is it supposed to be Ln? Anyways, whatever it means, maybe you could calculate each option's value and see which one is the same answer as the calculation I talked about above?

4 0

3 years ago

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