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Korvikt [17]
4 years ago
8

Explain how you solve a multiplication equation and a division equation.

Mathematics
1 answer:
kirill115 [55]4 years ago
6 0

Answer:

when solving for multiplication

for instance

2*2=4

2+2=4

4+2=8

as 2*4,=8 after 4 you can add 2 to get the next multiple of 2

Same goes to the rest of other multiplication tables

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HACTEHA [7]

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Step-by-step explanation:

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3 years ago
PLZ HELP I'M IN A TEST (will give brainliest)
olya-2409 [2.1K]

Answer:

The correct answer is C

Step-by-step explanation:

The height of the original triangle is 3 and the length is 6, so we have a ratio of 3:6. The only triangle that works is the one with a height of 5 and a length of 10 because both simplify to 1:2.

Hope this helped!

7 0
3 years ago
Read 2 more answers
The value of cos X is<br> 40/41<br> 42/41<br> 9/41<br> 35/41
DIA [1.3K]

Answer:

I think we're probably talking about a right triangle. So just find the adjacent side and the hypotenuse.

Step-by-step explanation:

Once we are considering x \in \mathbb{R}, the range of  \text{cos}(x) is [1, -1]. Therefore, $\frac{42}{41} $ is not true.

In this exercise, we may have

$\text{cos}(x)=\frac{x}{r} =\frac{x}{\sqrt{x^2 +y^2} } $

Considering the trigonometric functions beyond the unit circle.

Or considering the definition in the right triangle:

$\text{cos}(x)=\frac{\text{adjacent}}{\text{hypotenuse}} $

7 0
3 years ago
If 6 squared x =1 what is the value of x​
yarga [219]

Question: 6^x = 1 (finding x)

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4 years ago
Solve the equation below by completing the square and then solving for x. X2 - 6x + 1 = 17
blsea [12.9K]

The expression for the square of a binomial is

(a+b)^2 = a^2+2ab+b^2

Since your expression starts with x^2-6x +\ldots

It means that a = x, and so -6x must be twice the product of x and some number b. If you choose b = -3 you have

(x-3)^2 = x^2-6x+9

So, if you add 8 to both sides of your equation, you have

x^2-6x+1+8 = 17+8 \iff x^2-6x+9 = 25 \iff (x-3)^2=25 \iff x-3 = \pm 5

By choosing one sign at a time, we have the two solutions

x_1 \to x-3 = 5 \iff x = 8

x_2 \to x-3 = -5 \iff x = -2

7 0
3 years ago
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