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4 years ago

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]

Answer:

DAM BROOOOO UR SCREEN.

Step-by-step explanation:

5 0

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)

Answer: x = 0

Explanation: Any non-zero expression raised to the power of 0 equals 1. Since we are trying to find 1, 0 would be an appropriate vale for x.

3 0

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