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3 years ago

PLS HELP WILL MARK BRAINLIEST

Mathematics

1 answer:

gavmur [86]3 years ago

4 0

Answer:

1. Rolling a number less than 5

The numbers on a standard dice are 1, 2, 3, 4, 5, and 6. Therefore it is possible to roll a 6, but it is much more likely to be 1-5, simply because there are more of them.

2. Rolling a 2

Again, it is more likely that the result will be one of the other 5 numbers. But obviously a 2 is still an option, so it's not impossible.

3. Rolling a number less than 10

Every option is less than 10 so it is certain.

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A truck driver transported 748 boxes of goods at the first time. And the number of boxes of goods at the second time decreased t

dedylja [7]

Answer: 25% decrease in number of boxes of goods

Step-by-step explanation:

If you want to find a percent of a number you turn the percent into a decimal

Example:

Step 1

25% of 748

Step 2

0.25*748=561

8 0

3 years ago

How do I figure this out

ololo11 [35]

3 x 12 then divide the answer with 4

3 0

3 years ago

Read 2 more answers

Help please ITS OF TRIGONOMETRY<br>PROVE

Flauer [41]

Answer:

The equation is true.

Step-by-step explanation:

In order to solve this problem, one must envision a right triangle. A diagram used to represent the imagined right triangle is included at the bottom of this explanation. Please note that each side is named with respect to the angle is it across from.

Right angle trigonometry is composed of a sequence of ratios that relate the sides and angles of a right triangle. These ratios are as follows,

$sin(\theta)=\frac{opposite}{hypotenuse}\\\\cos(\theta)=\frac{adjacent}{hypotenuse}\\\\tan(\theta)=\frac{opposite}{adjacent}$

One is given the following equation,

$\frac{sin(A)+sin(B)}{cos(A) +cos(B)}+\frac{cos(A)-cos(B)}{sin(A)-sin(B)}=0$

As per the attached reference image, one can state the following, using the right angle trigonometric ratios,

$sin(A)=\frac{a}{c}\\\\sin(B)=\frac{b}{c}\\\\cos(A)=\frac{b}{c}\\\\cos(B)=\frac{a}{c}$

Substitute these values into the given equation. Then simplify the equation to prove the idenity,

$\frac{sin(A)+sin(B)}{cos(A) +cos(B)}+\frac{cos(A)-cos(B)}{sin(A)-sin(B)}=0$

$\frac{\frac{a}{c}+\frac{b}{c}}{\frac{b}{c}+\frac{a}{c}}+\frac{\frac{b}{c}-\frac{a}{c}}{\frac{a}{c}-\frac{b}{c}}=0$

$\frac{\frac{a+b}{c}}{\frac{a+b}{c}}+\frac{\frac{b-a}{c}}{\frac{a-b}{c}}$

Remember, any number over itself equals one, this holds true even for fractions with fractions in the numerator (value on top of the fraction bar) and denominator (value under the fraction bar).

$\frac{\frac{a+b}{c}}{\frac{a+b}{c}}+\frac{\frac{b-a}{c}}{\frac{a-b}{c}}$

$\frac{\frac{a+b}{c}}{\frac{a+b}{c}}+\frac{\frac{-(a-b)}{c}}{\frac{a-b}{c}}$