The expression represents the perimeter, in centimeters, of the triangle is 6q - 6r - 5s
<h3>What is the perimeter?</h3>
The formula for perimeter of a triangle is expressed as;
Perimeter = a + b + c
Where a , b and c are the lengths of its side
Now, let's substitute the values
Perimeter = (q + r) + (5q - 10s) + (5s - 7r)
expand the bracket
Perimeter = q + r + 5q - 10s + 5s - 7r
collect like terms
Perimeter = q + 5q + r - 7r -10s + 5s
Add like terms
Perimeter = 6q - 6r - 5s
Thus, the expression represents the perimeter, in centimeters, of the triangle is 6q - 6r - 5s
Learn more about perimeter here:
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This number is natural, whole, integer, and rational.
Whole numbers are numbers such as 0, 1, 2, ... This is a whole number.
Natural numbers can also be counting numbers. They are the whole numbers, but starting at 1, not 0. This is a natural number.
Integers are whole numbers with negatives. This is an integer.
Rational numbers are any numbers that can be written as a fraction. We can write this as 4563/1, so it is rational.
Answer:
<h3>
"The answer is the yellow lines in the attached figure.
</h3>
Step-by-step explanation: As shown in the attached figure, regular hexagon FGHIJK and square ABCD shares common centre on the co-ordinate plane and AB || FG.
We are to find the line across which the combined figure will reflect onto itself.
In the attached figure, we see two lines which are yellow in colour. We can easily detect that the figure will be reflected onto itself if these two lines acts as a mirror separately.
Hence these yellow lines are the required lines."
<h3><u>
Hope this helps!</u></h3>
Answer:
The integrals was calculated.
Step-by-step explanation:
We calculate integrals, and we get:
1) ∫ x^4 ln(x) dx=\frac{x^5 · ln(x)}{5} - \frac{x^5}{25}
2) ∫ arcsin(y) dy= y arcsin(y)+\sqrt{1-y²}
3) ∫ e^{-θ} cos(3θ) dθ = \frac{e^{-θ} ( 3sin(3θ)-cos(3θ) )}{10}
4) \int\limits^1_0 {x^3 · \sqrt{4+x^2} } \, dx = \frac{x²(x²+4)^{3/2}}{5} - \frac{8(x²+4)^{3/2}}{15} = \frac{64}{15} - \frac{5^{3/2}}{3}
5) \int\limits^{π/8}_0 {cos^4 (2x) } \, dx =\frac{sin(8x} + 8sin(4x)+24x}{6}=
=\frac{3π+8}{64}
6) ∫ sin^3 (x) dx = \frac{cos^3 (x)}{3} - cos x
7) ∫ sec^4 (x) tan^3 (x) dx = \frac{tan^6(x)}{6} + \frac{tan^4(x)}{4}
8) ∫ tan^5 (x) sec(x) dx = \frac{sec^5 (x)}{5} -\frac{2sec^3 (x)}{3}+ sec x
It is a reflection across the y axis, then a translation to the right 2 units.
I hope this helps! If you have more you want help with I can help.
