8x^3 -5x^2 + 8x + 9+5x^3 + 3x^2 - 5x + 4 =
8x^3+5x^3-5x^2 + 3x^2+ 8x - 5x+ 9<span>+ 4 = </span>
13x^3-2x^2+3x+13
There are 600 students including the seventh and eighth graders at the party.
This problem uses the concept of percentages to define the conditions that are laid in front of us.
Let the original number of students be S , and the number of seventh graders be = 0.60S
We know that percent is used to convey the mathematical term of a fraction multiplied by 100.
Total students after 20 eighth graders arrive = S + 20
And we have that
Number of seventh graders / total number of students = 58%
.60S / [ S + 20 ] = .58 we multiply both sides by S + 20
0.60S =0 .58 [ S + 20]
.60S = .58S + 11.6 we subtract 0.58S from both the sides
0.02S = 11.6 we divide both the sides by .02
S = 11/6 / 0.02 = 580
So the total number of students = 580 + 20 = 600 .
Hence there are 600 students at the party at that time.
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Using trigonometric ratio, the value of x is 63.6°
<h3>Trigonometric Ratio</h3>
This is the ratios of sides of a right-angled triangle with respect to any of its acute angles are known as the trigonometric ratios of that particular angle.
Trigonometric ratio are often coined as SOHCAHTOA
In the given triangle, we need to find the value of x using trigonomtric ratio.
Since we have the value of adjacent and hypothenuse, we definitely need to use cosine
cosθ = adjacent / hypothenuse
adjacent = 4
hypothenuse = 9
Substituting the values into the equation;
cos θ = 4 / 9
cos θ = 0.444
θ = cos⁻¹ 0.4444
θ = 63.6°
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