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2 years ago

Help me once more cause im in trouble

Mathematics

1 answer:

Sedaia [141]2 years ago

8 0

Answer: 36

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Explanation:

There are 35 girls and 50 boys. Making 35+50 = 85 students total.

Of the 85 students, 42% (see column 1, bottom row) bought their lunch from the cafeteria. So 0.42*85 = 35.7 which rounds to 36

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Find the range for these numbers. 12, 9, 24, 24, 37, 18, 9, 6, 24

svet-max [94.6K]

6, 9, 9, 12, 18, 24, 24, 24, 37

37-6= 31

5 0

2 years ago

Simplify each expression then factor the expression 14-28t-5(7+10t)

ludmilkaskok [199]

Answer:

Simplified: -21-98t

Factored: -7(14t+3)

Step-by-step explanation:

The given expresion is:

14-28t-5(7+10t)

We expand the parenthesis to get:

$14 - 28t -35 - 70t$

We group like terms to get:

$14 -35 - 28t- 70t$

We combine similar terms to get:

$- 21- 98t$

We now factor to obtain:

$- 21- 98t = - 7(14t + 3)$

5 0

3 years ago

Find the taylor polynomial t3(x) for the function f centered at the number

Vlad [161]

$e^{-3x}=\displaystyle\sum_{n=0}^\infty\frac{(-3x)^n}{n!}=1-3x+9x^2+\cdots$
$\sin2x=\displaystyle\sum_{n=0}^\infty\frac{(-1)^n(2x)^{2n+1}}{(2n+1)!}=2x-\dfrac{4x^3}3+\cdots$

$e^{-3x}\sin2x=\left(1-3x+9x^2+\cdots\right)\left(2x-\dfrac{4x^3}3+\cdots\right)$
$\approx T_3(x)=(1-3x+9x^2)\left(2x-\dfrac{4x^3}3\right)$
$T_3(x)=2x-6x^2+\left(18-\dfrac43\right)x^3$
$T_3(x)=2x-6x^2+\dfrac{50}3x^3$

5 0

2 years ago

Consider the following system of equations:

Delicious77 [7]

Answer:

The system has infinitely solutions

Step-by-step explanation:

we have

$-\frac{1}{3}x^{2}=-\frac{5}{6}+\frac{1}{3}y^{2}$

$\frac{1}{3}x^{2}+\frac{1}{3}y^{2}=\frac{5}{6}$

Multiply by 3 both sides

$x^{2}+y^{2}=\frac{5}{2}$ ----> equation A

The equation A is a circle centered at origin with radius $r=\sqrt{5/2}\ units$

and

$5y^{2} =\frac{25}{2}-5x^{2}$

$5x^{2}+5y^{2} =\frac{25}{2}$

Divide by 5 both sides

$x^{2}+y^{2} =\frac{5}{2}$ ----> equation B

The equation B is a circle centered at origin with radius $r=\sqrt{5/2}\ units$

Equation A and Equation B are the same

Therefore

The system has infinitely solutions

7 0

2 years ago

Read 2 more answers

Find the x and y intercepts of the quadratic equation f(x)=-x^2+4x-4

earnstyle [38]

X - Intercept
f(x) = -x² + 4x - 4
 0 = -x² + 4x - 4
 x = -(4) +/- √((4)² - 4(-1)(-4))
 2(-1)
 x = -4 +/- √(16 - 16)
 -2
 x = -4 +/- √(0)
 -2
 x = -4 +/- 0
 -2
 x = -4 + 0 x = -4 - 0
 -2 -2
 x = -4 x = -4
 -2 -2
 x = 2 x = 2

The solution to the problem is {2, 2}, or {2}. The x - intercept of the problem is (2, 0).

Y - Intercept
f(x) = -x² + 4x - 4
f(x) = -(0)² + 4(0) - 4
f(x) = -(0) + 0 - 4
f(x) = -0 + 0 - 4
f(x) = 0 - 4
f(x) = -4

The y - intercept of the problem is (0, -4).

8 0

3 years ago