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3 years ago

I need 9 and 10 plzzz!!

Mathematics

1 answer:

Alex73 [517]3 years ago

6 0

Number of students N, y years

N = 700 + 50y

N = 700 + 50 x 10 = 1200 (C)

Left not painted percentage = 100 - 60 = 40 %

Left = 40/100 x 45 = 18 (H)

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Enter number as an integer or decimal.

Strike441 [17]

Answer:

$x$

Step-by-step explanation:

$30-4x>6x-10$

Adding 10 to both sides.

$10+30-4x>6x-10+10$

$40-4x>6x$

Adding $4x$ to both sides.

$4x+40-4x>6x+4x$

$40>10x$

Dividing both sides by 10.

$\frac{40}{10}>\frac{10x}{10}$

$4>x$

∴ $x$

3 0

3 years ago

How do I divide 178.8 by 24?

Doss [256]

7.45 is the final answer

178.8/ 24 = about 7.45

4 0

3 years ago

Find the missing side length image below

barxatty [35]

Answer:

Step-by-step explanation:

Based on the Proportional Transversal Theorem, the three parallel lines hat intersects the two transversals, divides the transversal lines proportionally.

Therefore, we would have the following ratio:

28/35 = ?/50

Cross multiply

35*? = 50*28

35*? = 1,400

Divide both sides by 35

? = 1400/35

? = 40

4 0

3 years ago

Evaluate 5(x - 1) - 2 when x = 3. ОА. 12 ОВ. -4 Ос. о. O D. 8

Harlamova29_29 [7]

Answer:

Step-by-step explanation:

5(3-1)-2

5(2)-2

10-2

7 0

3 years ago

Read 2 more answers

Solve the equation on the interval [0,2π]

WARRIOR [948]

$\bf 16sin^5(x)+2sin(x)=12sin^3(x)
\\\\\\
16sin^5(x)+2sin(x)-12sin^3(x)=0
\\\\\\
\stackrel{common~factor}{2sin(x)}[8sin^4(x)+1-6sin^2(x)]=0\\\\
-------------------------------\\\\
2sin(x)=0\implies sin(x)=0\implies \measuredangle x=sin^{-1}(0)\implies \measuredangle x=
\begin{cases}
0\\
\pi \\
2\pi 
\end{cases}\\\\
-------------------------------$

$\bf 8sin^4(x)+1-6sin^2(x)=0\implies 8sin^4(x)-6sin^2(x)+1=0$

now, this is a quadratic equation, but the roots do not come out as integers, however it does have them, the discriminant, b² - 4ac, is positive, so it has 2 roots, so we'll plug it in the quadratic formula,

$\bf 8sin^4(x)-6sin^2(x)+1=0\implies 8[~[sin(x)]^2~]^2-6[sin(x)]^2+1=0
\\\\\\
~~~~~~~~~~~~\textit{quadratic formula}
\\\\
\begin{array}{lcccl}
& 8 sin^4& -6 sin^2(x)& +1\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array} 
\qquad \qquad 
sin(x)= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a}
\\\\\\
sin(x)=\cfrac{-(-6)\pm\sqrt{(-6)^2-4(8)(1)}}{2(8)}\implies sin(x)=\cfrac{6\pm\sqrt{4}}{16}
\\\\\\
sin(x)=\cfrac{6\pm 2}{16}\implies sin(x)=
\begin{cases}
\frac{1}{2}\\\\
\frac{1}{4}
\end{cases}$

$\bf \measuredangle x=
\begin{cases}
sin^{-1}\left( \frac{1}{2} \right)
sin^{-1}\left( \frac{1}{4} \right)
\end{cases}\implies \measuredangle x=
\begin{cases}
\frac{\pi }{6}~,~\frac{5\pi }{6}\\
----------\\
\approx~0.252680~radians\\
\qquad or\\
\approx~14.47751~de grees\\
----------\\
\pi -0.252680\\
\approx 2.88891~radians\\
\qquad or\\
180-14.47751\\
\approx 165.52249~de grees
\end{cases}$

3 0

3 years ago