HELP PLS I WILL MARK BRAINLYIiEST ASAP!!!!

1. The teacher assigns homework after 7/10

asambeis [7]

Answer:

B) Likely

Step-by-step explanation:

Unlikely = < 50%

Likely = > 50%

Certain = 100%

Equally likely = 50%

8 0

2 years ago

Look at the image for the problem :)

8_murik_8 [283]

Answer:

$\displaystyle V = 28 \pi \ ft^3$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Geometry

Volume of a Cylinder Formula: $\displaystyle V = \pi r^2h$

V is volume
r is radius
h is height

Step-by-step explanation:

Step 1: Define

Radius r = 2 ft

Height h = 7 ft

Step 2: Solve for V

Substitute in variables [Volume of a Cylinder Formula]: $\displaystyle V = \pi (2 \ ft)^2(7 \ ft)$
[Volume] Evaluate exponents: $\displaystyle V = \pi (4 \ ft^2)(7 \ ft)$
[Volume] Multiply: $\displaystyle V = 28 \pi \ ft^3$

5 0

2 years ago

Write the equation of the line with a slope of 0 and containing the point (- 6, 5).

defon

Y=mx+b
5=0(-6)+b
b=5
So equation
y=0x+5

Or

y=5

4 0

2 years ago

Identify the number of solutions. Y< -1/3x+5, y> -1/3x-1.

Elenna [48]

Answer: C. No solution. ( top shaded. Middle not. Bottom shaded.

Step-by-step explanation:

Given inequalities

y< -1/3x+5 and

y> -1/3x-1.

If we draw graph for y< -1/3x+5 equation, it would be a dotted line and we would shade down the dotted line because we have less than < symbol there.

Now, if we draw graph for second inequality y> -1/3x-1, it would also be a dotted line and we would shade up of dotted line because we have greater than > symbol there.

Now, we can see that slopes of both dotted lines are same that is -1/3.

So, there would not be any common shaded region.

Therefore, there would not by any solution of the system of inequalities.

And correct option is C option.

C. No solution. ( top shaded. Middle not. Bottom shaded.

8 0

3 years ago

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