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2 years ago

13

Karen stacks a set of cans as shown. The top can has a diameter of 2 inches, the middle can has a diameter of 5 inches, and the

bottom can has a diameter of 6 inches.

1 answer:

german2 years ago

4 0

Answer:

Unknown

Step-by-step explanation:

What is the question? You have the information listed but you forgot to ask the question.

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The temperature, T , of a given mass of gas varies inversely with its volume, V . The temperature of a certain gas with volume o

alekssr [168]

We know that

if t<span>he temperature T of a given mass of gas varies inversely with its volume V
</span>then
T=k/V

Step 1
Find the value of k
for T=30º C and V=105 cm³
we have
T=k/V--------> k=T*V--------> k=30*105=3150 °C*cm³

therefore
for V=84 cm³
T=3150/84=37.5 °C

the answer is 37.5 °C

3 0

3 years ago

Mr. price drives 123 miles each day on his bus route he gets reimbursed .56 per every mile he drives. How much will mr. price re

Andrei [34K]

he will be reimbursed 137.26 dollars for two days

6 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

What is a fraction box

vichka [17]

I don't know exactly what a fraction box is sorry

3 0

3 years ago

Consider the enlargement of the pentagon. A small pentagon has a bottom side length of x centimeters and left side length of 7 c

il63 [147K]

Answer:

3.3 cm

Step-by-step explanation:

From the above question, we can draw out a proportion

= Small pentagon = Large pentagon

= x/7 = 7/15

Cross Multiply

x × 15 = 7 × 7

15x = 49

x = 49/15

x = 3.2666666667 cm

Approximately= 3.3 cm

7 0

2 years ago

Read 2 more answers