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raketka [301]
3 years ago
9

Solve the equation. Solve the equation. q+59=16 q+5/9=1/6

Mathematics
1 answer:
Anika [276]3 years ago
3 0

Answer:

I keep getting no solution

Step-by-step explanation:

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Step-by-step explanation:

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Which country won<br><br> the last world cup​
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ABCD is a rhombus, where m∠AED = 5x – 10. Use the properties of a rhombus to determine the value of x. Question 10 options: A) x
fomenos

Answer:

A) x = 20

Step-by-step explanation:

ABCD is a rhombus, AC and BD are diagonals which intersect each other at point E.

Since, diagonals of a rhombus are perpendicular bisector.

\therefore \: m \angle AED = 90 \degree \\  \therefore \: (5x - 10) \degree = 90 \degree \\ \therefore \: 5x - 10 = 90  \\ \therefore \: 5x  = 90 + 10  \\ \therefore \: 5x  = 100  \\ \\  \therefore \: x  = \frac{100}{5}  \\  \\ \huge \red{ \boxed{ \therefore \: x  =20}}

5 0
3 years ago
A prism is completely filled with 96 cubes that have edge length of 1/2 cm.
ch4aika [34]
I know this is kinda late, but...

You can solve this easily as long as you remember how to find the volume of a cube. v=a^3, when a means length. The length is 1/2 cm, or .5 cm.
 So, v+.5^3, which equals .125 cm. Each cube has a that volume. Now, multiply that by how many actual cubes there are, 96.

96 * .125 = 12
8 0
3 years ago
Read 2 more answers
I will give lots of points please help
aalyn [17]

Answer:

a)  81π  in³

b)  27  in³

c)  divide the volume of the slice of cake by the volume of the whole cake

d)  10.6%

e)  see explanation

Step-by-step explanation:

<h3><u>Part (a)</u></h3>

The cake can be modeled as a <u>cylinder </u>with:

  • diameter = 9 in
  • height = 4 in

\sf Radius=\dfrac{1}{2}diameter \implies r=4.5\:in

\textsf{Volume of a cylinder}=\sf \pi r^2 h \quad\textsf{(where r is the radius and h is the height)}

\begin{aligned}\sf \implies \textsf{Volume of the cake} & =\pi (4.5)^2(4)\\ & = \sf \pi (20.25)(4)\\ & = \sf81 \pi \:\: in^3\end{aligned}

<h3><u>Part (b)</u></h3>

\begin{aligned}\textsf{Circumference of the cake} & = \sf \pi d\\& = \sf 9 \pi \:\:in\end{aligned}

If each slice of cake has an arc length of 3 in, then the volume of each slice is 3/9π of the entire volume of the cake.

\begin{aligned}\implies \textsf{Volume of slice of cake} & = \sf \dfrac{3}{9 \pi} \times \textsf{volume of cake}\\\\& = \sf \dfrac{3}{9 \pi} \times 81 \pi\\\\& = \sf \dfrac{243 \pi}{9 \pi}\\\\& = \sf 27\:\:in^3\end{aligned}

<h3><u>Part (c)</u></h3>

The volume of each slice of cake is 27 in³.

The volume of the whole cake is 81π in³.

To calculate the probability that the first slice of cake will have the marble, divide the volume of a slice by the volume of the whole cake:

\begin{aligned}\implies \sf Probability & = \sf \dfrac{27}{81 \pi}\\\\& = \sf 0.1061032954...\\\\ & = \sf 10.6\% \:\:(1\:d.p.)\end{aligned}

<h3><u>Part (d)</u></h3>

Probability is approximately 10.6%  (see above for calculation)

<h3><u>Part (e)</u></h3>

If the four slices of cake are cut and passed out <em>before </em>anyone eats or looks for the marble, the probability of getting the marble is the same for everyone. If one slice of cake is cut and checked for the marble before the next slice is cut, the probability will increase as the volume of the entire cake decreases, <u>until the marble is found</u>.  So it depends upon how the cake is cut and distributed as to whether Hattie's strategy makes sense.

3 0
2 years ago
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