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VladimirAG [237]
3 years ago
15

Costas and Sheila bought a raffle tickets. The grand prize is $100.00. Costas gave $3.50 and Sheila gave $1.50. If they win they

will share the money in the same proportion that they paid for. How much of the money should each person receive
Mathematics
1 answer:
Semenov [28]3 years ago
5 0

Answer: Costas gets $70 and Sheila gets $30

Step-by-step explanation:

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A cylinder and a cone have the same volume. The cylinder has a radius of 2 inches and a height of 3 inches. The cone has a radiu
Sloan [31]

Answer:

so lets get the cones volume first the formula for a cone is V=πr2/3

so V=πr2h

3=π·22·3

3≈12.56637

and then the height of the cylinder is 4

Step-by-step explanation:

7 0
3 years ago
In right ABC, AN is the altitude to the hypotenuse. FindBN, AN, and AC,if AB =2 5 in, and NC= 1 in.
Rama09 [41]

From the statement of the problem, we have:

• a right triangle △ABC,

,

• the altitude to the hypotenuse is denoted AN,

,

• AB = 2√5 in,

,

• NC = 1 in.

Using the data above, we draw the following diagram:

We must compute BN, AN and AC.

To solve this problem, we will use Pitagoras Theorem, which states that:

h^2=a^2+b^2\text{.}

Where h is the hypotenuse, a and b the sides of a right triangle.

(I) From the picture, we see that we have two sub right triangles:

1) △ANC with sides:

• h = AC,

,

• a = ,NC = 1,,

,

• b = NA.

2) △ANB with sides:

• h = ,AB = 2√5,,

,

• a = BN,

,

• b = NA,

Replacing the data of the triangles in Pitagoras, Theorem, we get the following equations:

\begin{cases}AC^2=1^2+NA^2, \\ (2\sqrt[]{5})^2=BN^2+NA^2\text{.}\end{cases}\Rightarrow\begin{cases}NA^2=AC^2-1, \\ NA^2=20-BN^2\text{.}\end{cases}

Equalling the last two equations, we have:

\begin{gathered} AC^2-1=20-BN^2.^{} \\ AC^2=21-BN^2\text{.} \end{gathered}

(II) To find the values of AC and BN we need another equation. We find that equation applying the Pigatoras Theorem to the sides of the bigger right triangle:

3) △ABC has sides:

• h = BC = ,BN + 1,,

,

• a = AC,

,

• b = ,AB = 2√5,,

Replacing these data in Pitagoras Theorem, we have:

\begin{gathered} \mleft(BN+1\mright)^2=(2\sqrt[]{5})^2+AC^2 \\ (BN+1)^2=20+AC^2, \\ AC^2=(BN+1)^2-20. \end{gathered}

Equalling the last equation to the one from (I), we have:

\begin{gathered} 21-BN^2=(BN+1)^2-20, \\ 21-BN^2=BN^2+2BN+1-20 \\ 2BN^2+2BN-40=0, \\ BN^2+BN-20=0. \end{gathered}

(III) Solving for BN the last quadratic equation, we get two values:

\begin{gathered} BN=4, \\ BN=-5. \end{gathered}

Because BN is a length, we must discard the negative value. So we have:

BN=4.

Replacing this value in the equation for AC, we get:

\begin{gathered} AC^2=21-4^2, \\ AC^2=5, \\ AC=\sqrt[]{5}. \end{gathered}

Finally, replacing the value of AC in the equation of NA, we get:

\begin{gathered} NA^2=(\sqrt[]{5})^2-1, \\ NA^2=5-1, \\ NA=\sqrt[]{4}, \\ AN=NA=2. \end{gathered}

Answers

The lengths of the sides are:

• BN = 4 in,

,

• AN = 2 in,

,

• AC = √5 in.

7 0
1 year ago
4c^2-8c-1=0 complete the square round to nearest hundredth
olga55 [171]
Move constant to other side
add 1
4c^2-8c=1
divide by 4 to make leading coeficient 1
c^2-2c=1/4
take 1/2 of linear coeficient and square it
-2/2=-1, (-1)^2=1
add that to both sides
c^2-2c+1=1/4+1
factor perfect squaer and add
(c-1)^2=5/4
square root both sides
c-1=+/-(√5)/2
add 1
c=1+/-(√5)/2

c=2.12 or -0.12
8 0
3 years ago
When using a vertical number line do the negative numbers go on the bottom or top?
pshichka [43]
Bottom Like They Go on the left horizontally
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3 years ago
Olive oil costs £3.50 for 1 litrehow much 200 ml of olive oil cost
Nookie1986 [14]
1000 ml costs £3.50
200 is 1/5 of 1000 so it will be £0.7
4 0
3 years ago
Read 2 more answers
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