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4 years ago

Last question for my homework

Mathematics

2 answers:

tensa zangetsu [6.8K]4 years ago

6 0

Answer:

A total of 15 cups of punch.

Step-by-step explanation:

20% ( or 0.20) of the punch is pineapple juice so if you have 3 cups of this juice, by proportion, you can make 3 / 0.20 = 15 cups

snow_lady [41]4 years ago

6 0

I think you can make 60%

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If A(4 -6) B(3 -2) and C (5 2) are the vertices of a triangle ABC fine the length of the median AD from A to BC. Also verify tha

Gnoma [55]

Answer:

a) The median AD from A to BC has a length of 6.

b) Areas of triangles ABD and ACD are the same.

Step-by-step explanation:

a) A median is a line that begin in a vertix and end at a midpoint of a side opposite to vertix. As first step the location of the point is determined:

$D (x,y) = \left(\frac{x_{B}+x_{C}}{2},\frac{y_{B}+y_{C}}{2} \right)$

$D(x,y) = \left(\frac{3 + 5}{2},\frac{-2 + 2}{2} \right)$

$D(x,y) = (4,0)$

The length of the median AD is calculated by the Pythagorean Theorem:

$AD = \sqrt{(x_{D}-x_{A})^{2}+ (y_{D}-y_{A})^{2}}$

$AD = \sqrt{(4-4)^{2}+[0-(-6)]^{2}}$

$AD = 6$

The median AD from A to BC has a length of 6.

b) In order to compare both areas, all lengths must be found with the help of Pythagorean Theorem:

$AB = \sqrt{(x_{B}-x_{A})^{2}+ (y_{B}-y_{A})^{2}}$

$AB = \sqrt{(3-4)^{2}+[-2-(-6)]^{2}}$

$AB \approx 4.123$

$AC = \sqrt{(x_{C}-x_{A})^{2}+ (y_{C}-y_{A})^{2}}$

$AC = \sqrt{(5-4)^{2}+[2-(-6)]^{2}}$

$AC \approx 4.123$

$BC = \sqrt{(x_{C}-x_{B})^{2}+ (y_{C}-y_{B})^{2}}$

$BC = \sqrt{(5-3)^{2}+[2-(-2)]^{2}}$

$BC \approx 4.472$

$BD = CD = \frac{1}{2}\cdot BC$ (by the definition of median)

$BD = CD = \frac{1}{2} \cdot (4.472)$

$BD = CD = 2.236$

$AD = 6$

The area of any triangle can be calculated in terms of their side length. Now, equations to determine the areas of triangles ABD and ACD are described below:

$A_{ABD} = \sqrt{s_{ABD}\cdot (s_{ABD}-AB)\cdot (s_{ABD}-BD)\cdot (s_{ABD}-AD)}$ , where $s_{ABD} = \frac{AB+BD+AD}{2}$

$A_{ACD} = \sqrt{s_{ACD}\cdot (s_{ACD}-AC)\cdot (s_{ACD}-CD)\cdot (s_{ACD}-AD)}$ , where $s_{ACD} = \frac{AC+CD+AD}{2}$

Finally,

$s_{ABD} = \frac{4.123+2.236+6}{2}$

$s_{ABD} = 6.180$

$A_{ABD} = \sqrt{(6.180)\cdot (6.180-4.123)\cdot (6.180-2.236)\cdot (6.180-6)}$

$A_{ABD} \approx 3.004$

$s_{ACD} = \frac{4.123+2.236+6}{2}$

$s_{ACD} = 6.180$

$A_{ACD} = \sqrt{(6.180)\cdot (6.180-4.123)\cdot (6.180-2.236)\cdot (6.180-6)}$

$A_{ACD} \approx 3.004$

Therefore, areas of triangles ABD and ACD are the same.

4 0

4 years ago

A science test, which is worth 100 points, consists of 24 questions. Each question is worth either 3 points or 5 points. If x is

marin [14]

A science test, which is worth 100 points, consists of 24 questions. Each question is worth either 3 points or 5 points. If x is the number of 3-point questions and y is the number of 5-point questions, the system shown represents this situation.

x + y = 24

3x + 5y = 100

What does the solution of this system indicate about the questions on the test?

The test contains 4 three-point questions and 20 five-point questions.
The test contains 10 three-point questions and 14 five-point questions.
The test contains 14 three-point questions and 10 five-point questions.
The test contains 20 three-point questions and 8 five-point questions.

5 0

3 years ago

Josie has $80. Addie has $60.

Anika [276]

Both of them will have the same amount of money in 2 months

8 0

3 years ago

Read 2 more answers

Someone help me ASAP!!!

Mila [183]

Given : Anne buys 600 packets of mints.

Given : Of these 600 packets of mints, 34% are small packets and 40% are large packets.

$:\implies$ Total % of small and large packets is (34 + 40) = 74%

It means : The remaining percentage is of the medium packets of mints which were brought by Anne.

$:\implies$ % of medium packets brought by Anne : (100 - 74) = 26%

$:\implies \textsf{Number of medium packets brought by Anne : $\left(\dfrac{26}{100} \times 600\right)$}$

$:\implies \textsf{Number of medium packets brought by Anne : $\left(26 \times 6\right)$ = 156}$

Given : Ben buys 400 packets of mints.

$\textsf{Given : Of these 400 packets of mints, $\dfrac{3}{10}$ are small and $\dfrac{1}{10}$ are large}$

$\implies \textsf{Total fraction of small and large packets brought by Ben :$\left(\dfrac{3}{10} + \dfrac{1}{10}\right)$}$

It means : The remaining fraction is of the medium packets of mints which were brought by Ben.

$\implies \textsf{Fraction of medium packets brought by Ben : $\left(1 - \dfrac{3}{10} - \dfrac{1}{10}\right)$}$

$\implies \textsf{Fraction of medium packets brought by Ben : $\left(1 - \dfrac{4}{10}\right)$ = $\dfrac{6}{10}$}$

$:\implies \textsf{Number of medium packets brought by Ben : $\left(\dfrac{6}{10} \times 400\right)$}$

$:\implies \textsf{Number of medium packets brought by Anne : $\left(6 \times 40\right)$ = 240}$

Given : Chas buys 210 small packets of mints.

Given : Number of small packets : number of medium packets = 3 : 4

$\implies \mathsf{\dfrac{210}{number \ of \ medium \ packets} = \dfrac{3}{4}}$

$\implies \mathsf{Number \ of \ medium \ packets = \left(210 \times \dfrac{4}{3}\right)}$

$\implies \textsf{Number of medium packets brought by Chas = 280}$

Total number of medium packets of mints brought by these three shop keepers (Anne, Ben, Chas) are :

$:\implies$ (156 + 240 + 280)

$:\implies$ 676

5 0

3 years ago

You are planning to take a group of friends to the movies for your birthday. The local movie theater charges $12.99 per person.

aleksley [76]

13 times 4 is 52
9 times 4 is 36 plus the 20 it takes for reservation
So the maximum amount would be 4

7 0

3 years ago

Read 2 more answers