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2 years ago

Help a broooo outttt

Mathematics

1 answer:

fiasKO [112]2 years ago

7 0

Answer:

Step-by-step explanation:

Let the missing length be represented as "x".

Based on the angle bisector theorem, we have the following equation:

x/21 = 30/15

x/21 = 2/1

Cross multiply

x*1 = 2*21

x = 42

Missing length = 42

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Cna i get some help with dis plz

Tomtit [17]

Let $b_1,b_2,\ldots,b_{20}$ be the 20 marks of the boys, and $g_1,g_2,\ldots,g_{10}$ be the 10 marks of the girls.

We know that the global mean was 70, meaning that

$\dfrac{b_1+b_2+\ldots+b_{20}+g_1+g_2+\ldots+g_{10}}{30}=70$

Multiplying both sides by 30 we deduce that the sum of the scores of the whole classroom is

$b_1+b_2+\ldots+b_{20}+g_1+g_2+\ldots+g_{10}=2100$

By the same logic, we work with the marks of the boys alone: we know the average:

$\dfrac{b_1+b_2+\ldots+b_{20}}{20}=62$

And we deduce the sum of the marks for the boys:

$b_1+b_2+\ldots+b_{20}=1240$

Which implies that the sum of the marks of the girls is $2100-1240=860$

And finally, the mean for the girls alone is

$\dfrac{860}{10}=86$

6 0

2 years ago

List all partitions of the set. (a) (1,2) (b) [a, b,c)

Alchen [17]

Answer: Hello mate!

The partition of a set is defined as a partition of the set into a nonempty subset, where the set itself is a subset of himself, then the set is a partition of himself.

a) in this we have a set of two objects; A = (1,2) the partitions of this set are: (∅), (1), (2) and (1,2). Where (∅) is the null set.

b) Now we have a set of three objects; B = (a,b,c) the partitions of this set are: (∅), (a), (b), (c), (a,b), (a,c), (b,c), (a,b,c)

3 0

2 years ago

A shopkeeper had 410 tangerines. He put some of them into 15 cartons containing

vladimir1956 [14]

Answer:

15*14 + 14*x = 410

x = 200/14 = 14 4/14 <===== 14*14 + 4 = 200

Step-by-step explanation:

8 0

2 years ago

A recipe requires 3/4 cup of nuts for one cake.

lana66690 [7]

10. How I would solve it, is I would convert them into decimals so it is 7.5/.75.

6 0

2 years ago

Read 2 more answers

Chose parameters h and k such that the system has a) a unique solution, b) many solutions, and c) no solution. 31+3x2= 4 2x1+kx2

Snezhnost [94]

Answer

a) k=7, h=9, the unique solution of the system is $(x_1,x_2)=(1,1)$

b) If k=6 and h=8 the system has infinite solutions.

c)If k=6 and h=9 the system has no solutions.

Step-by-step explanation:

I am assuming that the system is x1+3x2=4; 2x1+kx2=h

The augmented matrix of the system is $\left[\begin{array}{ccc}1&3&4\\2&k&h\end{array}\right]$ . If two times the row 1 is subtracted to row 2 we get the following matrix $\left[\begin{array}{ccc}1&3&4\\0&k-6&h-8\end{array}\right]$ .

Then

a) If k=7 and h=9, the unique solution of the system is $x_2=\frac{9-8}{7-6}=\frac{1}{1}=1$ and solviong for $x_1$ ,

$x_1+3x_2=4\\\\x_1=4-3(1)=1$

Then the solution is $(x_1,x_2)=(1,1)$

b) If k=6 and h=8 the system has infinite solutions because the echelon form of the matrix has a free variable.

c)If k=6 and h=9 the system has no solutions because the last equation of the system of the echelon form of the matrix is $0x_2=1$

8 0

3 years ago