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jok3333 [9.3K]
4 years ago
7

Divide: 7/8 ÷ 3/8 A) 21/64 B) 3/7 C) 64/21 D) 7/3

Mathematics
1 answer:
lozanna [386]4 years ago
5 0

Answer:

When you need to divide two fractions, reverse the second fraction, then multiply the top numbers by each other and the bottom numbers by each other.


7/8 / 3/8 =

7/8 x 8/3 =

(7 x 8) / (8 x 3) =

56/24 which reduces to 7/3

The answer is D) 7/3


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Please help me :(
stiv31 [10]

Answer:

Step-by-step explanation:

Cos \ D = \dfrac{adjacent \ side \ of angle \ D}{hypotenuse}\\\\\\ Cos \ D = \dfrac{72}{75}\\\\\\Cos D = \dfrac{24}{25}

5 0
3 years ago
Suppose the solutions of a homogeneous system of four linear equations in five unknowns are all multiples of one nonzero solutio
Akimi4 [234]

Consider a homogeneous machine of four linear equations in five unknowns are all multiples of 1 non-0 solution. Objective is to give an explanation for the gadget have an answer for each viable preference of constants on the proper facets of the equations.

Yes, it's miles true.

Consider the machine as Ax = 0. in which A is 4x5 matrix.

From given dim Nul A=1. Since, the rank theorem states that

The dimensions of the column space and the row space of a mxn matrix A are equal. This not unusual size, the rank of matrix A, additionally equals the number of pivot positions in A and satisfies the equation

rank A+ dim NulA = n

dim NulA =n- rank A

Rank A = 5 - dim Nul A

Rank A = 4

Thus, the measurement of dim Col A = rank A = five

And since Col A is a subspace of R^4, Col A = R^4.

So, every vector b in R^4 also in Col A, and Ax = b, has an answer for all b. Hence, the structures have an answer for every viable preference of constants on the right aspects of the equations.

8 0
3 years ago
Round 7.2352297363 to the nearest millionth.
Natasha2012 [34]
7.2352297363 rounded to the nearest millionth would be 7.235230 and all I did to get the answer was count from left to right and keep the first six numbers, and round 229 up one spot and make it 230, to get the answer.
7 0
3 years ago
If 3^2x+1 = 3^x+5 what is the value of X
cupoosta [38]

Answer:

4

Step-by-step explanation:

Since the base is the same, 3, you can just get rid of it using the principles of log. What you're left with is:

2x + 1 = x + 5

Now just solve for x:

2x - x = 5 - 1

x = 4

7 0
2 years ago
How many permutations of the 26 letters of the English alphabet do not contain any of the strings fish, rat, or bird
NARA [144]

The number of permutations of the 26 letters of the English alphabet that do not contain any of the strings fish, rat, or bird is 402619359782336797900800000

Let

\mathcal{E}=\{\text{All lowercase letters of the English Alphabet}\}\\\\B=\overline{\{b,i,r,d\}} \cup \{bird\}\\\\F=\overline{\{f,i,s,h\}} \cup \{fish\}\\\\R=\overline{\{r,a,t\}} \cup \{rat\}\\\\FR=\overline{\{f,i,s,h,r,a,t\}} \cup \{fish,rat\}

Then

Perm(\mathcal{E})=\{\text{All orderings of all the elements of } \mathcal{E}\}\\\\Perm(B)=\{\text{All orderings of all the elements of } \mathcal{E} \text{ containing bird}\}\\\\Perm(F)=\{\text{All orderings of all the elements of } \mathcal{E} \text{ containing fish}\}\\\\Perm(R)=\{\text{All orderings of all the elements of } \mathcal{E} \text{ containing rat}\}\\\\Perm(FR)=\{\text{All orderings of all the elements of } \mathcal{E} \text{ containing both fish and rat}\}\\

Note that since

F \cap R=\varnothing, Perm(F)\cap Perm(R)\ne \varnothing

But since

B \cap R \ne \varnothing, Perm(B)\cap Perm(R)= \varnothing

and

B \cap F \ne \varnothing , Perm(B)\cap Perm(F)= \varnothing

Since

|\mathcal{E} |=26 \text{, then, } |Perm(\mathcal{E})|=26! \\\\|B|=26-4+1=23 \text{, then, } |Perm(B)|=23!\\\\|F|=26-4+1=23 \text{, then, } |Perm(F)|=23!\\\\|R|=26-3+1=24 \text{, then, } |Perm(R)|=24!\\\\|FR|=26-7+2=21 \text{, then, } |Perm(FR)|=21!\\

where |Perm(X)|=\text{number of possible permutations of the elements of X taking all at once}

and

|Perm(F) \cup Perm(R)| = |Perm(F)| + |Perm(R)| - |Perm(FR)|\\= 23!+24!- 21! \text{ possibilities}

What we are looking for is the number of permutations of the 26 letters of the alphabet that do  not contain the strings fish, rat or bird, or

|Perm(\mathcal{E})|-|Perm(B)|-|Perm(F)\cup Perm(R)|\\= 26!-23!-(23!+24!- 21!)\\= 402619359782336797900800000 \text{ possibilities}

This link contains another solved problem on permutations:

brainly.com/question/7951365

6 0
3 years ago
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