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scoray [572]
3 years ago
13

Determine the quotient of 2 over 5 divided by 3 over 4. (2 points) PLS I NEED HELP FAST PLS

Mathematics
2 answers:
marshall27 [118]3 years ago
8 0

Answer:

8/15

Step-by-step explanation:

Bess [88]3 years ago
7 0

Answer:

8 over 15

Step-by-step explanation:

you should get a calculator.

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Is (1, 3) a solution to the system of equations listed below? (1 point) y= 6x -3 y = x - 2
ololo11 [35]

Answer:

The solution for given system of equation is: x=\frac{1}{5}\:and\:y=\frac{-9}{5} and ordered pair is: \mathbf{(\frac{1}{5},\frac{-9}{5})}

So, (1,3) is not solution to the given system of equations.

Step-by-step explanation:

we can solve the system of equations to find the value of x and y and then verify if (1,3) is a solution or not.

The system of equation given is:

y=6x-3\\y=x-2

Solving:

Let:

y=6x-3--eq(1)\\y=x-2--eq(2)

Put value of y from equation 2 into equation 1

y=6x-3\\Put\:y=x-2\\x-2=6x-3\\x-6x=-3+2\\-5x=-1\\x=\frac{-1}{-5}\\x=\frac{1}{5}

Now, put value of x in equation 2 to find value of y

y=x-2\\Put\:x=\frac{1}{5} \\y=\frac{1}{5} -2\\y=\frac{1-2*5}{5} \\y=\frac{1-10}{5}\\ y=\frac{-9}{5}

So, the solution for given system of equation is: x=\frac{1}{5}\:and\:y=\frac{-9}{5} and ordered pair is: \mathbf{(\frac{1}{5},\frac{-9}{5})}

So, (1,3) is not solution to the given system of equations.

7 0
2 years ago
When dividing 82 by 43, Linda estimated the quotient to be 2. Examine Linda’s work, and explain what
Vanyuwa [196]

Answer:

82/43 = 1 and 39/43

Step-by-step explanation:

When dividing 82 by 43 , the operation follows;

82/43 = 1 and 39/43

This means the quotient is 1 and reminder is 39

However;

82/43 = 1.907 ------nearest whole number is 2.

so the quotient can be approximated to the nearest whole number as 2.

8 0
3 years ago
(1) Let {v1,v2,v3} be a set of vectors in Rn . If u is Span {v1,v2,v3}, show that 3u is in Span {v1,v2,v3}.
Evgesh-ka [11]

Answer:

(1)

Multiplying by 3 both sides of the equality you get that

3u = 3c_1v_1+3c_2v_2+3c_3v_3

3u  is in the Span of the vectors \{v_1,v_2,v_3\}.

(2)

That's not true, consider the following counter example.

v_1 = (0,0,0,1)\\v_2 = (0,0,1,0)\\v_3 = (0,1,0,0)\\v_4 = (1,0,0,0)\\u = (0,1,1,1)

u is a linear combination of v_1,v_2,v_3 but is NOT a linear combination of v_1,v_2,v_3,v_4.

Step-by-step explanation:

(1)

As the hint indicates, you know that

u = c_1 v_1 + c_2v_2+c_3v_3

Then, if you multiply both sides of the equality by 3, you get that

3u = 3c_1v_1+3c_2v_2+3c_3v_3

And that's it. 3u  is in the Span of the vectors \{v_1,v_2,v_3\}

(2)

That's not true, consider the following counter example.

v_1 = (0,0,0,1)\\v_2 = (0,0,1,0)\\v_3 = (0,1,0,0)\\v_4 = (1,0,0,0)\\u = (0,1,1,1)

u is a linear combination of v_1,v_2,v_3 but is NOT a linear combination of v_1,v_2,v_3,v_4.

4 0
3 years ago
3/8 of the students and Mrs. Miles class rides a bus there are 24 students in the class how many students ride the bus
irina1246 [14]
9 students ride the bus.
3 0
3 years ago
Which of the following statements is always true about the function y=a•b^x
OLEGan [10]

Answer:

c. When a > 0 and b > 1, the function models exponential growth.

Step-by-step explanation:

You have exponential decay when a > 0 and 0 < b < 1.

5 0
3 years ago
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