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3 years ago

What is 2 4/5 divided by 1 1/3

Mathematics

1 answer:

marishachu [46]3 years ago

3 0

Answer:

2.1 or 21/10

Step-by-step explanation:

2 4/5--14/5 and 1 1/3--4/3

multiply by reciprocal--14/5 * 3/4--42/20=2.1

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A train travels from station A to C having distance of 285 miles in 6 hours. During the journey, it reaches intermediate station

Nesterboy [21]

Answer:

Distance between B and C = $3\times 55=165miles$

Step-by-step explanation:

Let the time taken to reach from station A to B be x hr

Let the time taken to reach from station B to C be y hr

Given total time from station A to C be 6 hr

∴ x + y = 6

⇒ y = 6 - x → (1)

Also total distance given is 285 miles

As $Distance=Speed\times Time$

∴ distance travelled between station A and B

= $40\times xmiles$

distance travelled between station B and station C

= $55\times ymiles$

So, 40x + 55y = 285

⇒ 40x + 55(6 -x) = 285 (From 1)

=> 40x + 330 - 55x = 285

=> -15x = 285 - 330

=> -15x = -45 ⇒ x=3

gives y = 6-3 = 3 hours

Distance between B and C = $3\times 55=165miles$

5 0

3 years ago

10. Given matrices A, B, and C below, perform the indicated operations if possible. If the operation is not possible, explain wh

Artemon [7]

Answer(a):

Order of matrix A = 3x3 {because number of rows=3, number of columns=3}

Order of matrix B = 3x3

Order of matrix C = 1x3

3A will be of order 3x3 because multiply by scalar doesn't change the order. Which is same as the order of matrix B.

So 3A+B is possible.

We simply multiply all elements of A with 3 and add them with corresponding elements of B.

$3\begin{bmatrix}2 & -1& 0\\ 0 & 5& 0.3\\ 1 & 4&10 \end{bmatrix}+\begin{bmatrix}5 & 0& 2\\ 1 & -3& 9\\ 2 & 0& 4\end{bmatrix}= \begin{bmatrix}11 & -3& 2\\ 1 & 12& 9.9\\ 5 & 12& 34\end{bmatrix}$

Answer(b):

2B will be of order 3x3 because multiply by scalar doesn't change the order. Which is NOT same as the order of matrix C.

So 2B+C is NOT possible.

Answer(c):

CA is possible only if

number of columns of C = number of columns of A

3=3

which is true hence CA is possible.

$\begin{bmatrix}1 &3 &5 \end{bmatrix}*\begin{bmatrix}2 & -1& 0\\ 0 & 5& 0.3\\ 1 & 4&10 \end{bmatrix}=\begin{bmatrix}1*2+3*0+5*1 & 1*-1+3*5+5*4 & 1*0+3*0.3+5*10\end{bmatrix}$

= $\begin{bmatrix}7 & 34& 50.9\end{bmatrix}$

If the matrix is not visible properly then look at the attached picture.

5 0

3 years ago

How do you solve for a and b showing your work

Mariulka [41]

Answer:

a) 4

b) $\frac{1}{2} \sqrt{y^5}$

Step-by-step explanation:

PART A

Simplify the expression by solving out the square roots.

$\frac{\sqrt{16} \sqrt{p} }{\sqrt{p}}$ = $\frac{4 \sqrt{p} }{\sqrt{p}}$

You can cancel out the $\sqrt{p}$ on the top and bottom.

So, a should be 4.

PART B

Solve out the square root. The square root is pretty much the 1/2 power, so you can divide the exponents by half.

$\frac{\sqrt{y^{10}} }{\sqrt{4y^{5}} }$ = $\frac{y^{5}}{2y^{\frac{5}{2}} }$

You can subtract exponents to divide them. So the y variable becomes $y^{5 - 5/2}$ , or $y^{5/2}$ .

Then, the overall expression is $\frac{1}{2} y^{\frac{5}{2} }$ , or you can turn that into $\frac{1}{2} \sqrt{y^5}$ .

5 0

4 years ago

Generate the nest three terms of each arithmetic sequence shown below.

o-na [289]

Answer:

A)2,6,10

B)2,-6,-18

C)-1,-3,-5

Step-by-step explanation:

A)a1=-2 and d=4

We know that the arithmetic sequence formula is

$a_{n}=a_{1}+(n-1)d$

Now

$a_{2}=a_{1}+(2-1)d$

$a_{2}=a_{1}+(1)d$

Substituting the given value we get

$a_{2}= -2+(1)4$

$a_{2}= -2+4$

$a_{2}= 2$

------------------------------------------

Similarly

$a_{3}=a_{1}+(3-1)d$

$a_{3}=a_{1}+(2)d$

Substituting the given value we get

$a_{3}= -2+(2)4$

$a_{3}= -2+8$

$a_{3}= 6$

-------------------------------------------------

$a_{4}=a_{1}+(4-1)d$

$a_{4}=a_{1}+(3)d$

Substituting the given value we get

$a_{4}= -2+(3)4$

$a_{4}= -2+16$

$a_{4}= 10$

-------------------------------------------------------------------------------------------

B $a_n=a_{(n-1)}-8$ with $a_1=10$

$a_2=a_{(2-1)}-8$

$a_2=a_{1}-8$

Substituting the given value

$a_2= 10-8$

$a_2=2$

---------------------------------------------------------------------

$a_3=a_{(3-1)}-8$

$a_3=a_{2}-8$

Substituting the value

$a_3=2-8$

$a_3= -6$

---------------------------------------------------------------------

$a_4=a_{(4-1)}-8$

$a_4=a_{3}-8$

Substituting the value

$a_4= -6-8$

$a_4= -14$

-------------------------------------------------------------------------------------------

C) $a_1=3, a_2=1$

Here the difference is -2

the arithmetic sequence formula is

$a_{n}=a_{1}+(n-1)d$

Now

$a_{3}=a_{1}+(3-1)d$

$a_{3}=a_{1}+(2)d$

Substituting the value we get

$a_{3}= 3+(2)-2$

$a_{3}= 3-4$

$a_{3}= -1$

------------------------------------------------------------------------------

$a_{4}=a_{1}+(4-1)d$

$a_{4}=a_{1}+(3)d$

Substituting the value we get

$a_{4}= 3+(3)-2$

$a_{4}= 3-6$

$a_{4}= -3$

----------------------------------------------------------------------------------

$a_{5}=a_{1}+(5-1)d$

$a_{5}=a_{1}+(4)d$

Substituting the value we get

$a_{5}= 3+(4)-2$

$a_{5}= 3-8$

$a_{5}= -5$

4 0

3 years ago

Plz help

Firdavs [7]

Answer

Step-by-step explanation:

y=2x+15

7=2(-4)+15

7=15-8

7=7

7x+2y=-14

7(-4)+2(7)=-14

-28+14=-14

-14=-14

3 0

3 years ago