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3 years ago

1 3/4 divided by 1/2-1 1/2 to the 3rd

Mathematics

1 answer:

Marina CMI [18]3 years ago

6 0

Answer:

-7/4

1/8

Step-by-step explanation:

1. If the problem is; (1 3/4) / (1/2 - 1 1/2)^(3)

(a). Convert to improper fractions;

(7/4) / (1/2 - 3/2)^3

(b). Simplify;

(7/4) / (-2/2)^3

(7/4) / (-1)^3

(7/4) / (-1)

-7/4

2. If the problem is; ((1 3/4) / (1/2)) - (1 1/2)^3

(a). Convert to improper fractions;

((7/4)/ (1/2)) - (3/2)^3

(b). Simplify;

((7/4)/ (1/2)) - (3/2)^3

((7/4)*(2/1)) - (27/8)

(7/2) - (27/8)

(28/8) - (27/8)

1/8

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3. The product of and 29 -26 3 is ...

eimsori [14]

Answer:

(29-26) * 3 = 9

Step-by-step explanation:

7 0

3 years ago

I need Help!!!! Please Solve and show WORK!!! 3(x+2)=15

ludmilkaskok [199]

Answer: x=3

Step-by-step explanation: First we do distributive property in the parentheses. Next we subtract both sides by 6. Lastly we divide both sides by 3. Algebraic Rules!!!

Work:

3(x+2)=15

3x+6=15

3x+6-6=15-6

3x=9

3x/3=9/3

x=3

4 0

3 years ago

Read 2 more answers

A rectangle has a perimeter of 26 units. The length of the rectangle is 2X+3 and the width is 3X. What is the width of the recta

densk [106]

Answer:

X = 2

Step-by-step explanation:

Divided 26 (total perimeter) by 4 = maximum side length possible = 26÷4 =.6.5

I used the trial and error technique

Guessed it was 2 working out to see if i was right or wrong:

Length: 2(2) + 3 = 4+3 = 7

Width: 3(2) = 6

7+7+6+6 = 26

Therefore X = 2

3 0

3 years ago

Find parametric equations and symmetric equations for the line. (Use the parameter t.) The line through (4, −5, 2) and parallel

Nataliya [291]

Answer:

Step-by-step explanation:

From the given information, the symmetric equations for the line pass through(4, -5, 2) i.e ( $x_o, y_o, z_o$ ) and are parallel to $\dfrac{x+5}{1} = \dfrac{y}{2}= \dfrac{z-3}{1}$

The parallel vector to the line i + zj+k = ai + bj + ck

Hence, the equation for the line is :

$x = x_o + at \\ \\ x = y_o + bt \\ \\ x = z_o + ct$

x = 4 + t

y = -5 + 2t

z = 2 + t

Thus, x, y, z = ( 4+t, -5+2t, 2+t )

The symmetric equation can now be as follows:

$\begin {vmatrix} x = 4+ t \\ \\ \dfrac{x-4}{1} = t \begin {vmatirx} \end {vmatrix}$ $\begin {vmatrix} y = - 5+2t \\ \\ \dfrac{y+5}{2} =t \end {vmatrix}$ $\begin {vmatrix} z =2+t \\ \\ \dfrac{z-2}{1} =t \end {vmatrix}$

∴

$\dfrac{x-4}{1}= \dfrac{y+5}{2}=\dfrac{z-2}{1}$

8 0

3 years ago

Two trains begin in Milford and end in Pinkerton, which is 300 miles away. Train A leaves Milford at 10:00 a.M. And travels at a

cluponka [151]

Answer:

B. Train B

Step-by-step explanation:

We are told that Milford and Pinkerton is 300 miles away.

Train A leaves Milford at 10:00 am. There are 3 hours between 10:00 am to 1:00 pm. So let us find distance traveled by train A in 3 hours.

$\text{Distance}=\text{Speed*Time}$

$\text{Distance traveled by train A}=90\frac{\text{ miles}}{\text{ hour}}\times 3\text{ hours}$

$\text{Distance traveled by train A}=90\times 3{\text{ miles}$

$\text{Distance traveled by train A}=270{\text{ miles}$

Since train A will cover 270 miles in 3 hours and distance between Milford and Pinkerton is 300 miles, therefore, train A will not arrive Pinkerton before 1:00 pm.

Since there are 5 hours between 8:00 am to 1:00 pm, so let us find distance traveled by train B in 5 hours.

$\text{Distance traveled by train B}=70\frac{\text{ miles}}{\text{ hour}}\times 5\text{ hours}$

$\text{Distance traveled by train B}=70\times 5 \text{ miles}$

$\text{Distance traveled by train B}=350\text{ miles}$

Since train B will cover 350 miles in 5 hours, therefore, train B will arrive Pinkerton before 1:00 pm and option B is the correct choice.

5 0

3 years ago