Writing algebraic expressions for: A number decreased by 5 is 12

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Bess [88]

$a = - 2$

Step-by-step explanation:

1) Subtract 8 from both sides.

$- 3a = 14 - 8$

2) Simplify 14 - 8 to 6.

$- 3a = 6$

3) Divide both sides by, -3.

$a = - \frac{6}{3}$

4) Simplify 6/3 to 2.

$a = - 2$

Therefor, the answer is, a = -2.

8 0

2 years ago

The answer needs to be a mixed number a integer or a proper fraction.

kobusy [5.1K]

Answer:

sorry sir i really dont even understand

7 0

2 years ago

Toby exercises 14 hours a week. John exercises 20% more than Toby and Jenny exercises two more hours than John. Which expression

Sedaia [141]

Answer: C. 18.8w

Step-by-step explanation:

Since Toby exercise 14hours a week, and John exercises 20% more than Toby

John increment in number of hours compared to Tobi will be;

20/100 ×14 = 2.8hrs

This shows John exercises 2.8hrs more than Tobi. Total number of hours exercised by John will become;

14+2.8 = 16.8hrs

Since Jenny exercises two more hours than John, Total number of hours exercised by Jenny will be;

16.8hrs+2hrs

= 18.8hrs/week

If Jenny exercise 18.8hrs in a week, it means she will exercise 18.8×w/1 in w weeks which gives 18.8w weeks.

5 0

2 years ago

Read 2 more answers

What is the answer to −(3x−9)

frutty [35]

Answer:

-3x + 9

Explanation:

Because these two integers that is - and - is said to be added and write the sign of of greater or bigger number.

3 0

1 year ago

A plane takes off from an airport and flies at a speed of 400km/h on a course of 120° for 2 hours. the plane then changes its co

butalik [34]

Answer:

Distance from the airport = 894.43 km

Step-by-step explanation:

Displacement and Velocity

The velocity of an object assumed as constant in time can be computed as

$\displaystyle \vec{v}=\frac{\vec{x}}{t}$

Where $\vec x$ is the displacement. Both the velocity and displacement are vectors. The displacement can be computed from the above relation as

$\displaystyle \vec{x}=\vec{v}.t$

The plane goes at 400 Km/h on a course of 120° for 2 hours. We can compute the components of the velocity as

$\displaystyle \vec{v_1}=$

$\displaystyle \vec{v_1}=\ km/h$

The displacement of the plane in 2 hours is

$\displaystyle \vec{x_1}=\vec{v_1}.t_1=.(2)$

$\displaystyle \vec{x_1}=km$

Now the plane keeps the same speed but now its course is 210° for 1 hour. The components of the velocity are

$\displaystyle \vec{v_2}=$

$\displaystyle \vec{v_2}=km/h$

The displacement in 1 hour is

$\displaystyle \vec{x_2}=\vec{v_2}.t_2=.(1)$

$\displaystyle \vec{x_2}=km$

The total displacement is the vector sum of both

$\displaystyle \vec{x_t}=\vec{x_1}+\vec{x_2}=+$

$\displaystyle \vec{x_t}=km$

$\displaystyle \vec{x_t}=$

The distance from the airport is the module of the displacement:

$\displaystyle |\vec{x_t}|=\sqrt{(-746.41)^2+492.82^2}$

$\displaystyle |\vec{x_t}|=894.43\ km$

8 0

3 years ago