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2 years ago

write three different division expressions, with fractional dividends and divisors, whose quotient is 5. pls help

Mathematics

2 answers:

Naddika [18.5K]2 years ago

7 0

Answers:

15/3, 5/1, 10/2

Step-by-step explanation:

All of these simplified equal 5.

Natali [406]2 years ago

6 0

Answer:

hope you understand:)

Step-by-step explanation:

2/3, 5/8, -3/14, -11/-5, 7/-9, 7/-15 and -6/-11 etc. Division of ordinary fractions is done by multiplying the first fraction by the reciprocal of the second fraction. For example, to divide, 4/3 ÷ 2/3, you simply find the product of the first fraction and the inverse of the second fraction; 4/3 x 3/2 = 2.

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5 + 3x - 2 for in the simplest form

emmasim [6.3K]

Answer:

3x +3

Step-by-step explanation:

5 + 3x - 2

Combine like terms

3x +5-2

3x +3

8 0

3 years ago

Read 2 more answers

Can someone explain this to me??? Are you able to see the picture?

Butoxors [25]

Answer:

P' = (3, 0)

Step-by-step explanation:

You are translating the point (x, y) of (5, -5) to the left 2 units and up 5 units. The x value changes when shifting left or right. So the x value shifts to the left 2 units, meaning it shifts left on the number line, so it reduces by 2 units. The y value changes when shifting up or down. So the y value shifts up 5 units, so it increases in value by 5 units

4 0

3 years ago

A man sells a sofa for rs 3210 making a profit of 7% what would have been his profit percent if he had sold it for rs 3360?

Digiron [165]

Let x be the original cost for the sofa.
X x (1 + 0.07) =3210
X = 3000
Now we know the cost for the sofa.
Let p be the profit percent.
3000 x (1 + p) = 3360
P = 0.12
So the percentage profit is 12%

7 0

3 years ago

Ben and Josh went to the roof of their 40-foot tall high school to throw their math books offthe edge.The initial velocity of Be

Taya2010 [7]

Answer

Josh's textbook reached the ground first

Josh's textbook reached the ground first by a difference of $t=0.6482$

Step-by-step explanation:

Before we proceed let us re write correctly the height equation which in correct form reads:

$h(t)=-16t^2 +v_{o}t+s$ Eqn(1).

Where:

$h(t)$ : is the height range as a function of time

$v_{o}$ : is the initial velocity

$s$ : is the initial heightin feet and is given as 40 feet, thus Eqn(1). becomes:

$h(t)=-16t^2 + v_{o}t + 40$ Eqn(2).

Now let us use the given information and set up our equations for Ben and Josh.

Ben:

We know that $v_{o}=60ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+60t+40$ Eqn.(3)

Josh:

We know that $v_{o}=48ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+48t+40$ Eqn. (4).

Now since we want to find whose textbook reaches the ground first and by how many seconds we need to solve each equation (i.e. Eqns. (3) and (4)) at $h(t)=0$ . Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one.

Thus we have for Ben, Eqn. (3) gives:

$h(t)=0-16t^2+60t+40=0$

Using the quadratic expression to find the roots of the quadratic we have:

$t_{1,2}=\frac{-b+/-\sqrt{b^2-4ac} }{2a} \\t_{1,2}=\frac{-60+/-\sqrt{60^2-4(-16)(40)} }{2(-16)} \\t_{1,2}=\frac{-60+/-\sqrt{6160} }{-32} \\t_{1,2}=\frac{15+/-\sqrt{385} }{8}\\\\t_{1}=4.3276 sec\\t_{2}=-0.5776 sec$