All categories

3 years ago

Someone please help me

Mathematics

2 answers:

ludmilkaskok [199]3 years ago

7 0

Answer:

Hundreds is 8, tenths is 5, and thousandths is 1!

Darya [45]3 years ago

5 0

Answer:

Hundreds: 8

Tenths: 5

Thousandths: 1

You might be interested in

The vet told Jake that his dog, Rocco, who weighed 55 pounds, needed to lose 10 pounds. Jake started walking Rocco every day and

madreJ [45]

Answer:it would take him five weeks to lose ten pounds

Step-by-step explanation:

6 0

4 years ago

Read 2 more answers

Can someone help me pleaseee?

ratelena [41]

Answer:

1. -9

2. -17

3. -4

4. -11

5. -17

6. -2

7. -20

8. 11

9. 22

10. 36

Have a good day! And do you mind marking me brainliest? much appreciated :)

6 0

3 years ago

Suppose the half-life of a material is 10 days. You have one 1 kg of the material today. How much of the material would you have

boyakko [2]

Answer:

0.50 kg of the material would be left after 10 days.

0.25 kg of the material would be left after 20 days.

Step-by-step explanation:

We have been given that the half-life of a material is 10 days. You have one 1 kg of the material today. We are asked to find the amount of material left after 10 days and 20 days, respectively.

We will use half life formula.

$A=a\cdot(\frac{1}{2})^{\frac{t}{h}}$ , where,

A = Amount left after t units of time,

a = Initial amount,

t = Time,

h = Half-life.

$A=1\text{ kg}\cdot(\frac{1}{2})^{\frac{10}{10}}$

$A=1\text{ kg}\cdot(\frac{1}{2})^{1}$

$A=1\text{ kg}\cdot\frac{1}{2}$

$A=0.5\text{ kg}$

Therefore, amount of the material left after 10 days would be 0.5 kg.

$A=1\text{ kg}\cdot(\frac{1}{2})^{\frac{20}{10}}$

$A=1\text{ kg}\cdot(\frac{1}{2})^{2}$

$A=1\text{ kg}\cdot\frac{1^2}{2^2}$

$A=1\text{ kg}\cdot\frac{1}{4}$

$A=0.25\text{ kg}$

Therefore, amount of the material left after 20 days would be 0.25 kg.

6 0

3 years ago

Gina bought one pound of cheese and used of it.

kaheart [24]

Answer: 1p/5

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

what is the y-intercept of the equation of the line that is perpendicular to the line y=3/5x+10 and passes through the point (15

ohaa [14]

The slope of the given line is the coefficient of x, 3/5. The slope of the perpendicular line will be the negative reciprocal of that: -1/(3/5) = -5/3.

The point-slope form of the equation for a line of slope m through point (h, k) can be written ...

... y = m(x -h) +k

For your point and the slope found above, this becomes

... y = (-5/3)(x -15) -5

When x=0, this is

... y = (-5/3)(-15) -5 = 20

The y-intercept is 20.

3 0

3 years ago