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2 years ago

What is 1,112 rounded to the nearest thousand?

Mathematics

1 answer:

DENIUS [597]2 years ago

7 0

Answer: 1000

because you round down

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Evaluate the expression for the given value(s) of the variables s=7 k=4 x=6 8 - k+ x

kow [346]

Answer: 9

Step-by-step explanation:

s = 7

k = 4

x = 6

7 - 4 + 6

3 + 6

8 0

2 years ago

Pls help answer. Brainliest to the first right answer

zavuch27 [327]

Answer

Is there any numbers or is it just like that?

8 0

2 years ago

Read 2 more answers

If a 16-foot silo casts a 12-foot shadow, what is the measure from the tip of the shadow to the top of the silo? 20 feet 400 fee

wlad13 [49]

Use the pythagorean theorm

16^2 + 12^2 = c^2
256 + 144 = c^2
400 = c^2
c = 20 feet or a

4 0

2 years ago

jonah paid $25 for dues to a cub last year. This year he paid $35 for dues. By what percent did his dues increase

4vir4ik [10]

Answer:

40 % increase

Step-by-step explanation:

percent increase = (new - original)/ original

percent increase = (35-25)/25

= 10/25

=.4

multiply by 100

40 percent

4 0

2 years ago

The length of a 200 square foot rectangular vegetable garden is 4feet less than twice the width. Find the length and width of th

inna [77]

Answer:

Length = 18.099 ft

Width = 11.049 ft

Step-by-step explanation:

let the length of the field be x ft

and the width be y ft

as per the condition given in problem

x=2y-4 -----------(A)

Also the area is given as 200 sqft

Hence

xy=200

Hence from A we get

y(2y-4)=200

taking 2 as GCF out

2y(y-2)=200

Dividing both sides by 2 we get

$y(y-2)=100$

$y^2-2y=100$

subtracting 100 from both sides

$y^2-2y-100=0$

Now we solve the above equation with the help of Quadratic formula which is given in the image attached with this for any equation in form

$ax^2+bx+c=0$

Here in our case

a=1

b=-1

c=-100

Putting those values in the formula and solving them for y

$y=\frac{-(-2)+\sqrt{(-2)^2-4 \times (-1) \times 100}}{2 \time 1}$

$y=\frac{-(-2)-\sqrt{(-2)^2-4 \times (-1) \times 100}}{2 \time 1}$

Solving first

$y=\frac{2+\sqrt{4+400}{2}$

$y=\frac{2+\sqrt{404}{2}$

$y=\frac{2+20.099}{2}$

$y=\frac{22.099}{2}$

$y=11.049$

Solving second one

$y=\frac{-(-2)-\sqrt{(-2)^2-4 \times (-1) \times 100}}{2 \time 1}$

$y=\frac{2-\sqrt{4+400}{2}$

$y=\frac{2-\sqrt{404}{2}$

$y=\frac{2-20.099}{2}$

$y=\frac{-18.99}{2}$

$y=-9.045$

Which is wrong as the width can not be in negative

Our width of the field is

y=11.099

Hence the length will be

x=2y-4

x=2(11.049)-4

x=22.099-4

x=18.099

Hence our length x and width y :

Length = 18.099 ft

Width = 11.049 ft

4 0

2 years ago