5\times 100+4\times 10+6\times 1+2\times \left(\frac{1}{10}\right)+8\times \left(\frac{1}{1000}\right)5×100+4×10+6×1+2×(

A package of 5 crackers contains 205 Calories. How many Calories are in one cracker using a double number line

natulia [17]

Answer:

Step-by-step explanation:

since there are 205 calories in 5 crackers we can represent this as

205/5

we need to find how many calories are in one cracker, so lets take "x" as the # of calories in 1 cracker

so

$\frac{x}{1} =\frac{205}{5}$

cross multiply

5x=205

divide by 5 on both sides

x=41

41 calories in 1 cracker

Number line:

7 0

3 years ago

Would anyone mind helping me

oksian1 [2.3K]

Answer:

29

Step-by-step explanation:

0.11v=3.19

$\frac{0.11x}{0.11} =\frac{3.19}{0.11}$

Divide:

3.19÷0.11=29

v=29

3 0

3 years ago

Read 2 more answers

Week 1 2 3 4 5 Water Level (inches) − 1 1 4 1 3 8 2 1 2 − 1 5 8 − 1 3 4 Between which two weeks did the water level change the m

suter [353]

Answer:

The water level change the most in week 3 and week 4

The change = -370 inches

Step-by-step explanation:

Given - Week 1 2 3 4 5

Water Level (inches) − 1 1 4 1 3 8 2 1 2 -1 5 8 -1 3 4

To find - Between which two weeks did the water level change the most? Calculate the change .

Proof -

The formula for change in water level with respect to week be-

Rate of Change in Water Level = Difference in water level / Difference in weeks

Now,

For week 1 and week 2

Rate of change in water level = $\frac{138 - 114}{2 - 1}$ = $\frac{24}{1}$ = 24 inches

Now,

For week 1 and week 3

Rate of change in water level = $\frac{212 - 114}{3 - 1}$ = $\frac{98}{2}$ = 49 inches

Now,

For week 1 and week 4

Rate of change in water level = $\frac{-158 - 114}{4 - 1}$ = $-\frac{272}{3}$ = -90.67 inches

Now,

For week 1 and week 5

Rate of change in water level = $\frac{-134 - 114}{5 - 1}$ = $-\frac{248}{4}$ = -62 inches

Now,

For week 2 and week 3

Rate of change in water level = $\frac{212 - 138}{3 - 2}$ = $\frac{74}{1}$ = 74 inches

Now,

For week 2 and week 4

Rate of change in water level = $\frac{-158 - 138}{4 - 2}$ = $-\frac{296}{2}$ = -148 inches

Now,

For week 2 and week 5

Rate of change in water level = $\frac{-134 - 138}{5 - 2}$ = $-\frac{272}{3}$ = -90.67 inches

Now,

For week 3 and week 4

Rate of change in water level = $\frac{-158 - 212}{4 - 3}$ = $-\frac{370}{1}$ = -370 inches

Now,

For week 3 and week 5

Rate of change in water level = $\frac{-134 - 212}{5 - 3}$ = $-\frac{346}{2}$ = 173 inches

Now,

For week 4 and week 5

Rate of change in water level = $\frac{-134 + 158}{5 - 4}$ = $\frac{24}{1}$ = 24 inches

∴ we get

The water level change the most in week 3 and week 4

The change = -370 inches

Note :

The highest change means which temperature goes the most change in water level. It can be negative also.

So, We have to take the modulus and then find the highest number.

Here, 370 > 173

So, Highest change occurs in Week 3 and Week 4 but not in Week 3 and Week 5.

7 0

3 years ago

1. Subtract (7 – 19i) from (–6 + 12i). A. –13 – 31i B. –13 + 31i C. 13 – 31i D. 13 + 31i

pav-90 [236]

Answer:

Step-by-step explanation:

(-6+12i)-(7-19i)=

-6+12i-7+19i=

-13+31i

(b)

6 0

3 years ago

Read 2 more answers

A bag contains 50 blue marbles and 66 red marbles. If 88 more red marbles are added to the bag, how many more blue marbles must

vaieri [72.5K]

Answer:

60

Step-by-step explanation:

After adding 88 red marbles, there are 50 blue marbles out of the 204 marbles total.

Let the number of blue marbles to be added be x.

$\frac{50+x}{204+x}=\frac{5}{12} \\ \\ 600+12x=1020+5x \\ \\ 420=7x \\ \\ x=60$

7 0

1 year ago

​ 5\times 100+4\times 10+6\times 1+2\times \left(\frac{1}{10}\right)+8\times \left(\frac{1}{1000}\right)5×100+4×10+6×1+2×(

5\times 100+4\times 10+6\times 1+2\times \left(\frac{1}{10}\right)+8\times \left(\frac{1}{1000}\right)5×100+4×10+6×1+2×(