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3 years ago

What value represents the number 1 in the number 4,105.8

Mathematics

2 answers:

Jlenok [28]3 years ago

8 0

Answer:

Hundredths place

.....

BlackZzzverrR [31]3 years ago

8 0

The 1 is in the hundredths place, so it represents 100

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X/10 + 6 >= 8<br> What is the solution to this inequality

GuDViN [60]

Answer:

x >= 20

Step-by-step explanation:

Solve for x:

Start to isolate x by subtracting the 6 from both sides.

Then multiply 10 from each side.

After that, you would get x >= 20

8 0

2 years ago

Identify the volume of a cone with base area 64π m2 and a height 4 m less than 3 times the radius, rounded to the nearest tenth.

melomori [17]

V=basearea times 1/3 times height
basearea=6<span>4π m2
hmm, they try to make it difficult

basearea=circle=pir^2
pir^2=64pi
divide by pi
r^2=64
sqrt
r=9

h is 4 les than 3 time r
h=-4+3(8)
h=-4+24
h=20

v=1/3*64pi*20=1280pi/3 m^3=1350.4

C
</span>

4 0

3 years ago

Fill in the blank. If 9/24 = 3/x , then x is _____. PLS HELP!!

lina2011 [118]

Answer:

Step-by-step explanation:

9 /3 3

-- --

24 /3 8

3 0

3 years ago

Andrea gets a monthly allowance. She first spends a fourth of her money on Target. Then, she spends $5 on food. She has $12 left

motikmotik

Answer:

Step-by-step explanation:

12 + 5 = 17

17 x 4 = 68

Andrea's monthly allowance is $68.

5 0

2 years ago

Read 2 more answers

PLEASE HELP

lisabon 2012 [21]

Step-by-step explanation:

The figure below shows a portion of the graph of the function $j\left(x\right) \ = \ 4^{x-2}$ , hence the average rate of change (slope of the blue line) between the $x$ and $x+h$ is

$\text{Average rate of change} \ = \ \displaystyle\frac{\Delta y}{\Delta x} \\ \\ \rule{3.7cm}{0cm} = \dsiplaystyle\frac{f\left(x+h\right) \ - \ f\left(x\right)}{\left(x \ + \ h \right) \ - \ x} \\ \\ \\ \rule{3.7cm}{0cm} = \displaystyle\frac{f\left(x + h\right) \ - \ f\left(x\right)}{h} \\ \\ \\ \rule{3.7cm}{0cm} = \displaystyle\frac{4^{x+h-2} \ - \ 4^{x-2}}{h} \\ \\ \\ \rule{3.7cm}{0cm} = \displaystyle\frac{4^{x-2+h} \ - \ 4^{x-2}}{h}$

$\\ \\ \\ \rule{3.7cm}{0cm} = \displaystyle\frac{\left(4^{x-2}\right)\left(4^{h}\right) \ - \ 4^{x-2}}{h} \\ \\ \\ \rule{3.7cm}{0cm} = \displaystyle\frac{\left(4^{x-2}\right)\left(4^{h} \ - \ 1 \right)}{h}$

7 0

1 year ago