All categories

2 years ago

What number is in the tenth's place? 41.53

Mathematics

2 answers:

zysi [14]2 years ago

6 0

Answer: The number in the tenth's place is 5

Step-by-step explanation:

There is no ones place in decimals

41.53

the tenth's place come first then the hundredth's and so on.

larisa86 [58]2 years ago

4 0

Answer:

Step-by-step explanation:

4 = tens

1 = ones

5 = tenths

3 = hundredths

You might be interested in

Given the following sets.

laila [671]

B = {a, b, c, d}

C = {0, a, 2, b}

B ∪ C = {a, b, c, d, 0, 2}

<h3>Answer: E)</h3>

Everything from the set B and everything from the set C give to one set. Duplicated elements are written only once.

8 0

3 years ago

Read 2 more answers

A TV is on sale at 30% off for $420. What was the original price of the TV?

liberstina [14]

"A TV is on sale at 30% off for $420"

What was the original price of the TV?


Well if you start to calculate first off you are sooner or later going to figure out that in order for it to be 420 the answer has to be bigger than 1,000. If you get to 1,200 and calculate 30 percent off of it it the answer is 360 so a ballpark quesstiminate would be to go two more hundredths up. Therefore 30 percent of 1,400 is 420

4 0

3 years ago

Subtracting 5 from each score in a distribution with a mean equal to 40 and a standard deviation equal to 3 will result in a sta

LenaWriter [7]

Adding or subtracting a constant from each value of a distribution changes the mean, but not the standard deviation.

The standard deviation will equal 3.

6 0

2 years ago

Please help solve for x..

aalyn [17]

Answer:

x = 1 or x = -1

Step-by-step explanation:

Given equation:

$x^{100}-4^x \cdot x^{98}-x^2+4^x=0$

Factor out -1:

$\implies -1(-x^{100}+4^x \cdot x^{98}+x^2-4^x)=0$

Divide both sides by -1:

$\implies -x^{100}+4^x \cdot x^{98}+x^2-4^x=0$

Rearrange the terms:

$\implies 4^x \cdot x^{98}-4^x-x^{100}+x^2=0$

$\textsf{Apply exponent rule} \quad a^{b+c}=a^b \cdot a^c \quad \textsf{to }x^{100}:$

$\implies 4^x \cdot x^{98}-4^x-x^{98}x^2+x^2=0$

Factor the first two terms and the last two terms separately:

$\implies 4^x(x^{98}-1)-x^2(x^{98}-1)=0$

Factor out the common term $(x^{98}-1)$ :

$\implies (4^x-x^2)(x^{98}-1)=0$

Zero Product Property: If a ⋅ b = 0 then either a = 0 or b = 0 (or both).

Using the Zero Product Property, set each factor equal to zero and solve for x (if possible):

$\begin{aligned}x^{98}-1 & = 0 & \quad \textsf{or} \quad \quad4^x-x^2 & = 0 \\x^{98} & =1 & 4^x & = x^2 \\x & = 1, -1 & \textsf{no}& \textsf{ solutions for } x \in \mathbb{R}\end{aligned}$