All categories

3 years ago

What is the value of the digit 1 in the number 0.413

Mathematics

2 answers:

Akimi4 [234]3 years ago

4 0

.01 is the answer to that or one one hundredth

hoa [83]3 years ago

3 0

0.010 tens.......☺☺☺☺☺

You might be interested in

What can you say about the quotient, when you divide 23 by something equal to 1?

Gwar [14]

Answer: the quotient will be equal to 23

The equation to calculate what divided by 23 equals 1 is as follows:

X/23 = 1

Where X is the answer. When we solve the equation by multiplying each side by 23, you get get:

X = 23

Therefore, the answer to what divided by 23 equals 1 is 23

so the quotient will be equal to 23

Step-by-step explanation:

8 0

3 years ago

Multiply.

Reptile [31]

Answer:

correct answer is D

Step-by-step explanation:

7 0

3 years ago

Fast easy question!<br> The figures below are similar. What is the value of x

Ivahew [28]

Answer:

Step-by-step explanation:

6 0

3 years ago

Use mathematical induction to prove

Alex17521 [72]

$Prove\ that\ the\ assumption \is \true for\ n=1\\1^3=\frac{1^2(1+1)^2}{4}\\ 1=\frac{4}{4}=1\\$

Formula works when n=1

Assume the formula also works, when n=k.

Prove that the formula works, when n=k+1

$1^3+2^3+3^3...+k^3+(k+1)^3=\frac{(k+1)^2(k+2)^2}{4} \\\frac{k^2(k+1)^2}{4}+(k+1)^3=\frac{(k+1)^2(k+2)^2}{4} \\\frac{k^2(k^2+2k+1)}{4}+(k+1)^3=\frac{(k^2+2k+1)(k^2+4k+4)}{4} \\\frac{k^4+2k^3+k^2}{4}+k^3+3k^2+3k+1=\frac{k^4+4k^3+4k^2+2k^3+8k^2+8k+k^2+4k+4}{4}\\\\\frac{k^4+2k^3+k^2}{4}+k^3+3k^2+3k+1=\frac{k^4+6k^3+13k^2+12k+4}{4}\\\frac{k^4+2k^3+k^2}{4}+\frac{4k^3+12k^2+12k+4}{4}=\frac{k^4+6k^3+13k^2+12k+4}{4}\\\frac{k^4+6k^3+13k^2+12k+4}{4}=\frac{k^4+6k^3+13k^2+12k+4}{4}\\$

Since the formula has been proven with n=1 and n=k+1, it is true. $\square$

7 0

2 years ago

Please help! enter a range of values for x:

myrzilka [38]

Answer:

Step-by-step explanation:

using sine formula

let ∠ADB=y

then ∠CDB=180-y

∠DAB =180-(35+y)

$\frac{AB}{sin~y} =\frac{12}{sin ~35}} \\AB=\frac{12 sin y}{sin 35} \\\frac{BC}{sin(180-y)} =\frac{4x-4}{sin 20 } \\\frac{BC}{sin y} =\frac{4x-4}{sin 20} \\BC=\frac{(4x-4)sin y}{sin 20} \\But~AB=BC (given)\\\frac{12 sin y}{sin 35} =\frac{(4x-4)sin y}{sin 20} \\\frac{4x-4}{sin 20} =\frac{12}{sin 35} \\4(x-1)=\frac{12 sin 20}{sin35} \\x-1=\frac{3 sin 20}{sin35} \approx 1.789\\x \approx 2.789$

3 0

3 years ago