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2 years ago

100 times 100 times 100 times 100 times 100 times 4000

Mathematics

2 answers:

strojnjashka [21]2 years ago

5 0

Answer:

40,000,000,000,000

Step-by-step explanation:

Because 100 times 100 is 10,000 times 100 is 1 million 1,000,000 times 100 is 100,000,000 times 100 is 10,000,000,000 times 4000 is 40,000,000,000,000

Molodets [167]2 years ago

4 0

(100^5) * 4000 = 10000000000 * 4000 = 4 * 10^13

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Consider the triangle below. Determine the angle measure for the bottom two angles in order for the triangle to be classified as

FinnZ [79.3K]

Answer:

The angle measure for the bottom two angles in order for the triangle to be classified as an isosceles triangle are 30 degrees each

Explanation:

An isosceles triangle has two of its angles equal.

Given that one angle is 120 degrees, assuming the remaining angles are equal, since the sum of the angles in a triangle is 180 degrees, we have:

120 + x + x = 180

120 + 2x = 180

2x = 180 - 120 = 60

x = 60/2 = 30

Each of the remaining angles measure 30 degrees

6 0

9 months ago

Andy estimated that he would need 79 feet of lumber for a tree house project. He later found that the actual amount of lumber ne

Radda [10]

Answer:

Step-by-step explanation:

79 divided by 68 is your answer

You can do the math

5 0

3 years ago

A bank advertises that it compounds money quarterly and that it will take Double your money in 10 years. what is the interest-ra

S_A_V [24]

The interest rate is 6.992%, if a bank advertises that it compounds money quarterly and that it will take Double your money in 10 years.

Step-by-step explanation:

The given is,

Compounds money quarterly

Double your money in 10 years

Step:1

Formula to calculate future investment with compounded quarterly,

$A =P(1+\frac{r}{n} )^{nt}$ ...............................(1)

Where, A - Future amount

P - Initial investment\

r - Rate of interest

n - No. of compounding in a year

t - No. of years

Step:2

Let, P = X

A = 2X ( Double your money )

From given, n - 4 ( for compounding quarterly )

t - 10 years

From equation (1)

$2X =X(1+\frac{r}{4} )^{(4)(10)}$

$\frac{2X}{X} =(1+\frac{r}{4} )^{(4)(10)}$

$2 =(1+\frac{r}{4} )^{40}$

Take root $40^{th}$ root on both side,

$\sqrt[40]{2} = (1+\frac{r}{4} )$

$1.017479692 = (\frac{4+r}{4} )$

$(1.017479692)(4) = (r+4)$

$4.06992=(r+4)$

$r = 4.06992 -4$

$r =( 0.06992)(100)$

r = 6.992 %

Result:

The interest rate is 6.992%, if a bank advertises that it compounds money quarterly and that it will take Double your money in 10 years.

3 0

2 years ago

The sum of a rational number and an even integer is rational. A) Always True B) Sometimes True C) Usually True D) Never True

SOVA2 [1]

I think that the sum will always be a rational number
let's prove that

any rational number can be represented as a/b where a and b are integers and b≠0

and an integer is the counting numbers plus their negatives and 0
so like -4,-3,-2,-1,0,1,2,3,4....

so, 2 rational numbers can be represented as

a/b and c/d (where a,b,c,d are all integers and b≠0 and d≠0)

their sum is
a/b+c/d=
ad/bd+bc/bd=
(ad+bc)/bd

1. the numerator and denominator will be integers
2. that the denominator does not equal 0

alright
1.
we started with that they are all integers
ab+bc=?
if we multiply any 2 integers, we get an integer
like 3*4=12 or -3*4=-12 or -3*-4=12, etc.
even 0*4=0, that's an integer
the sum of any 2 integers is an integer
like 4+3=7, 3+(-4)=-1, 3+0=3, etc.
so we have established that the numerator is an integer

now the denominator
that is just a product of 2 integers so it is an integer

2. we originally defined that b≠0 and d≠0 so we're good

therefore, the sum of any 2 rational numbers will always be a rational number is the correct answer.

6 0

2 years ago

Can someone solve this with steps cause idk how to solve the radicals

IceJOKER [234]

$\bf 343^{\frac{2}{3}}+36^{\frac{1}{2}}-256^{\frac{3}{4}}\qquad \begin{cases}
343=7\cdot 7\cdot 7\\
\qquad 7^3\\
36=6\cdot 6\\
\qquad 6^2\\
256=4\cdot 4\cdot 4\cdot 4\\
\qquad 4^4
\end{cases}\\\\\\ (7^3)^{\frac{2}{3}}+(6^2)^{\frac{1}{2}}-(4^4)^{\frac{3}{4}}
\\\\\\
\sqrt[3]{(7^3)^2}+\sqrt[2]{(6^2)^1}-\sqrt[4]{(4^4)^3}\implies \sqrt[3]{(7^2)^3}+\sqrt[2]{(6^1)^2}-\sqrt[4]{(4^3)^4}
\\\\\\
7^2+6-4^3\implies 49+6-64\implies -9$

to see what you can take out of the radical, you can always do a quick "prime factoring" of the values, that way you can break it in factors to see who is what.

8 0

3 years ago