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3 years ago

46x−23=46x+23 how many solutions

Mathematics

2 answers:

11Alexandr11 [23.1K]3 years ago

8 0

Step-by-step explanation:

no solutions because when divide it by 23 it will be like this

2x-1=2x+1

now let assume x=1

2-1=2+1

so 1cant be equal to 3

Gala2k [10]3 years ago

7 0

Answer:

zero solutions

Step-by-step explanation:

46x−23=46x+23

46x - 46x = 23 + 23

0 = 46 (this cannot be a valid solution because 0 is obviously not equal to 46, hence there are no solutions)

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Two bonds are available on the market as follows: Bond 1: Face value $250, 5 years to maturity at a simple interest rate of 5%.

Lera25 [3.4K]

Answer:

5.95%

Step-by-step explanation:

Bond 1:

I = 250(.05)5

I = $62.50

Bond 2:

62.5 = 350(3)r

62.5 = 1050r

r = 62.5/1050

r = 5.95%

4 0

3 years ago

Read 2 more answers

A 1,200g block of phosphorus-32, which has a half life of 14.3 days, is stored for 100.1 days. At the end of this period, how mu

Valentin [98]

Answer:

The amount of phosphorus-32 left after 100.1 days is 9.3 g.

Step-by-step explanation:

Given:

Initial amount of Phosphorus-32 is, $N_0=1200\ g$

Time period of decay is, $t=100.1\ days$

Half life of the block is, $t_{1/2}=14.3\ days$

Now, final amount left is, $N=?$

We know that, the decay equation for a radioactive material is given as:

$N=N_0e^{-kt}\\k\to decay\ constant$

The value of the decay constant is given as:

$k=\frac{\ln 2}{t_{1/2}}\\\\k=\frac{0.693}{14.3}=0.0485$

Now, plug in all the given values and calculate 'N'. This gives,

$N=(1200)e^{(-0.0485\times 100.1)}\\\\N=9.349\approx 9.3\ g$

Therefore, the amount of phosphorus-32 left after 100.1 days is 9.3 g

5 0

3 years ago

Not sure how to solve this

Margarita [4]

Answer:

-3/17

Step-by-step explanation:

(7,1) and (-10,4)

slope formula: m = (y₂ - y₁) / (x₂ - x₁)

plug in data: m = (4 - 1) / (-10 - 7)

solve: 3 / -17

simplify: -3/17 (written as a fraction)

8 0

3 years ago

For how many positive three-digit integers is the product of their three digits 90?

Masteriza [31]

So just find the factors of 90
90=2 times 3 times 3 times 5
so the factors could be
2,3,15
2,9,5
6,3,5
10,3,3
there are 4our

4 0

4 years ago

What is the remainder when (3x4 + 2x3 − x2 + 2x − 19) ÷ (x + 2)? 0 5 10 15

Elanso [62]

Answer: Second option is correct.

Explanation:

Since we have given that

$f(x)=3x^4+2x^3-x^2+2x-19$

And $g(x)=x+2$

To find the remainder, we use the Remainder Theorem, which states that when f(x) is divided by (x-c) then the f(c) is the required remainder.

So,

Here, we have,

$x+2=0\\\\x=-2$

So, we will find f(-2) which will give us the required remainder,

$f(-2)=3(-2)^4+2(-2)^3-(-2)^2+2(-2)-19\\\\f(-2)=(3\times 16)-(2\times 8)- 4-4-19\\\\f(-2)=48-16-36-4-19\\\\f(-2)=5$

Hence, Second option is correct.

7 0

3 years ago