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3 years ago

Solve for x: ✓72 +95 – 1 = 8 What would X be?

Mathematics

2 answers:

Flauer [41]3 years ago

7 0

2 mil. That would finish the process

lesya [120]3 years ago

3 0

2 MILLIONNNNNNNNNNNNNNNNNN

You might be interested in

Which expression can be expanded using the binomial theorem?

kirill [66]

Answer:

The third choice: (x² - 1)³

Step-by-step explanation:

That is the only expression that is a binomial. Binomial means "two numbers". For this expression, x² is one number, and -1 is the other. Choice 1 is a monomial (one number) and choices 2 and 4 are trinomials (three numbers)

7 0

2 years ago

Hey Bestie! 50 pts Pls help! Real answers only pls <3

frosja888 [35]

Answer:

$1. \: 3i$

2. option D

3. option C

4. option D

5. option C

6. option B

7. option C

8. option D

9. option C

10. option C

Step-by-step explanation:

$\sqrt{ - 1(9)}$

$\sqrt{ - 1} \times \sqrt{ 9}$

$i \times \sqrt{9}$

$i \times \sqrt{ {3}^{2} }$

$i \times 3$

$3i$

$\sqrt{ - 1(8)}$

$\sqrt{ - 1} \times \sqrt{8}$

$i \times \sqrt{8}$

$i \times \sqrt{ {2}^{2} \times 2 }$

$2i \sqrt{2}$

$\sqrt{ - 1} \times \sqrt{80}$

$i \times \sqrt{80}$

$4i \: \sqrt{5}$

$\sqrt{ - 1} \times \sqrt{75}$

$i \times \sqrt{75}$

$i \times \sqrt{ {5}^{2} \times 3 }$

$5i \sqrt{3}$

$\sqrt{ - 1} \times \sqrt{72}$

$i \times \sqrt{72}$

$i \times ( {6}^{2} \times 2)$

$6i \sqrt{2}$

$\sqrt{ - 1} \times \sqrt{20}$

$i \times \sqrt{20}$

$i \times \sqrt{ {2}^{2} \times 5}$

$2i \sqrt{5}$

$\sqrt{ - 1} \times \sqrt{27}$

$i \times \sqrt{27}$

$i \times \sqrt{ {3}^{2} \times 3}$

$3i \sqrt{3}$

$\sqrt{ - 1 \times 12}$

$i \times \sqrt{12}$

$i \times \sqrt{4(3)}$

$2i \sqrt{3}$

$\sqrt{ - 1} \times \sqrt{125}$

$i \times \sqrt{ {5}^{2} \times 5 }$

$5i \sqrt{5}$

$\sqrt{ - 1} \times \sqrt{180}$

$i \times \sqrt{ {6}^{2} \times 5 }$

$6i \sqrt{5}$

<h3>Hope it is helpful...</h3>

5 0

2 years ago

Calculer x si (x+1/2)^2= 4/9

-Dominant- [34]

Answer:

$x=-\frac{7}{6}\\x=\frac{1}{6}$

Step-by-step explanation:

The equation to solve is:

$(x+\frac{1}{2})^2=\frac{4}{9}$

To get rid of the "square", we need to take square root of both sides:

$\sqrt{(x+\frac{1}{2})^2}=\sqrt{\frac{4}{9}}\\x+\frac{1}{2}=\frac{\sqrt{4}}{\sqrt{9}}$

Then we use algebra to find the value(s) of x. Remember, when we take square root, we have to add up a "+-" (on the right side). Shown below:

$x+\frac{1}{2}=+-\frac{\sqrt{4}}{\sqrt{9}}\\x+\frac{1}{2}=+-\frac{2}{3}\\x=\frac{2}{3}-\frac{1}{2}=\frac{1}{6}\\x=-\frac{2}{3}-\frac{1}{2}=-\frac{7}{6}$

So these are 2 answers for x.

5 0

2 years ago

For this exercise assume that all matrices are ntimesn. Each part of this exercise is an implication of the form "If "statement

inna [77]

Answer:

C. True; by the Invertible Matrix Theorem if the equation Ax=0 has only the trivial solution, then the matrix is invertible. Thus, A must also be row equivalent to the n x n identity matrix.

Step-by-step explanation:

The Invertible matrix Theorem is a Theorem which gives a list of equivalent conditions for an n X n matrix to have an inverse. For the sake of this question, we would look at only the conditions needed to answer the question.

There is an n×n matrix C such that CA= $I_n$ .
There is an n×n matrix D such that AD= $I_n$ .
The equation Ax=0 has only the trivial solution x=0.
A is row-equivalent to the n×n identity matrix $I_n$ .
For each column vector b in $R^n$ , the equation Ax=b has a unique solution.
The columns of A span $R^n$ .

Therefore the statement:

If there is an n X n matrix D such that AD=I, then there is also an n X n matrix C such that CA = I is true by the conditions for invertibility of matrix:

The equation Ax=0 has only the trivial solution x=0.

A is row-equivalent to the n×n identity matrix $I_n$ .

The correct option is C.

5 0

3 years ago

PLEASE HELP! I'LL GIVE BRAINLIEST!!!!!!!

daser333 [38]

Answer:

1.6x10^-18 gram

Step-by-step explanation:

(5.3 x 10⁻²³ gram/molecule) x (20,000 molecule)

= (5.3 x 10⁻²³ x 2 x 10⁴) gram

= (10.6 x 10⁻²³⁺⁴) gram

= (1.06 x 10⁻²³⁺⁵) gram

= 1.06 x 10⁻¹⁸ gram

We need to find the mass of 20,000 molecules of oxygen. It can be calculated using unitary method. Here, we can multiply 20,000 molecules by the mass of one oxygen molecule.

7 0

2 years ago