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3 years ago

65 Divided by 3459 what is it?

Mathematics

2 answers:

telo118 [61]3 years ago

8 0

Answer: 53.2

Step-by-step explanation:

sladkih [1.3K]3 years ago

6 0

Answer:

0.01879

Step-by-step explanation:

You might be interested in

Using 7 four times and 1 one time getting to 100

777dan777 [17]

Answer:

So the answer is $(7+\frac{1}{7} )\times (7+7)=100$

Step-by-step explanation:

Given,

Using $7$ four times and $1$ one time getting to $100$

If we are write this way then easily get the answer;

$(7+\frac{1}{7} )\times (7+7)$

Add together $7+\frac{1}{7}$ by making a common denominator of $7$ and also add $(7+7)$ in the second set of parenthesis to yield $14$

$(\frac{49+1}{7} )\times 14$

$\frac{50}{7} \times 14$

$50\times 2$ (By Cross-reduce and multiply the fractions)

$100$

∴ $(7+\frac{1}{7} )\times (7+7)=100$

4 0

3 years ago

Simplify the following polynomial expression. (5x^4 - 9x^3 + 7x - 1) + (-8x^4 + 4x^2 - 3x + 2) - ( -4x^3 + 5x - 1)( 2x - 7)

Lady bird [3.3K]

Answer: $5x^{4}-37x^{3}-6x^{2} +41x-6$

Step-by-step explanation:

1. You must apply the Distributive property. You need to remember the product property of exponents: If you multiply powers with the same base, you must add the exponents. Them you have:

$=5x^{4}-9x^{3}+7x-1-8x^{4}+4x^{2}-3x+2-(8x^{4}+28x^{3}+10x^{2}-35x-2x+7)\\=5x^{4}-9x^{3}+7x-1-8x^{4}+4x^{2}-3x+2-8x^{4}-28x^{3}-10x^{2}+35x+2x-7$

2. Now, you must add the like terms. Then, you obtain the following result:

$=5x^{4}-37x^{3}-6x^{2} +41x-6$

7 0

3 years ago

Is 4x+2y=9 a linear equation

ExtremeBDS [4]

Yes it is a linear equation

3 0

3 years ago

What is a=16 b= 28 and c=? What would c be?

sergiy2304 [10]

A^2+b^2= c^2
16^2+ 28^2= c^2
square root of 256+784 = square root of 1,040
32.25=32.25

C= 32.25

3 0

3 years ago

In Exercises 40-43, for what value(s) of k, if any, will the systems have (a) no solution, (b) a unique solution, and (c) infini

svet-max [94.6K]

Answer:

If k = −1 then the system has no solutions.

If k = 2 then the system has infinitely many solutions.

The system cannot have unique solution.

Step-by-step explanation:

We have the following system of equations

$x - 2y +3z = 2\\x + y + z = k\\2x - y + 4z = k^2$

The augmented matrix is

$\left[\begin{array}{cccc}1&-2&3&2\\1&1&1&k\\2&-1&4&k^2\end{array}\right]$

The reduction of this matrix to row-echelon form is outlined below.

$R_2\rightarrow R_2-R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\2&-1&4&k^2\end{array}\right]$

$R_3\rightarrow R_3-2R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&3&-2&k^2-4\end{array}\right]$

$R_3\rightarrow R_3-R_2$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&0&0&k^2-k-2\end{array}\right]$

The last row determines, if there are solutions or not. To be consistent, we must have k such that

$k^2-k-2=0$

$\left(k+1\right)\left(k-2\right)=0\\k=-1,\:k=2$

Case k = −1:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-1-2\\0&0&0&(-1)^2-(-1)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-3\\0&0&0&-2\end{array}\right]$

If k = −1 then the last equation becomes 0 = −2 which is impossible.Therefore, the system has no solutions.

Case k = 2:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&2-2\\0&0&0&(2)^2-(2)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&0\\0&0&0&0\end{array}\right]$

This gives the infinite many solution.

5 0

3 years ago