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3 years ago

Double my tens digit to get my ones digit. Double me and I less than 50

Mathematics

2 answers:

frez [133]3 years ago

8 0

It could be anything with a tens digit half of ones digit
Happy to Help!

pickupchik [31]3 years ago

6 0

One example of this number is 24, or maybe 12, just follow the rules and you'll have a lot of answers.

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Two opposite sides of the square are increased by 8 inches and the other two sides are decreased by 5 inches. the area of the ne

Iteru [2.4K]

Answer:

The length of a side of the square would be 8.

Step-by-step explanation:

7 0

2 years ago

A straight line has equation 3y-3x=4. Write down the equation of another straight line parallel to it

Sophie [7]

Y = 3x + 4/3
............

8 0

3 years ago

A gigantic 25.6 ft. anaconda was seen in a cave by workers at a construction site in Brazil.

Arada [10]

Answer:

12 30

Step-by-step explanation:

7 0

2 years ago

I’m confused can someone explain?

Novay_Z [31]

Answer:

It is 150 percent

Step-by-step explanation:

You can find this value by dividing the numerator by the denominator, summing this value to the integer part and multiplying the result by 100%, so:

1 1/2 in percent = (1 + 1/2) × 100% =

(1 + 0.5) × 100% = 1.5 × 100% = 150%

Hope this helps! Sorry if I'm wrong.

3 0

3 years ago

Read 2 more answers

The polynomial x^2+3x-1 is a factor of x^4+3x^3-2x^2-3x+1 true or false?

pashok25 [27]

Answer:

Yes, it is true that $x^2+3x-1$ is a factor of $x^4+3x^3-2x^2-3x+1$ .

Step-by-step explanation:

Let us try to factorize $x^4+3x^3-2x^2-3x+1$

$x^4+3x^3-2x^2-3x+1\\\Rightarrow x^4-2x^2+1-3x+3x^3$

Let us try to make a whole square of the given terms:

$\Rightarrow (x^2)^2-2\times x^2 \times 1+1^2+3x^3-3x\\\Rightarrow (x^2-1)^2+3x^3-3x\\$

--------------

Formula used above:

$a^{2} -2 \times a \times b +b^{2} = (a-b)^2$

In the above equation, we had $a = x, b = 1$ .

--------------

Further solving the above equation, taking $3x$ common out of $3x^3-3x$

$\Rightarrow (x^2-1)^2+3x(x^2-1)\\$

Taking $(x^{2} -1)$ common out of the above term:

$\Rightarrow (x^2-1)((x^2-1)+3x)\\\Rightarrow (x^2-1)(x^2+3x-1)$

So, the two factors are $(x^2-1)\ and\ (x^2+3x-1)$ .

$\therefore$ The statement that $x^2+3x-1$ is a factor of $x^4+3x^3-2x^2-3x+1$ is <em>True.</em>

7 0

3 years ago