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swat32
3 years ago
9

I am a pattern block that has 2 fewer sides than a hexagon. I have 2 pairs of parallel sides a 4 right angles. Which shape am I

Mathematics
1 answer:
never [62]3 years ago
8 0

Answer:

you are a square

Step-by-step explanation:

it says " i have 2 pairs of parallel sides and 4 right angles", and typically 4 right angles make a square

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NO LINKS!! Please help me with these notes. Part 1a​
erik [133]

Answers:

When we evaluate a logarithm, we are finding the exponent, or <u>    power   </u>  x, that the  <u>   base   </u> b, needs to be raised so that it equals the <u>  argument   </u> m. The power is also known as the exponent.

5^2 = 25 \to \log_5(25) = 2

The value of b must be <u>   positive    </u> and not equal to <u>   1   </u>

The value of m must be <u>   positive   </u>

If 0 < m < 1, then x < 0

A <u>   logarithmic  </u>    <u>   equation  </u> is an equation with a variable that includes one or more logarithms.

===============================================

Explanation:

Logarithms, or log for short, basically undo what exponents do.

When going from 5^2 = 25 to \log_5(25) = 2, we have isolated the exponent.

More generally, we have b^x = m turn into \log_b(m) = x

When using the change of base formula, notice how

\log_b(m) = \frac{\log(m)}{\log(b)}

If b = 1, then log(b) = log(1) = 0, meaning we have a division by zero error. So this is why b \ne 1

We need b > 0 as well because the domain of y = log(x) is the set of positive real numbers. So this is why m > 0 also.

5 0
2 years ago
I really need it to be sold in imaginary numbers
Yuliya22 [10]
Solving a 5th grade polynomial

We want to find the answer of the following polynomial:

x^5+3x^4+3x^3+19x^2-54x-72=0

We can see that the last term is -72

We want to find all the possible numbers that can divide it. Those are:

{±1, ±2, ±3, ±4, ±6, ±8, ±9, ±12, ±18, ±36, ±72}

We want to factor this polynomial in order to find all the possible x-values. In order to factor it we will have to find some binomials that can divide it using the set of divisors of -72.

We know that if

(x - z) is a divisor of this polynomial then z might be a divisor of the last term -72.

We will verify which is a divisor using synthetic division. If it is a divisor then we can factor using it:

Let's begin with

(x-z) = (x - 1)

We want to divide

\frac{(x^5+3x^4+3x^3+19x^2-54x-72)}{x-1}

Using synthetic division we have that if the remainder is 0 it will be a factor

We can find the remainder by replacing x = z in the polynomial, when it is divided by (x - z). It is to say, that if we want to know if (x -1) is a factor of the polynomial we just need to replace x by 1, and see the result:

If the result is 0 it is a factor

If it is different to 0 it is not a factor

Replacing x = 1

If we replace x = 1, we will have that:

\begin{gathered} x^5+3x^4+3x^3+19x^2-54x-72 \\ \downarrow \\ 1^5+3\cdot1^4+3\cdot1^3+19\cdot1^2-54\cdot1-72 \\ =1+3+3+19-54-72 \\ =-100 \end{gathered}

Then the remainder is not 0, then (x - 1) is not a factor.

Similarly we are going to apply this until we find factors:

(x - z) = (x + 1)

We replace x by -1:

\begin{gathered} x^5+3x^4+3x^3+19x^2-54x-72 \\ \downarrow \\ (-1)^5+3\cdot(-1)^4+3\cdot(-1)^3+19\cdot(-1)^2-54\cdot(-1)-72 \\ =-1+3-3+19+54-72 \\ =0 \end{gathered}

Then, (x + 1) is a factor.

Using synthetic division we have that:

Then:

x^5+3x^4+3x^3+19x^2-54x-72=(x+1)(x^4+2x^3+x^2+18x-72)

Now, we want to factor the 4th grade polynomial.

Let's remember our possibilities:

{±1, ±2, ±3, ±4, ±6, ±8, ±9, ±12, ±18, ±36, ±72}

Since we verified ±1, let's try with ±2 as we did before.

(x - z) = (x - 2)

We want to divide:

\frac{x^4+2x^3+x^2+18x-72}{x-2}

We replace x by z = 2:

\begin{gathered} x^4+2x^3+x^2+18x-72 \\ \downarrow \\ 2^4+2\cdot2^3+2^2+18\cdot2-72 \\ =16+16+4+36-72 \\ =0 \end{gathered}

Then (x - 2) is a factor. Let's do the synthetic division:

Then,

x^4+2x^3+x^2+18x-72=(x-2)(x^3+4x^2+9x+36)

Then, our original polynomial is:

\begin{gathered} x^5+3x^4+3x^3+19x^2-54x-72 \\ =\mleft(x+1\mright)\mleft(x^4+2x^3+x^2+18x-72\mright) \\ =(x-1)(x-2)(x^3+4x^2+9x+36) \end{gathered}

Now, let's prove if (x +2) is a factor, using the new 3th grade polynomial.

(x - z) = (x + 2)

We replace x by z = -2:

\begin{gathered} x^3+4x^2+9x+36 \\ \downarrow \\ (-2)^3+4(-2)^2+9(-2)+36 \\ =-8+16-18+36 \\ =26 \end{gathered}

Since the remainder is not 0, (x +2) is not a factor.

All the possible cases are:

{±1, ±2, ±3, ±4, ±6, ±8, ±9, ±12, ±18, ±36, ±72}

let's prove with +4

(x - z) = (x + 4)

We want to divide:

\frac{x^3+4x^2+9x+36}{x+4}

Let's replace x by z = -4 in order to find the remainder:

\begin{gathered} x^3+4x^2+9x+36 \\ \downarrow \\ (-4)^3+4(-4)^2+9(-4)+36 \\ =-64+64-36+36 \\ =0 \end{gathered}

Then (x + 4) is a factor. Let's do the synthetic division:

Then,

x^3+4x^2+9x+36=(x+4)(x^2+9)

Since

x² + 9 cannot be factor, we have completed our factoring:

\begin{gathered} x^5+3x^4+3x^3+19x^2-54x-72 \\ =(x-1)(x-2)(x^3+4x^2+9x+36) \\ =(x-1)(x-2)(x+4)(x^2+9) \end{gathered}

Now, we have the following expression:

(x-1)(x-2)(x+4)(x^2+9)=0

Then, we have five posibilities:

(x - 1) = 0

or (x - 2) = 0

or (x + 4) = 0

or (x² + 9) = 0

Then, we have five solutions;

x - 1 = 0 → x₁ = 1

x - 2 = 0 → x₂ = 2

x + 4 = 0 → x₃ = -4

x² + 9 = 0 → x² = -9 → x = ±√-9 = ±√9√-1 = ±3i

→ x₄ = 3i

→ x₅ = -3i

<h2><em>Answer- the solutions of the polynomial are: x₁ = 1, x₂ = 2, x₃ = -4, x₄ = 3i and x₅ = -3i</em></h2>

7 0
1 year ago
F(x)= -x over x^2+5 on the domain [0,4]
BartSMP [9]
If you're asking for extrema, like the previous posting

well

\bf f(x)=\cfrac{-x}{x^2+5}&#10;\\\\\\&#10;\cfrac{df}{dx}=\cfrac{(x^2+5)+2x^2}{(x^2+5)^2}\implies \cfrac{df}{dx}=\cfrac{5+3x^2}{(x^2+5)^2}\impliedby &#10;\begin{array}{llll}&#10;using\ the\\&#10;quotient\ rule&#10;\end{array}

like the previous posting, since this rational is identical, just that the denominator is negative, the denominator yields no critical points

and the numerator, yields no critical points either, so the only check you can do is for the endpoints, of 0 and 4

f(0) = 0        <---- only maximum, and thus absolute maximum

f(4) ≈ - 0.19  <----  only minimum, and thus absolute minimum
5 0
3 years ago
What are the perimeter and the area of the square? <br>80<br>200<br>400<br>8√5<br>16√5<br>32√5​
Deffense [45]

Answer:

680

8√5=17.8

16√5=75.7

32√5​=71.5

Step-by-step explanation:

8 0
3 years ago
Mike got a 45% discount when he bought a new jacket. Which of the following is NOT equivalent to 45%?
MariettaO [177]

Answer:

4/5 is NOT equivalent to 45%

Step-by-step explanation:

4/5 = 80%

4 0
2 years ago
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