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Molodets [167]
3 years ago
7

I have no clue what I’m doing can someone help me with number 5

Mathematics
1 answer:
marin [14]3 years ago
5 0

a) 35w+55-10w = 25w+55

  28w+75-5w+12 = 23w +87

b)  25w+55 + 23w +87

=48w+142

they will be short by 2(24w+71)

c) 50w+25 = 25(2w+1)

answer: =98w+167

i think

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HELP FASTTT 50 POINTS.... NO FOOLING
Kobotan [32]

Answer:

She beat 6 levels last week

Step-by-step explanation:

She beat 6 levels last week because

7 1/2 ÷ 1 1/4 = 6

6 0
3 years ago
Read 2 more answers
What is the relationship between area in the scale drawing and area on the on the actual tennis court if the scale is 1in.: 5ft.
sveticcg [70]

Given:

Length of the scale = 15.6 in.

Width of the scale = 7.2 in.

Scale of drawing = 1 in. : 5ft.

To find:

The ratio of area of the actual court to the area of the drawing (as a unit rate).

And to check whether it is the same as the ratio of length of the actual court to the length of the drawing.

Step-by-step explanation:

We have,

1 in. = 5ft.

Now, using this scale we get

15.6 in. = (15.6 × 5) ft =78 ft.

7.2 in. = (7.2 × 5) ft = 36 ft.

So, the actual length and width of tennis court are 78 ft and 36 ft respectively.

Area of actual tennis court is

Area=length\times width

Area=78\times 36

Area=2808\text{ ft}^2

The area of drawing is

Area=15.6\times 7.2

Area=112.32\text{ in.}^2

Now, ratio of area of the actual court to the area of the drawing (as a unit rate) is

\dfrac{2808}{112.32}=\dfrac{25}{1}=25:1

Ratio of area is 25:1 and ratio of length is 5:1 both area not same.

Therefore, the ratio of area of the actual court to the area of the drawing (as a unit rate) is 25:1.

7 0
3 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
Help please I need help
kirza4 [7]
Move the constant to the right by adding it’s opposite to both sides
x+2-2= 7 squared - 2
Next remove the opposites
x= 7 squared -2 is your answer
7 0
3 years ago
Can someone help me with question e please ? ​
MAXImum [283]

Answer:

Step-by-step explanation:

a) The domain is R- all real numbers.

b) The domain is R- all real numbers.

c) we need

x^{2}-4\neq  0\\x^{2} \neq 4\\x\neq 2\\and\\x\neq -2

So the domain is R\{-2, 2}

d)

we need

x^{2}+x-2\neq  0\\x\neq 1\\and\\x\neq -2

So the domain is R\{-2, 1}

e)

we need: x-1>0 or x> 1

So the domain is (1,+∞)

6 0
3 years ago
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