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3 years ago

Need some help! Please

Mathematics

1 answer:

Genrish500 [490]3 years ago

3 0

this is confusing where is angle b , a, and c

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the room charge for a hotel is 230$ each day. the estevez family spends 4 days on vacation at the hotel . at the end of their st

DochEvi [55]

10% of 230 is 23 so
230
+23 =
253
hope i helped:)

3 0

3 years ago

Read 2 more answers

What is the product of the polynomial (7x+7) (x+2)

kolbaska11 [484]

Answer: 7x^2+21x+14

Step-by-step explanation:

(7x+7)(x+2)

Multiply each term in the first parentheses by each term in the second parentheses (FOIL)

7x×x+7x×2+7x+7×2

↘ ↙

7x×x calculate product

7x^2+7x×2+7x+7×2

↘ ↙

7x×2 calculate product

7x^2+14x+7x+7×2

↘↙

7×2 multiply numbers

7x^2+14x+7x+14

↘ ↙

21x collect like terms

7x^2+21x+14 is your end result.

3 0

3 years ago

What is the rule used to transform ABC to its image

Karo-lina-s [1.5K]

Answer:

we don't see an image?

Step-by-step explanation:

8 0

2 years ago

Suppose that F(x) = x2 and G(X) = 2/3 x^2. Which statement best compares the graph of G(x) with the graph of F(x)?

masha68 [24]

<h2>Answer:The graph of $G(x)$ is the graph of $F(x)$ compressed vertically.</h2>

Step-by-step explanation:

Given that $F(x)=x^{2}$ and $G(x)=\frac{2}{3}x^{2}$

$F(x)$ is always positive because $x^{2}$ is always positive.

$G(x)$ is always positive because $\frac{2}{3}x^{2}$ is always positive.

So,both are always positive.

So,there is no flipping over x-axis.

In $F(x)$ ,the height of a point at $x_{0}$ is $x_{0}^{2}$

In $G(x)$ ,the height of a point at $x_{0}$ is $\frac{2}{3}x_{0}^{2}$

So,height of any point has less height in $G(x)$ than $F(x)$

So,the graph of $G(x)$ is the graph of $F(x)$ compressed vertically.

3 0

2 years ago

Solve the following equation. X cubed minus 6X squared plus 6X equals zero

anyanavicka [17]

We have to solve this equation:

$x^3-6x^2+6x=0$

Third degree polynomials like this one are not easily solved, but this one has a root at x = 0. The let us factorize this polynomial as x times a second degree polynomial:

$\begin{gathered} x^3-6x^2+6x=0 \\ x(x^2-6x+6)=0 \end{gathered}$

Now we can find the roots of the quadratic polynomial as:

$\begin{gathered} x=\frac{-(-6)\pm\sqrt[]{(-6)^2-4\cdot1\cdot6}}{2\cdot1} \\ x=\frac{6\pm\sqrt[]{36-24}}{2} \\ x=\frac{6\pm\sqrt[]{12}}{2} \\ x=\frac{6\pm\sqrt[]{4\cdot3}}{2} \\ x=\frac{6\pm2\sqrt[]{3}}{2} \\ x=3\pm\sqrt[]{3} \\ x_1=3-\sqrt[]{3} \\ x_2=3+\sqrt[]{3} \end{gathered}$

Then, the solutions to the equation are:

x = 0

x = 3 - √3

x = 3 + √3

4 0

1 year ago