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Sidana [21]
3 years ago
5

Ḧ̤ë̤l̤̈p̤̈ p̤̈l̤̈ë̤ä̤s̤̈ë̤

Mathematics
1 answer:
monitta3 years ago
4 0
Well, Gina makes quite a bit to be a waitress...but let's focus on the math!
We can make an equation based on the information given to us...

$6.00 per hour
PLUS tips

Now, notice how <u>6.00 is a constant</u>, it never changes. 
Her tips and her hours do change, however, so they must be made into variables.

We can form the equation 6(times how many hours) + (tips)
OR
<em>6x + y = z</em>
X - hours
Y - tips
Z - total money earned

Now we know what the variables mean, let's put these into action!

6(37) + 43 = Z?
Simplify: 6 * 37 = $222.00.

222 + 43 = Z
Simplify: 222 + 43 = $265.00

The total amount she earned altogether is $265.00!

Hope I could help you out! If my answer is incorrect, or I provided an answer you were not looking for, please let me know.
However, if my answer is explained well and correct, please consider marking my answer as Brainliest!  :)

Have a good one.
God bless!
You might be interested in
In right ABC, AN is the altitude to the hypotenuse. FindBN, AN, and AC,if AB =2 5 in, and NC= 1 in.
Rama09 [41]

From the statement of the problem, we have:

• a right triangle △ABC,

,

• the altitude to the hypotenuse is denoted AN,

,

• AB = 2√5 in,

,

• NC = 1 in.

Using the data above, we draw the following diagram:

We must compute BN, AN and AC.

To solve this problem, we will use Pitagoras Theorem, which states that:

h^2=a^2+b^2\text{.}

Where h is the hypotenuse, a and b the sides of a right triangle.

(I) From the picture, we see that we have two sub right triangles:

1) △ANC with sides:

• h = AC,

,

• a = ,NC = 1,,

,

• b = NA.

2) △ANB with sides:

• h = ,AB = 2√5,,

,

• a = BN,

,

• b = NA,

Replacing the data of the triangles in Pitagoras, Theorem, we get the following equations:

\begin{cases}AC^2=1^2+NA^2, \\ (2\sqrt[]{5})^2=BN^2+NA^2\text{.}\end{cases}\Rightarrow\begin{cases}NA^2=AC^2-1, \\ NA^2=20-BN^2\text{.}\end{cases}

Equalling the last two equations, we have:

\begin{gathered} AC^2-1=20-BN^2.^{} \\ AC^2=21-BN^2\text{.} \end{gathered}

(II) To find the values of AC and BN we need another equation. We find that equation applying the Pigatoras Theorem to the sides of the bigger right triangle:

3) △ABC has sides:

• h = BC = ,BN + 1,,

,

• a = AC,

,

• b = ,AB = 2√5,,

Replacing these data in Pitagoras Theorem, we have:

\begin{gathered} \mleft(BN+1\mright)^2=(2\sqrt[]{5})^2+AC^2 \\ (BN+1)^2=20+AC^2, \\ AC^2=(BN+1)^2-20. \end{gathered}

Equalling the last equation to the one from (I), we have:

\begin{gathered} 21-BN^2=(BN+1)^2-20, \\ 21-BN^2=BN^2+2BN+1-20 \\ 2BN^2+2BN-40=0, \\ BN^2+BN-20=0. \end{gathered}

(III) Solving for BN the last quadratic equation, we get two values:

\begin{gathered} BN=4, \\ BN=-5. \end{gathered}

Because BN is a length, we must discard the negative value. So we have:

BN=4.

Replacing this value in the equation for AC, we get:

\begin{gathered} AC^2=21-4^2, \\ AC^2=5, \\ AC=\sqrt[]{5}. \end{gathered}

Finally, replacing the value of AC in the equation of NA, we get:

\begin{gathered} NA^2=(\sqrt[]{5})^2-1, \\ NA^2=5-1, \\ NA=\sqrt[]{4}, \\ AN=NA=2. \end{gathered}

Answers

The lengths of the sides are:

• BN = 4 in,

,

• AN = 2 in,

,

• AC = √5 in.

7 0
1 year ago
Help me plss need help with theses three
fiasKO [112]
interior angle of a regular 18-gon.
It is easier to calculate the exterior angle of a regular polygon of n-sides (n-gon) by the relation
exterior angle = 360/n
For a 18-gon, n=18, so exterior angle = 360/18=20 &deg;
The value of each interior angle is therefore the supplement, or
Interior angle = 180-20=160 degrees.

Naming of a 9-gon
A polygon with 9 vertices is called a nonagon (in English) or enneagon (French ennéagone, but the English version is sometimes used)
You had a good start with the correct answer.

Exterior angle of a 15-gon
The exterior angle of a 15-gon can be calculated using the relation given in the first paragraph, namely
Exterior angle = 360/15=24 degrees
4 0
2 years ago
Step 1: (6×10^3) ÷ (2×10^2)
Nookie1986 [14]

Answer:

Error is in between the second and third step

the exponents must be subtracted

Step-by-step explanation:

8 0
2 years ago
Solve the equation by completing the square
Andrews [41]

Answer:

b1=2\sqrt{46} +10

b2=10-2\sqrt{46}

Step-by-step explanation:

4 0
3 years ago
The triangle T has vertices at (-2, 1), (2, 1) and (0,-1). (It might be an idea to
Firdavs [7]

Rewrite the boundary lines <em>y</em> = -1 - <em>x</em> and <em>y</em> = <em>x</em> - 1 as functions of <em>y </em>:

<em>y</em> = -1 - <em>x</em>  ==>  <em>x</em> = -1 - <em>y</em>

<em>y</em> = <em>x</em> - 1  ==>  <em>x</em> = 1 + <em>y</em>

So if we let <em>x</em> range between these two lines, we need to let <em>y</em> vary between the point where these lines intersect, and the line <em>y</em> = 1.

This means the area is given by the integral,

\displaystyle\iint_T\mathrm dA=\int_{-1}^1\int_{-1-y}^{1+y}\mathrm dx\,\mathrm dy

The integral with respect to <em>x</em> is trivial:

\displaystyle\int_{-1}^1\int_{-1-y}^{1+y}\mathrm dx\,\mathrm dy=\int_{-1}^1x\bigg|_{-1-y}^{1+y}\,\mathrm dy=\int_{-1}^1(1+y)-(-1-y)\,\mathrm dy=2\int_{-1}^1(1+y)\,\mathrm dy

For the remaining integral, integrate term-by-term to get

\displaystyle2\int_{-1}^1(1+y)\,\mathrm dy=2\left(y+\frac{y^2}2\right)\bigg|_{-1}^1=2\left(1+\frac12\right)-2\left(-1+\frac12\right)=\boxed{4}

Alternatively, the triangle can be said to have a base of length 4 (the distance from (-2, 1) to (2, 1)) and a height of length 2 (the distance from the line <em>y</em> = 1 and (0, -1)), so its area is 1/2*4*2 = 4.

6 0
3 years ago
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