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2 years ago

Alana ate breakfast at a restaurant. The bill came to $79.75. If she left 20% tip, how much was the tip?

Mathematics

1 answer:

Lana71 [14]2 years ago

3 0

$79.75*0.2= 15.95
$79.75-15.95= $63.80
63.80+15.95= $79.75

Answer= $15.95

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ASAP The tee for the fourth hole on a golf course is 300 yards from the tee. On that hole, Marsha hooked her ball to the left, a

andrey2020 [161]

Answer:

C. 97.5 yd

Step-by-step explanation:

The distance between Marsha, the ball and the tee for the fourth hole, forms a triangle.

Thus, the distance between Marsha's ball and the hole can be calculated using the Cosine formula below:

c² = a² + b² - 2abcos(C)

Where,

c = distance between Marsha's ball and the hole = ?

a = distance between the tee for the fourth hole and the tee = 300 yd

b = distance between tee and the hole = 255 yd

C = 18°

c² = 300² + 255² - 2(300)(255)cos(18)

c² = 90000 + 65025 - 153000 × 0.951

c² = 155025 - 145503

c² = 9522

c = √9522 ≈ 97.5

c = distance between Marsha’s ball and the hole = 97.5 yd (nearest tenth)

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2 years ago

Describe and correct the error in evaluating the expression.<br> 9 + 2 x 3 = 11 x 3 = 33

Vinil7 [7]

Answer:

so basically is right how you write but multiplication goes first

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2 years ago

Write the equation for the hyperbola with foci (–12, 6), (6, 6) and vertices (–10, 6), (4, 6).

fomenos

Answer:

$\frac{(x--3)^2}{49} -\frac{(y-6)^2}{32}=1$

Step-by-step explanation:

The standard equation of a horizontal hyperbola with center (h,k) is

$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$

The given hyperbola has vertices at (–10, 6) and (4, 6).

The length of its major axis is $2a=|4--10|$ .

$\implies 2a=|14|$

$\implies 2a=14$

$\implies a=7$

The center is the midpoint of the vertices (–10, 6) and (4, 6).

The center is $(\frac{-10+4}{2},\frac{6+6}{2}=(-3,6)$

We need to use the relation $a^2+b^2=c^2$ to find $b^2$ .

The c-value is the distance from the center (-3,6) to one of the foci (6,6)

$c=|6--3|=9$

$\implies 7^2+b^2=9^2$

$\implies b^2=9^2-7^2$

$\implies b^2=81-49$

$\implies b^2=32$

We substitute these values into the standard equation of the hyperbola to obtain:

$\frac{(x--3)^2}{7^2} - \frac{(y-6)^2}{32}=1$

$\frac{(x+3)^2}{49} -\frac{(y-6)^2}{32}=1$

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2 years ago

Which of the following is the solution to the following system of inequalities?

jek_recluse [69]

Answer:

Step-by-step explanation:

for example point (2,0) is one of the solutions.

=> x + 2y <4

2+ 2(0) = 2 => 2< 4 (true)

=> 3x – y > 2

3(2)-0= 6 => 6>2 (true)

so, the option D is the correct one

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2 years ago

Read 2 more answers

M2/n2 is equivalent to _____.

tankabanditka [31]

Answer:

$\frac{m^2}{n^2}$ is equivalent to $m^2 \cdot n^{-2}$

Step-by-step explanation:

Using exponent rules:

$a^{-n}=\frac{1}{a^n}$

Given the expression:

$\frac{m^2}{n^2}$

⇒ $\frac{m^2}{n^2} = m^2 \cdot \frac{1}{n^2}$

Apply the exponent rule, we have;

$m^2 \cdot n^{-2}$

Therefore, the given expression $\frac{m^2}{n^2}$ is equivalent to $m^2 \cdot n^{-2}$

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2 years ago