Ask question

Ask question

All categories

FromTheMoon [43]

3 years ago

12

Nayeli brought Jamba Juice smoothies for herself and Evelyn after school one day the smoothies cost 4.95 each plus 8.5% tax how

much charge did she receive from a $20 bill

1 answer:

Anna11 [10]3 years ago

3 0

Answer: $9.26

Step-by-step explanation:

The total cost for each of their smoothies was:

= 4.95 * 2

= $9.90

Including taxes it will rise to:

= 9.90 * (1 + tax)

= 9.90 * (1 + 8.5%)

= $‭10.74

Change she will receive is:

= 20 - 10.74

= $9.26

You might be interested in

Triangle ΔABC has side lengths of a = 18, <img src="https://tex.z-dn.net/?f=b%3D18%5Csqrt%7B3%7D" id="TexFormula1" title="b=18\s

ozzi

The measure of angle A is 30 degree, the value of tan A is 1/√3 and the area of triangle ABC is 280.6 squared inches.

<h3 /><h3>What is the law of cosine?</h3>

When the three sides of a triangle is known, then to find any angle, the law of cosine is used.

It can be given as,

$\angle A=\cos^{-1}\left(\dfrac{b^2+c^2-a^2}{2bc}\right) \\\angle B=\cos^{-1}\left(\dfrac{a^2+c^2-b^2}{2ac}\right) \\\angle C=\cos^{-1}\left(\dfrac{a^2+b^2-c^2}{2ab}\right)$

Here, a,b and c are the sides of the triangle and A,B and C are the angles of the triangle.

Triangle ΔABC has side lengths of a = 18, b=18√3 and c = 36 inches.

Part A: Determine the measure of angle A

Put the value, in the cosine law, the measure of angle A.

$\angle A=\cos^{-1}\left(\dfrac{b^2+c^2-a^2}{2bc}\right) \\\angle A=\cos^{-1}\left(\dfrac{(18\sqrt{3})^2+36^2-18^2}{2(18\sqrt{3})(36)}\right) \\\angle A=0.5236\rm\; rad\\\angle A=30^o\rm\; degree\\$

Part B: Show how to use the unit circle to find tan A

Using the chart of unit circle, the value of tangent A can be found out. The tangent A is,

$\tan A=\tan 30^o\\\tan A=\dfrac{1}{\sqrt{3}}$

Part C: Calculate the area of ΔABC.

Use the following formula to find area of ΔABC.,

$A=\dfrac{ab.\sin C}{2}\\A=\dfrac{18\times18\sqrt{3}.\sin C}{2}\\A=280.6\rm\; in^2$

Thus, the measure of angle A is 30 degree, the value of tan A is 1/√3 and the area of triangle ABC is 280.6 squared inches.

Learn more about the law of cosine here;

brainly.com/question/4372174

#SPJ1

7 0

2 years ago

Find the volume and the total surface area of a cuboide whose dimension 2m, 80cm and 60cm

DENIUS [597]

Answer:

Volume= 0.96m^3

TSA= 6.56m^2

Step-by-step explanation:

Given data

Length = 2m

Width= 80cm= 0.8m

Height= 60cm= 0.6m

Volume= L*W*H

Volume= 2*0.8*0.6

Volume= 0.96m^3

TSA=2lw+2lh+2hw

TSA= 2*2*0.8+2*2*0.6+2*0.6*0.8

TSA= 3.2+ 2.4+ 0.96

TSA= 6.56m^2

5 0

3 years ago

Can someone please answer?

Whitepunk [10]

Answer:

Step-by-step explanation:

another way to express 52+24 is 4(13+6)

7 0

2 years ago

Read 2 more answers

Tems11 [23]

QUESTION A

The given multiplication problem is

$\frac{39}{64} \times \frac{8}{13}$

Factor each term to obtain;

$\frac{13\times 3}{8\times8} \times \frac{8}{13}$

Cancel out the common factors to obtain;

$\frac{1\times 3}{8\times1} \times \frac{1}{1}$

Simplify to get;

$\frac{3}{8}$

QUESTION B

The given multiplication problem is

$\frac{2}{3}\times \frac{1}{5}\times \frac{4}{7}$

This the same as

$\frac{2\times 1\times 4}{3\times 5\times 7}$

This simplifies to;

$\frac{8}{105}$

QUESTION C

The given problem is

$\frac{3}{5}\times \frac{10}{12} \times \frac{1}{2}$

This is the same as

$\frac{3}{5}\times \frac{5}{6} \times \frac{1}{2}$

$=\frac{1}{1}\times \frac{1}{2} \times \frac{1}{2}$

This simplifies to

$=\frac{1}{4}$

QUESTION D.

The given expression is

$\frac{4}{9}\times 54$

Factor the 54 to obtain;

$\frac{4}{9}\times 9\times 6$

Cancel the common factors to get;

$\frac{4}{1}\times 1\times 6$

This simplifies to;

$=24$

QUESTION E

The given problem is

$20\times 3\frac{1}{5}$

Convert the mixed numbers to improper fraction to obtain;

$=20\times \frac{16}{5}$

$=4\times5 \times \frac{16}{5}$

Cancel the common factors to get;

$=4\times1 \times \frac{16}{1}$

$=64$

QUESTION F

The multiplication problem is

$11 \times 2 \frac{7}{11}$

Convert the mixed numbers to improper fractions to obtain;

$11 \times \frac{29}{11}$

Cancel out the common factors to get;

$=1 \times \frac{29}{1}$

Simplify;

$=29$

QUESTION G

The given problem is

$5\frac{1}{3}\times 5\frac{1}{8}$

Convert to improper fractions;

$=\frac{16}{3}\times \frac{41}{8}$

Cancel out the common factors to get;

$=\frac{2}{3}\times \frac{41}{1}$

$=\frac{82}{3}$

Convert back to mixed numbers

$=27\frac{1}{3}$

QUESTION H

The given expression is

$10\frac{2}{3} \times 1\frac{3}{8}$

Convert to improper fraction to get;

$\frac{32}{3} \times \frac{11}{8}$

Cancel common factors to get;

$=\frac{4}{3} \times \frac{11}{1}$

Simplify

$=\frac{44}{3}$

Convert back to mixed numbers;

$=14\frac{2}{3}$

7 0

3 years ago

-6h plus 4 - 3plus h equals 11

olga55 [171]

$- 6h + 4 - 3 + h = 11$
So we start by combining like terms;
$- 6h + h = - 5h$
$4 - 3 = 1$
Now we have;
$- 5h + 1 = 11$
We then subtract 1 from both sides;
$- 5h + 1 = 11$
$- 5h = 10$
Then we divide both sides by 5;
$- 5h = 10$
$- 5h \div - 5h = h$
$10 \div - 5h = - 2$
Leaving the final answer to be h = -2

5 0

3 years ago