Ask question

Ask question

All categories

3 years ago

14

The list price on slacks is $22, and the list price on jumpers is $37. If Petit’s Clothing Store orders 30 pairs of slacks and 4

0 jumpers at a discount rate of 11%, what is the trade discount on the purchase?

1 answer:

Marrrta [24]3 years ago

8 0

Answer: $235.4

Step-by-step explanation:

Given

Price list on slacks is $22

Price list on jumpers is $37

Store ordered 30 pairs of slacks and 40 Jumpers

Total price becomes

$\Rightarrow 22\times 30+37\times 40\\\Rightarrow \$2140$

for a discount of 11%

Trade discount is $2140\times 11\%$

$\Rightarrow 2140\times 0.11\\\Rightarrow \$235.4$

You might be interested in

If A and B are two angles in standard position in Quadrant I, find cos( A +B ) for the given function values. sin A = 8/17 and c

horsena [70]

Answer:

Part 1) $cos(A + B) = \frac{140}{221}$

Part 2) $cos(A - B) = \frac{153}{185}$

Part 3) $cos(A - B) = \frac{84}{85}$

Part 4) $cos(A + B) = -\frac{36}{85}$

Part 5) $cos(A - B) = \frac{63}{65}$

Part 6) $cos(A+ B) = -\frac{57}{185}$

Step-by-step explanation:

<u><em>the complete answer in the attached document</em></u>

Part 1) we have

$sin(A)=\frac{8}{17}$

$cos(B)=\frac{12}{13}$

Determine cos (A+B)

we know that

$cos(A + B) = cos(A) cos(B)-sin(A) sin(B)$

step 1

Find the value of cos(A)

Remember that

$cos^2(A)+sin^2(A)=1$

substitute the given value

$cos^2(A)+(\frac{8}{17})^2=1$

$cos^2(A)+\frac{64}{289}=1$

$cos^2(A)=1-\frac{64}{289}$

$cos^2(A)=\frac{225}{289}$

$cos(A)=\pm\frac{15}{17}$

The angle A belong to the I quadrant, the cosine is positive

$cos(A)=\frac{15}{17}$

step 2

Find the value of sin(B)

Remember that

$cos^2(B)+sin^2(B)=1$

substitute the given value

$sin^2(B)+(\frac{12}{13})^2=1$

$sin^2(B)+\frac{144}{169}=1$

$sin^2(B)=1-\frac{144}{169}$

$sin^2(B)=\frac{25}{169}$

$sin(B)=\pm\frac{25}{169}$

The angle B belong to the I quadrant, the sine is positive

$sin(B)=\frac{5}{13}$

step 3

Find cos(A+B)

substitute in the formula

$cos(A + B) = \frac{15}{17} \frac{12}{13}-\frac{8}{17}\frac{5}{13}$

$cos(A + B) = \frac{180}{221}-\frac{40}{221}$

$cos(A + B) = \frac{140}{221}$

Part 2) we have

$sin(A)=\frac{3}{5}$

$cos(B)=\frac{12}{37}$

Determine cos (A-B)

we know that

$cos(A - B) = cos(A) cos(B)+sin(A) sin(B)$

step 1

Find the value of cos(A)

Remember that

$cos^2(A)+sin^2(A)=1$

substitute the given value

$cos^2(A)+(\frac{3}{5})^2=1$

$cos^2(A)+\frac{9}{25}=1$

$cos^2(A)=1-\frac{9}{25}$

$cos^2(A)=\frac{16}{25}$

$cos(A)=\pm\frac{4}{5}$

The angle A belong to the I quadrant, the cosine is positive

$cos(A)=\frac{4}{5}$

step 2

Find the value of sin(B)

Remember that

$cos^2(B)+sin^2(B)=1$

substitute the given value

$sin^2(B)+(\frac{12}{37})^2=1$

$sin^2(B)+\frac{144}{1,369}=1$

$sin^2(B)=1-\frac{144}{1,369}$

$sin^2(B)=\frac{1,225}{1,369}$

$sin(B)=\pm\frac{35}{37}$

The angle B belong to the I quadrant, the sine is positive

$sin(B)=\frac{35}{37}$

step 3

Find cos(A-B)

substitute in the formula

$cos(A - B) = \frac{4}{5} \frac{12}{37}+\frac{3}{5} \frac{35}{37}$

$cos(A - B) = \frac{48}{185}+\frac{105}{185}$

$cos(A - B) = \frac{153}{185}$

Part 3) we have

$sin(A)=\frac{15}{17}$

$cos(B)=\frac{3}{5}$

Determine cos (A-B)

we know that

$cos(A - B) = cos(A) cos(B)+sin(A) sin(B)$

step 1

Find the value of cos(A)

Remember that

$cos^2(A)+sin^2(A)=1$

substitute the given value

$cos^2(A)+(\frac{15}{17})^2=1$

$cos^2(A)+\frac{225}{289}=1$

$cos^2(A)=1-\frac{225}{289}$

$cos^2(A)=\frac{64}{289}$

$cos(A)=\pm\frac{8}{17}$

The angle A belong to the I quadrant, the cosine is positive

$cos(A)=\frac{8}{17}$

step 2

Find the value of sin(B)

Remember that

$cos^2(B)+sin^2(B)=1$

substitute the given value

$sin^2(B)+(\frac{3}{5})^2=1$

$sin^2(B)+\frac{9}{25}=1$

$sin^2(B)=1-\frac{9}{25}$

$sin^2(B)=\frac{16}{25}$

$sin(B)=\pm\frac{4}{5}$

The angle B belong to the I quadrant, the sine is positive

$sin(B)=\frac{4}{5}$

step 3

Find cos(A-B)

substitute in the formula

$cos(A - B) = \frac{8}{17} \frac{3}{5}+\frac{15}{17} \frac{4}{5}$

$cos(A - B) = \frac{24}{85}+\frac{60}{85}$

$cos(A - B) = \frac{84}{85}$

Part 4) we have

$sin(A)=\frac{15}{17}$

$cos(B)=\frac{3}{5}$

Determine cos (A+B)

we know that

$cos(A + B) = cos(A) cos(B)-sin(A) sin(B)$

step 1

Find the value of cos(A)

Remember that

$cos^2(A)+sin^2(A)=1$

substitute the given value

$cos^2(A)+(\frac{15}{17})^2=1$

$cos^2(A)+\frac{225}{289}=1$

$cos^2(A)=1-\frac{225}{289}$

$cos^2(A)=\frac{64}{289}$

$cos(A)=\pm\frac{8}{17}$

The angle A belong to the I quadrant, the cosine is positive

$cos(A)=\frac{8}{17}$

step 2

Find the value of sin(B)

Remember that

$cos^2(B)+sin^2(B)=1$

substitute the given value

$sin^2(B)+(\frac{3}{5})^2=1$

$sin^2(B)+\frac{9}{25}=1$

$sin^2(B)=1-\frac{9}{25}$

$sin^2(B)=\frac{16}{25}$

$sin(B)=\pm\frac{4}{5}$

The angle B belong to the I quadrant, the sine is positive

$sin(B)=\frac{4}{5}$

step 3

Find cos(A+B)

substitute in the formula

$cos(A + B) = \frac{8}{17} \frac{3}{5}-\frac{15}{17} \frac{4}{5}$

$cos(A + B) = \frac{24}{85}-\frac{60}{85}$

$cos(A + B) = -\frac{36}{85}$

4 0

4 years ago

Jennie is paid $10 an hour for regular time work and $20 an hour for overtime

lyudmila [28]

Answer:

Jennie earned $300

Step-by-step explanation:

We are already given the amount of time that she works and the rate. We just need to multiply the time by the corresponding rate. She worked 22 regular hours and gets $10 for each one. That means she earned $220 just from the regular hours.

Now we need to add the overtime hours. She worked 4 hours and made $20 for each one. In total, that's $80.

220+80=300

Jennie earned $300

8 0

4 years ago

A man used 10 gallons of gasoline on a 180-mile trip. How many

Stolb23 [73]

Answer:

Step-by-step explanation:

gs used in 180 mile=10 gallons

gas used in 1 mile=10/180=1/18

gas used in 450 mile=1/18×450=25 gallons

4 0

3 years ago

Read 2 more answers

Plz solve this question<br>Its too much urgent<br>

Sergeu [11.5K]

Standard form of the polynomial is $x^{7}-2 x^{6}-3 x^{5}+x^{2}+\sqrt{2} x+4$ .

Degree of the polynomial is 7.

Solution:

Given polynomial:

$x^{7}-3 x^{5}+\sqrt{2} x+x^{2}-2 x^{6}+4$

<u>To write this polynomial in standard form:</u>

$x^{7}-3 x^{5}+\sqrt{2} x+x^{2}-2 x^{6}+4$

Standard form of the polynomial is writing x from highest power to lowest power.

Therefore, standard form of the polynomial is

$x^{7}-2 x^{6}-3 x^{5}+x^{2}+\sqrt{2} x+4$

<u>Degree of the polynomial:</u>

The highest power of x in the polynomial is the degree of the polynomial.

Here, the highest power is 7.

Hence the degree of the polynomial is 7.

3 0

3 years ago

A line goes through the point (–4,–2) and has an x-intercept of –1.

Nady [450]

Answer:

(5,4)

Step-by-step explanation:

thanks to the second information we know that a certain second point of the line is (-1,0)

Now we can use a formula to find the equation of the line, that is

(y-y1)/(y2-y1) = (x-x1)/(x2-x1) where x and y 1 and 2 are the coordinates of the points.

(y+2)/2 = (x+4)/(-1+4)

y+2/2 = (x+4)/ 3

3y+6 = 2x+8

2x-3y+2 = 0

now we have to find the point that make true the equation (we have to sobstitute x and y with the coordinates of the given points)

1) -12 +15 + 2 = 0 ; 3 = 0 (wrong)

2) 6-4+2 = 0 ; 4 = 0 (wrong)

3) 8-15+2 = 0 ; -5 = 0 (wrong)

4) 10-12+2 = 0 ; 0=0 (true)

so (5,4) is the correct answer

4 0

3 years ago

Read 2 more answers