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Ostrovityanka [42]

3 years ago

15

Steve usually buys four books a month for the same price. Every August, his favorite bookstore holds a back-to-school sale and e

verything in the store is priced at 10 percent off. Steve takes advantage of the sale by buying six books in August. What happened to the amount of money that Steve spends on books in August, compared to other months?

1 answer:

Arturiano [62]3 years ago

3 0

Answer:

The amount of money spent by Steve in August is 135% of the amount spent in other months

Step-by-step explanation:

Let the price of each book bought be x

Therefore amount spent on books by Steve in every other month = 4x

In August price of books is 0.9x and Steve bought 6 books.

Therefore amount spent by Steve in August = 5.4x

Therefore difference between amount spent by Steve in August & every other month = 5.4x - 4x = 1.4x

Therefore percentage increase of amount spent in August

(1.4x ÷ 4x) × 100 = 35%

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solmaris [256]

Answer:

$\frac{68}{112}\approx 0.61=61\%$

Step-by-step explanation:

Gráfico recuperado mostra os seguintes dados:

Fundamental Incompleto: 14

Fundamental: 16

Médio Incompleto: 14

Médio: 54

Superior Incompleto: 14

Os funcionário que concluíram o Ensino Médio são 54 (Médio) +14 (Superior Incompleto. Basta somar (54+14)e dividir pelo total de funcionários (112).

$\frac{68}{112}\approx 0.61=61\%$

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3 years ago

Hello help me with this question thanks in advance

Ede4ka [16]

$\bold{\huge{\green{\underline{ Solutions }}}}$

<h3><u>Answer </u><u>1</u><u>1</u><u> </u><u>:</u><u>-</u></h3>

<u>We </u><u>have</u><u>, </u>

$\sf{HM = 5 cm }$

<u>In </u><u>square </u><u>all </u><u>sides </u><u>of </u><u>squares </u><u>are </u><u>equal </u>

<u>The </u><u>perimeter </u><u>of </u><u>square </u>

$\sf{ = 4 × side }$

$\sf{ = 4 × 5 }$

$\sf{ = 20 cm }$

Thus, The perimeter of square is 20 cm

Hence, Option C is correct .

<h3><u>Answer </u><u>1</u><u>2</u><u> </u><u>:</u><u>-</u></h3>

<u>We </u><u>have</u><u>, </u>

$\sf{MX = 3.5 cm }$

<u>In </u><u>square</u><u>,</u><u> </u><u>diagonals </u><u>are </u><u>equal </u><u>and </u><u>bisect </u><u>each </u><u>other </u><u>at </u><u>9</u><u>0</u><u>°</u>

<u>Here</u><u>, </u>

$\sf{MX = MT/2}$

$\sf{MT = 2 * 3.5 }$

$\sf{MT = 7 cm}$

Thus, The MT is 7cm long

Hence, Option C is correct .

<h3><u>Answer </u><u>1</u><u>3</u><u> </u><u>:</u><u>-</u><u> </u></h3>

<u>We </u><u>have </u><u>to </u><u>find </u><u>the </u><u>measure </u><u>of </u><u>Ang</u><u>l</u><u>e</u><u> </u><u>MAT</u>

<u>All </u><u>angles </u><u>of </u><u>square </u><u>are </u><u>9</u><u>0</u><u>°</u><u> </u><u>each </u>

<u>From </u><u>above </u>

$\sf{\angle{MAT = 90° }}$

Thus, Angle MAT is 90°

Hence, Option B is correct .

<h3><u>Answer </u><u>1</u><u>4</u><u> </u><u>:</u><u>-</u></h3>

<u>We </u><u>know </u><u>that</u><u>, </u>

<u>All </u><u>the </u><u>angles </u><u>of </u><u>square </u><u>are </u><u>equal </u><u>and </u><u>9</u><u>0</u><u>°</u><u> </u><u>each </u>

<u>Therefore</u><u>, </u>

$\sf{\angle{MHA = }}{\sf{\angle{ MHT/2}}}$

$\sf{\angle{MHA = 90°/2}}$

$\sf{\angle {MHA = 45°}}$

Thus, Angle MHA is 45°

Hence, Option A is correct

<h3><u>Answer </u><u>1</u><u>5</u><u> </u><u>:</u><u>-</u><u> </u></h3>

Refer the above attachment for solution

Hence, Option A is correct

<h3><u>Answer </u><u>1</u><u>6</u><u> </u><u>:</u><u>-</u><u> </u></h3>

Both a and b

<u>The </u><u>median </u><u>of </u><u>isosceles </u><u>trapezoid </u><u>is </u><u>parallel </u><u>to </u><u>the </u><u>base</u>
<u>The </u><u>diagonals </u><u>are </u><u>congruent </u>

Hence, Option C is correct

<h3><u>Answer </u><u>1</u><u>7</u><u> </u><u>:</u><u>-</u></h3>

In rhombus PALM,

<u>All </u><u>sides </u><u>and </u><u>opposite </u><u>angles </u><u>are </u><u>equal </u>

Let O be the midpoint of Rhombus PALM

<u>In </u><u>Δ</u><u>OLM</u><u>, </u><u>By </u><u>using </u><u>Angle </u><u>sum </u><u>property </u><u>:</u><u>-</u>

$\sf{35° + 90° + }{\sf{\angle{ OLM = 180°}}}$

$\sf{\angle{OLM = 180° - 125°}}$

$\sf{\angle{ OLM = 55° }}$

<u>Now</u><u>, </u>

$\sf{\angle{OLM = }}{\sf{\angle{OLA}}}$

<u>OL </u><u>is </u><u>the </u><u>bisector </u><u>of </u><u>diagonal </u><u>AM</u>

<u>Therefore</u><u>, </u>

$\sf{\angle{ PLA = 55° }}$

Thus, Angle PLA is 55° .

Hence, Option C is correct

8 0

3 years ago

Evaluate the expression when n=6<br> n² – 5n-2

eduard

Answer:

4

Step-by-step explanation:

(6^2) -5(6) -2 =4

that is the answer

8 0

3 years ago

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At the start of the month, the value of an investment was $73.42. By the end of the month, the value of the investment changed b

Aleonysh [2.5K]

Answer:

0.8%.

Step-by-step explanation:

Given:

At the start of the month, the value of an investment was $73.42.

By the end of the month, the value of the investment changed by a loss of $13.53.

Question asked:

What was the value, in dollars, of the investment at the end of the month?

What was the percent loss?

Solution:

At the start of the month, the value of an investment = $73.42.

Loss = $13.53.

<u><em>The value, of the investment at the end of the month = The value of an investment at the start of the month - loss</em></u>

The value, of the investment at the end of the month = $73.42 - $13.53

= $59.89

Thus, the value, of the investment at the end of the month is $59.89.

Now, we will find percent loss:-

$Percent\ loss = \frac{Loss}{value\ of \ investment\ at \ start}$

$=\frac{59.89}{73.42} \\=0.81\%$

Thus, percent loss on investment by the end of the month is 0.8%.

7 0

3 years ago

Someone help me asap please

tiny-mole [99]

Its 2,9 because thats the highest it gets which is the maximum

8 0

3 years ago

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