The square below represents one whole. Express the shaded area as a fraction, a decimal, and a percent of the whole.

Mathematics

2 answers:

lozanna [386]3 years ago

8 0

Answer as a fraction = 1/2

Answer as a decimal = 0.5

Answer as a percent = 50%

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Explanation:

There are 5 blue rectangles out of 10 total. This forms the fraction 5/10 and that reduces to 1/2 after dividing both parts by 5. Therefore, the fractions 5/10 and 1/2 are the same. Notice that exactly half of the entire figure is shaded blue.

Use long division or your calculator to find that 1/2 = 0.5 which converts to 50% when you move the decimal point 2 spots to the right.

stira [4]3 years ago

4 0

Answer:

5/10 , 0.5 , 50%

Step-by-step explanation:

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A carpenter has less than 120 min. to spend painting furniture each day. Today, he has spent 30.5 min. painting a desk. Now he w

FinnZ [79.3K]

Answer:

Hence, a carpenter can paint a maximum of 7 chairs.

Hence, option A is correct.

Step-by-step explanation:

A carpenter has less than 120 min. to spend painting furniture each day.

Today, he has spent 30.5 min. painting a desk.

Now he will paint x chairs, each of which takes 12.5 min.

i.e. he will take a total of 12.5x minutes in painting 'x' chairs.

Also total time spent on painting furniture each day will be equal to time spent on painting a desk and all of the chairs.

Hence,

30.5+12.5x<120 (since the total time he spends on painting is less than 120 mins.)

⇒ 12.5 x<120-30.5

⇒ 12.5 x<89.5

⇒ x<7.16 ( on dividing both side of the inequality with 12.5)

Hence, a carpenter can paint a maximum of 7 chairs.

6 0

3 years ago

Read 2 more answers

I need help w/ this!! thank you

Fittoniya [83]

<h3>Answer:</h3>

$\boxed{\pink{\sf \leadsto Yes \ there \ is \ a \ solution \ of \ the \ given \ inequality .}}$

<h3>Step-by-step explanation:</h3>

A inequality is given to us and we need to convert it into standard form and see whether if it has a solution . So let's solve the inequality.

The inequality given to us is :-

$\bf\implies |2y + 3 | - 1 \leq 0 \\\\\bf\implies |2y+3|\leq 1 \\\\\bf\implies (|2y+3|)^2 \leq 1^2 \\\\\bf\implies (2y+3)^2 \leq 1 \\\\\bf\implies (2y)^2+3^2+2(2y)(3) \leq 1 \\\\\bf\implies 4y^2+9+12y - 1 \leq 0 \\\\\bf\implies 4y^2+12y+8 \leq 0 \\\\\bf\implies 4( y^2 + 3y + 2 ) \leq 0 \\\\\bf\implies y^2+3y +2 \leq 0 \:\:\bigg\lgroup \purple{\bf Standard \ form \ of \ inequality }\bigg\rgroup \\\\\bf\implies y^2y+2y+y+2 \leq 0 \\\\\bf\implies y(y+2)+1(y+2)\leq 0 \\\\\bf\implies ( y+2)(y+1)\leq 0 \\\\\bf\implies \boxed{\red{\bf y \leq (-2) , (-1) }}$