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3 years ago

Estimate the sum. 11One-fifth + 3.98

Mathematics

2 answers:

jekas [21]3 years ago

7 0

Answer:

15.18

Step-by-step explanation:

lozanna [386]3 years ago

6 0

Answer:

Step-by-step explanation:

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3 4/7 minus 2 1/5 by rounding to the nearest half

mylen [45]

Answer:

$1 \frac{13}{35}$

Step-by-step explanation:

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

you can solve for the fractions first, but you have to give them a common denominator

$\frac{20}{35}- \frac{7}{35}$

$= \frac{13}{35}$ which is in simplest form

then solve for the whole numbers

3-2=1

therefore, the answer is $1 \frac{13}{35}$

6 0

3 years ago

What are factors of 27?

Artyom0805 [142]

The answer would be 3^33

6 0

3 years ago

Read 2 more answers

-8x+5-2x-4+5x when x=2

julia-pushkina [17]

Answer:

-9

Step-by-step explanation:

Just substitute in 2 for x

-8(2) + 5 - 2(2) - 4 + 5(2)

-16 + 5 - 4 - 4 + 10

-11 - 4 - 4 + 10

-15 - 4 + 10

-19 + 10

-9

Hope this helps!!!

-Unicorns110504

*Please mark brainliest*

7 0

2 years ago

Read 2 more answers

A 12 1/2 mile stretch of road needs repaving . A company bids on the repaving project saying they could replace 2 1/4 miles of r

Nataly [62]

2 1/4 miles => 1 week

1 mile => 1 ÷ 9/4

( 9/4 is an improper fraction of 2 1/4 )

12 1/2 miles => ( 1 ÷ 9/4 ) × 25/2

= 50/9

= 5 5/9 weeks

( 25/2 is an improper fraction of 12 1/2 )

Therefore, it will take 5 5/9 weeks for the company to replace the stretch of road.

( you can round it off to 6 weeks if the qn specify to give to the nearest whole number )

8 0

3 years ago

You want to wrap a gift shaped like the regular triangular prism shown. How many

kotegsom [21]

Answer:

$697.6 \ in^2$

Step-by-step explanation:

-Given the dimensions of the box is 8.7 in, 14 in and 10 in.

Let length=14 in, width=10 in and height =8.7 in

#The surface area of a rectangular prism is calculated as:

$Surface \ Area=2(wl+hl+hw)\\\\l=length, w=width, h=height\\\\\therefore Surface \ Area=2(14\times 10+10\times 8.7+14\times 8.7)\\\\=697.6 \ in^2$

Hence, the wrapping paper needed has an area of $697.6 \ in^2$

8 0

3 years ago