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N76 [4]
3 years ago
12

A teacher bought 5 boxes of markers. Each box contained 28 markers. Estimate how many markers she bought by rounding the number

of markers in a box to the nearest ten.
Mathematics
2 answers:
Alona [7]3 years ago
6 0

Answer:

150

Step-by-step explanation:

The answer is actually 150 because you first round 28 to 30, then you do 30x5=150

Hope this helps.

alisha [4.7K]3 years ago
4 0

Answer:

140 markers

Step-by-step explanation:

From the question;

  • The teacher bought 5 boxes of markers
  • Markers in 1 box = 28 markers

We are required to determine the total number of markers she bought

  • Number of markers = Number of boxes × number of markers per box

Therefore;

Number of markers bought = 5 boxes × 28 markers/box

                                              = 140 markers

Thus, the teacher bought 140 markers

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The table shows how Mura spends her free time on a typical Saturday.if she has 6hours of free time how many hours does she spend
ycow [4]

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3 0
1 year ago
Help I have to turn this in. In a couple hours
Gelneren [198K]

28. Surface Area

This is some sort of house-like model so for every face we see there's a congruent one that's hidden.  We'll just double the area we can see.

Area = 2 × ( [14×9 rectangle] + 2[15×9 rectangle]+[triangle base 14, height 6] )

Let's separate the area into the area of the front and the sides; the front will help us for problem 29.

Front =  [14×9 rectangle] + [triangle base 14, height 6]

          = 14×9 + (1/2)(14)(6) = 14(9 + 3) = 14×12 = 168 sq ft

OneSide =  2[15×9 rectangle] = 30×9 = 270 sq ft

Surface Area = 2(168 + 270) = 876 sq ft

Answer: D) 876 sq ft

29.  Volume of an extruded shape is area of the base, here the front, times the height, here 15 feet.  

Volume = 168 * 15 = 2520 cubic ft

Answer: D) 2520 cubic ft

6 0
3 years ago
What is the range of possible sizes for side x? _ < x < _
qwelly [4]

Answer:

2.8 < x < 5.8

Step-by-step explanation:

We must apply the Triangle Inequality Theorem which states that for any triangle with sides a, b, and c:

a + b > c

b + c > a

c + a > b

Here, let's arbitrarily denote a as 4.1, b as 1.3, and c as x. So, let's plug these values into the 3 inequalities listed above:

a + b > c  ⇒  4.1 + 1.3 > x  ⇒  5.8 > x

b + c > a  ⇒  1.3 + x > 4.1  ⇒  x > 2.8

c + a > b  ⇒  x + 4.1 > 1.3  ⇒  x > -2.8

Look at the last two: clearly if x is greater than 2.8 (from the second one), then it will definitely be greater than -2.8 (from the third), so we can just disregard the last inequality.

Thus, the range of possible sizes for x are:

2.8 < x < 5.8

<em>~ an aesthetics lover</em>

8 0
3 years ago
Read 2 more answers
Which of the following are solutions to the equation below?
scoundrel [369]

Answer:

The correct options for the solution values are:

  • x=2\sqrt{2}-5
  • x=-2\sqrt{2}-5

Step-by-step explanation:

Given the expression

x^2+10x+25=8

Subtract 25 from both sides

x^2+10x+25-25=8-25

Simplify

x^2+10x=-17

Add 25 or 5² to both sides

x^2+10x+5^2=8

as

x^2+10x+5^2=\left(x+5\right)^2

so the expression becomes

\left(x+5\right)^2=8

\mathrm{For\:}f^2\left(x\right)=a\mathrm{\:the\:solutions\:are\:}f\left(x\right)=\sqrt{a},\:-\sqrt{a}

solve

x+5=\sqrt{8}

Subtract 5 from both sides

x+5-5=2\sqrt{2}-5

x=2\sqrt{2}-5

solve

x+5=-\sqrt{8}

Subtract 5 from both sides

x+5-5=-2\sqrt{2}-5

x=-2\sqrt{2}-5

Therefore, the solution to the equation

x=2\sqrt{2}-5,\:x=-2\sqrt{2}-5

Hence, the correct options for the solution values are:

  • x=2\sqrt{2}-5
  • x=-2\sqrt{2}-5

6 0
3 years ago
Find the value of x in x+y=8 and x-y=6
Diano4ka-milaya [45]

Answer: x=-y+8 and x=y+6

Step-by-step explanation:

7 0
3 years ago
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