All categories

2 years ago

Describe different ways that you can show a story problem

1 answer:

3 0

Once you know your basic operations addition, subtraction,multiplication, division you will encounter story problems also known as word problems that require you to read a problem and decide which operation to perform in order to get the answer. There are key words here that often indicate which operation you will use. We will give you a list of them, but remember that for many problems there may not be a key word and you’ll have to use your best judgment in order to figure out what to do.

You might be interested in

An ice chest contains six cans of apple juice, eight cans of grape juice, four cans of orange juice, and two cans of mango ju

Katena32 [7]

6+8+4+2=20
2/20 to grab mango can = 1/10
8/19 (remove one for previous) fro grape juice
6/18 for apple juice = 1/3
1/10*8/19*1/3
=
8/570
=
4/285

8 0

2 years ago

Solve for x. (9x – 16)+(3x + 11)+(7x – 5)=180

sergey [27]

Answer:

Step-by-step explanation:

See the picture attached.

8 0

2 years ago

Read 2 more answers

What number is midway between 2/5 and 1?

jeka57 [31]

4/10 and 10/10:

7/10.

5 0

3 years ago

Read 2 more answers

Help me! This is Algebra problem

aleksklad [387]

Answer:

They paid 16 3/4% tax

Step-by-step explanation:

9%+1 3/4%+6%=16 3/4%

7 0

2 years ago

Given:

diamong [38]

Answer:

10-5 $\sqrt{2}$

Step-by-step explanation:

As per the attached figure, right angled $\triangle MDL$ has an inscribed circle whose center is $I$ .

We have joined the incenter $I$ to the vertices of the $\triangle MDL$ .

Sides MD and DL are equal because we are given that $\angle M = \angle L = 45 ^\circ.$

Formula for area of a $\triangle = \dfrac{1}{2} \times base \times height$

As per the figure attached, we are given that side a = 10.

Using pythagoras theorem, we can easily calculate that side ML = 10 $\sqrt{2}$

Points P,Q and R are at $90 ^\circ$ on the sides ML, MD and DL respectively so IQ, IR and IP are heights of $\triangle$ MIL, $\triangle$ MID and $\triangle$ DIL.

Also,

$\text {Area of } \triangle MDL = \text {Area of } \triangle MIL +\text {Area of } \triangle MID+ \text {Area of } \triangle DIL$

$\dfrac{1}{2} \times 10 \times 10 = \dfrac{1}{2} \times r \times 10 + \dfrac{1}{2} \times r \times 10 + \dfrac{1}{2} \times r \times 10\sqrt2\\\Rightarrow r = \dfrac {10}{2+\sqrt2} \\\Rightarrow r = \dfrac{5\sqrt2}{\sqrt2+1}\\\text{Multiplying and divinding by }(\sqrt2 +1)\\\Rightarrow r = 10-5\sqrt2$

So, radius of circle = $10-5\sqrt2$

8 0

2 years ago