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stich3 [128]
3 years ago
6

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

Mathematics
2 answers:
olasank [31]3 years ago
4 0

Answer:

fraction = 57/100

decimal = 0.57

percent = 57%

Step-by-step explanation:

o-na [289]3 years ago
3 0
Fraction: 57/100
Percent: 57%
Decimal: 0.57
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(ab² + b²a) - (-2b+a²) = ?​
nlexa [21]

Answer:

Step-by-step explanation:

2AB² + 2B - A³

7 0
3 years ago
Note: Enter your answer and show all the steps that you use to solve this problem Find the area. The figure is not drawn to scal
fgiga [73]

Answer:

area=13.8cm^2

Step-by-step explanation:

6 0
4 years ago
4) Solve ∆ABC, a = 2.5 cm, c = 3.6 cm, and ∠A = 43°. Begin by sketching and labelling
mezya [45]

The length of b and angle B and C are 3cm, 45  degrees and 79 degrees respectively.

<h3>How to determine the parameters</h3>

To determine the angles and length of sides, we use the sine rule

The sine rule is thus:

\frac{sin A}{a} = \frac{sin B}{b} = \frac{sin C}{c}

Given;

  • a = 2. 5cm
  • c = 3. 6cm
  • ∠A = 43°

Let's find angle C

\frac{sin 43}{2. 5} = \frac{sin C}{3. 6}

cross multiply

0. 682 × 3. 6 = sin C × 2. 5

sin C = 2. 4552/ 2. 5

C = sin^-^1(0. 982)

C = 79°

To find length of b

b= \sqrt{c^2 - a^2}

substitute the values

b = \sqrt{3.6^2 - 2. 5^2}

b = \sqrt{6. 71}

b = 2. 59 cm

b = 3cm

To find angle B, we have

\frac{sin 43}{2. 5} = \frac{sin B}{2. 59}

cross multiply

0. 682 × 2. 59= sin B × 2. 5

sin B = 0. 7065

B = sin^-^1(0. 7065)

B = 45°

Hence, the length of b and angle B and C are 3cm, 45  degrees and 79 degrees respectively.

Learn more about sine rule here:

brainly.com/question/12827625

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8 0
2 years ago
What is 108:200 as a fraction and a decimal
belka [17]

Answer:

108/200 as a fraction, I don't know how to do it as a decimal

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Write an inequality to represent the following statement.
harina [27]
X<2.5

Don't forget to vote brainliest and follow me!!!
5 0
3 years ago
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