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Vika [28.1K]
3 years ago
13

Pls help asappppppppppp

Mathematics
2 answers:
dedylja [7]3 years ago
8 0

Answer:

see below

Step-by-step explanation:

for 7, about 520/20 = 26

for 8, around 2,000/100= 20

Alika [10]3 years ago
8 0

Answer:

31 for 7 and 8 it would be 22

for 5 it would be 9 and 6: 1

Step-by-step explanation:

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Clarise evaluated this expression. (66.3 – 14.62) ÷ 0.6 – 0.22 (51.68) ÷ 0.6 – 0.22 (51.68) ÷ 0.42 51.68 ÷ 0.16 32.3 Which error
Gennadij [26K]
Let's go through this problem step by step while bearing in mind the concept of PEMDAS.

Step 1: (66.3-14.62)/0.6-0.2^{2}
Declaration of the expression. Nothing wrong here yet.

Step 2: (51.68)/0.6-0.2^{2}
Clarise evaluated what's inside the parenthesis first - which was the right thing to do! There are no mistakes in this step.

Step 3: (51.68)/0.4^{2}
In this step she subtracted first. This should not be the case! PEMDAS tells us that the exponents and division gets higher priority than subtraction. This is therefore the first mistake Clarise makes.

Step 4: (51.68)/0.16
In this step Clarise evaluates the exponent. This does not violate any rules (relative to the previous expression) since PEMDAS tells us that exponents take higher priority than division.

Step 5: 3.23 (Clarise's final answer)
In Clarise's final step, she manages to get the wrong answer! Dividing 51.68 by 0.16 would give us 323. This is another mistake of Clarise.

Looking at the choices, we can now identify what mistakes Clarise made:
-She subtracted before evaluating the exponents
-She subtracted before she divided
-She divided incorrectly
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3 years ago
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What is the area of a 3.3in circle​
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Answer:

3.3in

Step-by-step explanation:

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Find all values of k for which the equation 3x^2−4x+k=0 has no solutions.
dusya [7]

The given equation has no solution when K is any real number  and k>12

We have given that

3x^2−4x+k=0

△=b^2−4ac=k^2−4(3)(12)=k^2−144.

<h3>What is the condition for a solution?</h3>

If Δ=0, it has 1 real solution,

Δ<0 it has no real solution,

Δ>0 it has 2 real solutions.

We get,

Δ=k^2−144 here Δ is not zero.

It is either >0 or <0

Δ<0 it has no real solution,

Therefore the given equation has no solution when K is any real number.

To learn more about the solution visit:

brainly.com/question/1397278

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