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kogti [31]
3 years ago
8

Help with the question please​

Mathematics
1 answer:
galina1969 [7]3 years ago
5 0

Step-by-step explanation:

since all 4 sides are equal in length, the figure is a square.

let the length of the square be lm.

529 = l x l

l = 23m

perimeter = 23 x 4 = 92m

Topic: mensuration

If you like to venture further, do check out my insta (learntionary) where I regularly post useful math tips! Thank you!

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What is the value of |-7| - |3|
Rudik [331]

Answer:

4 is the correct answer

Step-by-step explanation:

|-7| = 7

|3| = 3

7 - 3 = 4

3 0
3 years ago
??? times 4/9 is negative 8
Nitella [24]
-2/9 x 4/9 is -8
so the answer is -2/9
5 0
3 years ago
Read 2 more answers
What was the original price? Explain how you arrived at this answer.
Bumek [7]

Answer:

180

Step-by-step explanation:

120 divide by 2

then times by 3

4 0
3 years ago
Natalle is a softball coach. She has a pitching private lesson with Kasey. Kasey's mom pays Natalle $11 per hour plus a $6 tip.
kirill [66]
11t\text{ + 6 = 39}

where t is the number of hours worked by Natalle

Here, we want to write an equation that will represent the given scenario

Let the number of hours for which she earned the pay be t

From the question, she earns $11 per hour

So for t hours, the amount earned excluding the tip will be;

(11\times t)\text{ = \$11t}

Now, let us add the tip and equate to the total amount earned

We have this as;

11t\text{ + 6 = 39}

3 0
1 year ago
Given that the line y=c-2x is tangent to the curve y^2=kx where c and k are non-zero constants,express k in terms of c
Viktor [21]

Answer:

k = -8c.

Step-by-step explanation:

y^2 = kx

Find the slope of the tangent by differentiating:

y' 2y = k

y' = k / 2y = the slope of the tangent.

The given equation of the tangent is y = -2x + c so the slope = -2.

Therefore  k/2y =  -2  so k = -4y and y = -k/4

y^2 = kx so k^2/16 = kx giving x = k/16.

Substituting for x and y in y = -2x + c:

-k/4 = -2 k/16 + c

-k/4 = -k/8  + c

c = -k/4 + k/8 = -k/8.

So -k = 8c

k = -8c.

3 0
4 years ago
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