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marta [7]
3 years ago
13

Triangle $ABC$ is isosceles with $AB = BC.$ If $AC = 20$ and $[ABC] = 240,$ then find the perimeter of triangle $ABC.$

Mathematics
2 answers:
gtnhenbr [62]3 years ago
5 0

The perimeter of triangle ABC is 84.

ollegr [7]3 years ago
3 0

Answer:

C is the Answer

Step-by-step explanation:

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Need help with this problem.
Galina-37 [17]

Answer:

18

Step-by-step explanation:

To find this we must divide 45 by 20 which gives the answer of 2.25. As we know this we now must multiply that by 8 resulting in to 18.

Therefore the answer is 18

5 0
3 years ago
Which of the following are monomials?
Leviafan [203]

Answer:

A. x^11 and B. 8x.

Step-by-step explanation:

A monomial is a polynomial with just one term. That would be Options A and B.

Options C and E are trinomials.

Options D and F are binomials.

Hope this helps!

7 0
3 years ago
Rafeeq bought a field in the form of a quadrilateral (ABCD)whose sides taken in order are respectively equal to 192m, 576m,228m,
Valentin [98]

Answer:

a. 85974 m²

b. 17,194,800 AED

c. 18,450 AED

Step-by-step explanation:

The sides of the quadrilateral are given as follows;

AB = 192 m

BC = 576 m

CD = 228 m

DA = 480 m

Length of a diagonal AC = 672 m

a. We note that the area of the quadrilateral consists of the area of the two triangles (ΔABC and ΔACD) formed on opposite sides of the diagonal

The semi-perimeter, s₁,  of ΔABC is found as follows;

s₁ = (AB + BC + AC)/2 = (192 + 576 + 672)/2 = 1440/2 = 720

The area, A₁, of ΔABC is given as follows;

Area\, of \, \Delta ABC = \sqrt{s_1\cdot (s_1 - AB)\cdot (s_1-BC)\cdot (s_1 - AC)}

Area\, of \, \Delta ABC = \sqrt{720 \times (720 - 192)\times  (720-576)\times  (720 - 672)}

Area\, of \, \Delta ABC = \sqrt{720 \times 528 \times  144 \times  48} = 6912·√(55) m²

Similarly, area, A₂, of ΔACD is given as follows;

Area\, of \, \Delta ACD= \sqrt{s_2\cdot (s_2 - AC)\cdot (s_2-CD)\cdot (s_2 - DA)}

The semi-perimeter, s₂,  of ΔABC is found as follows;

s₂ = (AC + CD + D)/2 = (672 + 228 + 480)/2 = 690 m

We therefore have;

Area\, of \, \Delta ACD = \sqrt{690 \times (690 - 672)\times  (690 -228)\times  (690 - 480)}

Area\, of \, \Delta ACD = \sqrt{690 \times 18\times  462\times  210} = \sqrt{1204988400} = 1260\cdot \sqrt{759} \ m^2

Therefore, the area of the quadrilateral ABCD = A₁ + A₂ = 6912×√(55) + 1260·√(759) = 85973.71 m² ≈ 85974 m² to the nearest meter square

b. Whereby the cost of 1 meter square land = 200 AED, we have;

Total cost of the land = 200 × 85974 = 17,194,800 AED

c. Whereby the cost of fencing 1 m = 12.50 AED, we have;

Total perimeter of the land = 576 + 192 + 480 + 228 = 1,476 m

The total cost of the fencing the land = 12.5 × 1476 = 18,450 AED

4 0
3 years ago
A school bought 10³ erasers as part of an order for supplies. The total cost of the erasers was $30. What was the cost of each e
tresset_1 [31]

Answer:

0.03

Step-by-step explanation:

divide 3 by 10^3 to obtain the answer.

8 0
2 years ago
When a ribbon is cut into 3 equal pieces,each piece is _______ m long (correct to 2 decimal places).
inn [45]

Answer:

How long was the ribbon before it was cut?

Step-by-step explanation:

3 0
3 years ago
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