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2 years ago

PLEASE HELP WILL GIVE BRAINLIEST I don't understand this at all

Mathematics

1 answer:

castortr0y [4]2 years ago

4 0

Answer:

y= -x(x+2)(x-1)².

Step-by-step explanation:

the graph and equation is in the attached picture.

The zeros of function y are: -2; 0 and 1 - they help to determine the required equation.

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What is the vertex of y<∣x−3∣+5

Tcecarenko [31]

Answer:

(3, 5)

Step-by-step explanation:

The graph is is the standard y=|x| except the values tells you that x shifts 3 (within the absolute value or parentheses x does the opposite) to the right and the y value shifts 5 up (numbers outside parentheses affects y and does what it says). You can try using a table of values then graphing to check your answer.

3 0

3 years ago

What is 6 divide by 5

lara31 [8.8K]

Hey there!

6 divided by 5 = 1.2

Hope this helps!
Have a great day! (:

7 0

3 years ago

Witch is greater -15.26 or 15 13/50?

frozen [14]

Answer:

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

If j and k are nonzero integers, which pair of points must lie in the same quadrant?

Marizza181 [45]

<h3>Answer: Choice D. (3j, 3k) and (3/j, 3/k)</h3>

The slash indicates a fraction.

=============================================

Proof:

We'll need to consider 4 different cases.

-----------------------

Case (1): j > 0 and k > 0

If j > 0, then 3j > 0 and 3/j > 0

If k > 0, then 3k > 0 and 3/k > 0

The two points (3j, 3k) and (3/j, 3/k) are both in quadrant 1.

-----------------------

Case (2): j > 0 and k < 0

If j > 0, then 3j > 0 and 3/j > 0

If k < 0, then 3k < 0 and 3/k < 0

Points (3j, 3k) and (3/j, 3/k) are both in quadrant 4.

------------------------

Case (3): j < 0 and k > 0

If j < 0, then 3j < 0 and 3/j < 0

If k > 0, then 3k > 0 and 3/k > 0

Points (3j, 3k) and (3/j, 3/k) are in quadrant 3.

------------------------

Case (4): j < 0 and k < 0

If j < 0, then 3j < 0 and 3/j < 0

If k < 0, then 3k < 0 and 3/k < 0

Points (3j, 3k) and (3/j, 3/k) are in quadrant 4.

------------------------

For nonzero integers j and k, we've shown that Points (3j, 3k) and (3/j, 3/k) are in the same quadrant. This concludes the proof.

8 0

3 years ago

Read 2 more answers

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