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3 years ago

√x<4 A.20 B.12 C.2 D.-1 E.5. F. No solution

1 answer:

5 0

We know that for √x to be meaningful, we must have x ≥ 0.

If we square both sides of the inequality, we have
x < 4²
x < 16

Thus the whole solution is
0 ≤ x < 16

The following choices are valid solutions to the inequality:
B. 12
C. 2
E. 5

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Jill is thinking of a number. Twenty less than one forth of the number is -45. Find Jill's number.

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Answer:

The number is -100, Jill can think of something else, now.

Step-by-step explanation:

x = the number

x/4 - 20 = -45 {twenty less than one-fourth of the number is -45}

x/4 = -25 {added 20 to each side}

x = -100 {multiplied each side by 4}

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Which statement is true about the solution ?

Harrizon [31]

Answer:

There are no real solutions to this equation

Step-by-step explanation:

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2 years ago

How would the expression x^3+64 be rewritten using sum off cubes

Mamont248 [21]

$\bf \textit{difference of cubes}
\\ \quad \\
a^3+b^3 = (a+b)(a^2-ab+b^2)\qquad
(a+b)(a^2-ab+b^2)= a^3+b^3 \\ \quad \\
a^3-b^3 = (a-b)(a^2+ab+b^2)\qquad
(a-b)(a^2+ab+b^2)= a^3-b^3\\\\
-------------------------------\\\\
\boxed{64=4^3}\qquad x^3+64\implies x^3+4^3\implies (x+4)(x^2-x4+4^2)
\\\\\\
(x+4)(x^2-4x+16)$

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2 years ago

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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

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3 years ago