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2 years ago

Which of the following points satisfies the inequality x ≥ -3? (select all that apply) a(0, -2) b(-3, 4) c(-4, 2) d(5, -5)

Mathematics

1 answer:

VladimirAG [237]2 years ago

8 0

Answer:

a, b, d

Step-by-step explanation:

For the inequality x ≥ -3 . . .

a(0, -2) . . . 0 ≥ -3, true

b(-3, 4) . . . -3 ≥ -3, true

c(-4, 2) . . . -4 ≥ -3, FALSE

d(5, -5) . . . 5 ≥ -3, true

Points a, b, d satisfy the inequality.

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Sliva [168]

Answer:

Step-by-step explanation:

a² - b² = (a+ b)(a - b)

1) (2n-4/2n) ÷ (n^2-4/n)

$=\frac{2n-4}{2n}*\frac{n}{n^{2}-4}\\\\=\frac{2n-2*2}{2n}*\frac{n}{n^{2}-2^{2}}\\\\=\frac{2*(n-2)}{2n}*\frac{n}{(n+2)*(n-2)}\\\\=\frac{1}{n+2}$

2) [y^2-36/y^2-49] ÷[ y+6/y-7]

$=\frac{y^{2}-36}{y^{2}-49}*\frac{y-7}{y+6}\\\\=\frac{y^{2}-6^{2}}{y^{2}-7^{2}}*\frac{y-7}{y+6}\\\\=\frac{(y+6)*(y-6)}{(y+7)*(y-7)}*\frac{y-7}{y+6}\\\\=\frac{y-6}{y+7}\\$

3) [m^2-1/ m^2-m] ÷ [m^2-7m-8/3m ]

$=\frac{m^{2}-1}{m^{2}-m}*\frac{3m}{m^{2}-7m-8}\\\\=\frac{(m+1)*(m-1)}{m*(m-1)}*\frac{3m}{(m-8)*(m+1)}\\\\=\frac{3}{m-8}$

Hint : m² - 7m - 8

sum = -7

Product = -8

Factor = (-8), 1

m² - 7m - 8 =m² - 8m + m - 8

= m*(m - 8) + (m-8)

= (m - 8)(m +1)

7 0

2 years ago

PLLSSS REALLY URGENT <br> If g(x)=3x+4, determine n that g(-3)=g^-1(n).

grigory [225]

Answer:

we have:

g(-3) = 3.(-3) + 4 = -5

$g {}^{ - 1} (n) = \frac{1}{3n + 4}$

because

$g( - 3) = g {}^{ - 1} (n)$

$- 5 = \frac{1}{3n + 4} \\ \\ < = > - 5(3n + 4) = 1 \\ \\ < = > - 15n - 20 = 1 \\ \\ < = > n = \frac{ - 21}{15} \\ \\ < = > n = \frac{ - 7}{5}$

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2 years ago

Round the number to the place indicated.<br> Round to the nearest tenth: 1.4511

Evgen [1.6K]

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2 years ago

How to solve -5 + 13y = -7

zubka84 [21]

Answer:

y = -2/13

Step-by-step explanation:

-5 + 13y = -7

Add 5 to each side

-5+5 + 13y = -7+5

13y = -2

Divide by 13

13y/13 = -2/13

y = -2/13

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2 years ago

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