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2 years ago

What is the domain of the function y=2 /X-6? EXCO o osx

Mathematics

2 answers:

maks197457 [2]2 years ago

8 0

I believe the answers for this is D

kondaur [170]2 years ago

3 0

Answer:

last option

$6 \leqslant x < \infty$

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PLEASE ANSWER!!!!!!!!!!

vredina [299]

Answer:

C. 350

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

A total of $20,000 is to be invested, some in bonds and some in certificates of deposit (CDs). If the amount invested in bonds i

Brilliant_brown [7]

Answer:

$8,500 in CDs; $11,500 in bonds

Step-by-step explanation:

Let c = amount invested in CDs.

Let b = amount invested in bonds.

"A total of $20,000 is to be invested, some in bonds and some in certificates of deposit (CDs)."

From the total investment, we get this equation.

c + b = 20000

"If the amount invested in bonds is to exceed that in CDs by $3000"

From the details of the amounts, we get this equation.

b = c + 3000

We have a system of equations in two unknowns.

c + b = 20000 Eq. 1

b = c + 3000 Eq. 2

Since Eq. 2 is already solved for b, we will use the substitution method to solve the system of equations.

Substitute b in Eq. 1 with c + 3000.

c + c + 3000 = 20000

2c + 3000 = 20000

2c = 17000

c = 8500

Now substitute c = 8500 in Eq. 2 and solve for b.

b = c + 3000

b = 8500 + 3000

b = 11500

Answer: $8,500 in CDs; $11,500 in bonds

6 0

3 years ago

Charlie saved $130 from $260 he earned. What percent did he save?

JulsSmile [24]

Answer: 50%

Step-by-step explanation:

130/260 = 0.5 = 50%

5 0

3 years ago

If (2i/2+i)-(3i/3+i)=a+bi, then a=<br>A. 1/10<br>B. -10<br>C. 1/50<br>D. -1/10

Verizon [17]

Answer:

Option A is correct.

Step-by-step explanation:

We are given:

$\frac{2i}{2+i}-\frac{3i}{3+i} = a+bi$

We need to find the value of a.

The LCM of (2+i) and (3+i) is (2+i)(3+i)

$=\frac{2i(3+i)}{(2+i)(3+i)}-\frac{3i(2+i)}{(2+i)(3+i)}\\=\frac{6i+2i^2}{(2+i)(3+i)}-\frac{6i+3i^2}{(2+i)(3+i)}\\=\frac{6i+2i^2-(6i+3i^2)}{(2+i)(3+i)}\\=\frac{6i+2i^2-6i-3i^2)}{5+5i}\\=\frac{-i^2}{5+5i}\\i^2=-1\\=\frac{-(-1)}{5+5i}\\=\frac{1}{5+5i}$

Now rationalize the denominator by multiplying by 5-5i/5-5i

$=\frac{1}{5+5i}*\frac{5-5i}{5-5i} \\=\frac{5-5i}{(5+5i)(5-5i)}\\=\frac{5-5i}{(5+5i)(5-5i)}\\(a+b)(a-b)= a^2-b^2\\=\frac{5(1-i)}{(5)^2-(5i)^2}\\=\frac{5(1-i)}{25+25}\\=\frac{5(1-i)}{50}\\=\frac{1-i}{10}\\=\frac{1}{10}-\frac{i}{10}$