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2 years ago

15

Select all equations that have infinitely many solutions.

1 answer:

9966 [12]2 years ago

6 0

Answer:

C And D

Step-by-step explanation:

HAVE A GREAT DAY!

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The rectangular room shown has a length of 20 ft, width of 48 ft, and height of 10 ft. Use the Pythagorean Theorem to find lengt

Vlad [161]

For the diagonal of the floor:
20^2 + 48^2= c^2
400+2304= c^2
c=52
For the diagonal of the cube:
52^2 + 10^2 = c^2
2704+100=c^2
c=52.95
Rounded= 53
Tell me if this helps!!!

4 0

3 years ago

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Simplify 12 to the 16th power over 12 to the 4th power

dexar [7]

$\dfrac{12^{16}}{12^4} = 12^{16 - 4} = 12^{12}$

6 0

2 years ago

Financial Algebra 1. Unit 1: Simple and Compound Interest Quiz A..

Nikolay [14]

Answer:

$56,558.1

Step-by-step explanation:

This is a question on compound interest.

The formula to calculate the Total Amount based on compound interest is given as:

A = P( 1 + r/n) ^nt

A = Total or Final amount in the account after t years

P = Principal/ Initial amount invested=$35,000

r = Interest rate = 12%

n = compounding Frequency = daily = using 30 days in a month = 30 × 12 = 360 days

t = time in years = 4

A = $35,000( 1 + 0.12/360)^360 × 4

A = $56558.08

Approximately to the nearest cent ≈ A = $56,558.1

Therefore, Priscilla should be expecting $56,558.1 in the account after 4 years.

8 0

2 years ago

[tex]cos {}^{4} α+sin {}^{4} α= \frac{1}{4} (3+cos4α)<br>Prove:<br>

asambeis [7]

Given:

$\cos^4 \alpha+\sin^4\alpha=\dfrac{1}{4}(3+\cos 4 \alpha)$

To prove:

The given statement.

Proof:

We have,

$\cos^4 \alpha+\sin^4\alpha=\dfrac{1}{4}(3+\cos 4 \alpha)$

$LHS=\cos^4 \alpha+\sin^4\alpha$

$LHS=(\cos^2 \alpha)^2+(\sin^2 \alpha)^2$

$LHS=(\cos^2 \alpha+\sin^2\alpha)^2-2\sin ^2\alpha\cos^2 \alpha$ $[\because a^2+b^2=(a+b)^2-2ab]$

$LHS=(1)^2-2(1-\cos^2 \alpha)\cos^2 \alpha$ $[\because \cos^2 \alpha+\sin^2\alpha=1]$

$LHS=1-2\cos^2 \alpha+2\cos^4 \alpha$

Now,

$RHS=\dfrac{1}{4}(3+\cos 4 \alpha)$

$RHS=\dfrac{1}{4}[3+(2\cos^2 2\alpha-1)]$ $[\because \cos 2\theta=2\cos^2\theta -1]$

$RHS=\dfrac{1}{4}[2+2\cos^2 2\alpha]$

$RHS=\dfrac{1}{4}[2+2(2\cos^2 \alpha-1)^2]$ $[\because \cos 2\theta=2\cos^2\theta -1]$

$RHS=\dfrac{1}{4}[2+2(4\cos^4 \alpha-4\cos \alpha+1)]$ $[\because (a-b)^2=a^2-2ab+b^2]$

$RHS=\dfrac{1}{4}[2+8\cos^4 \alpha-8\cos \alpha+2]$

$RHS=\dfrac{1}{4}[4+8\cos^4 \alpha-8\cos \alpha]$

$RHS=1+2\cos^4 \alpha-2\cos \alpha$

$RHS=1-2\cos^2 \alpha+2\cos^4 \alpha$

$LHS=RHS$

Hence proved.

8 0

2 years ago

A researcher posts a magazine advertisement offering $30 in exchange for participation in a short study. The researcher accepts

matrenka [14]

The answer to your question is A

7 0

2 years ago

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