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2 years ago

The sum of two numbers is twenty-five.One number is five less than the other number. Find the larger number

1 answer:

3 0

The larger number is 30.

Set up your equation.

x + y = 25

The question tells us one number is 5 less than the other.

The equation now becomes
x -5 = 25

Add 5 to each side to isolate x.

x=30

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Anyone know the answer to this algebra 2 question?

KATRIN_1 [288]

Answer:

The answer to your question is: I(dB) = 320

Step-by-step explanation:

Data

I(dB) = 10log( $\frac{I}{Io}$ )

I = intensity of a given sound

Io = hearing intensity

if I = 10³²(Io) find I(dB)

Process

Substitute the value of I in the equation given and simplify using logarithms rules.

l(dB) = $10 log \frac{10^{32}Io }{Io}$

Eliminate Io

I(dB) = 10 log (10³²)

Use the rule of powers in logarithms

I(dB) = 10(32)log(10)

l(dB) = 320(1)

I(dB) = 320

7 0

3 years ago

What are the solutions of the equation x6 + 6x3 + 5 = 0? Use factoring to solve.

lapo4ka [179]

Consider the equation $x^6+6x^3+5=0$ .

First, you can use the substitution $t=x^3$ , then $x^6=(x^3)^2=t^2$ and equation becomes $t^2+6t+5=0$ . This equation is quadratic, so

$D=6^2-4\cdot 5\cdot 1=36-20=16=4^2,\ \sqrt{D}=4,\\ \\ t_{1,2}=\dfrac{-6\pm 4}{2} =-5,-1$ .

Then you can factor this equation:

$(t+5)(t+1)=0$ .

Use the made substitution again:

$(x^3+5)(x^3+1)=0$ .

You have in each brackets the expression like $a^3+b^3$ that is equal to $(a+b)(a^2-ab+b^2)$ . Thus,

$x^3+5=(x+\sqrt[3]{5})(x^2-\sqrt[3]{5}x+\sqrt[3]{25}) ,\\x^3+1=(x+1)(x^2-x+1)$

and the equation is

$(x+\sqrt[3]{5})(x^2-\sqrt[3]{5}x+\sqrt[3]{25})(x+1)(x^2-x+1)=0$ .

Here $x_1=-\sqrt[3]{5} , x_2=-1$ and you can sheck whether quadratic trinomials have real roots:

1. $D_1=(-\sqrt[3]{5}) ^2-4\cdot \sqrt[3]{25}=\sqrt[3]{25} -4\sqrt[3]{25} =-3\sqrt[3]{25}$ .

2. $D_2=(-1)^2-4\cdot 1=1-4=-3$ .

This means that quadratic trinomials don't have real roots.

Answer: $x_1=-\sqrt[3]{5} , x_2=-1$

If you need complex roots, then

$x_{3,4}=\dfrac{\sqrt[3]{5}\pm i\sqrt{3\sqrt[3]{25}}}{2} ,\\ \\x_{5,6}=\dfrac{1\pm i\sqrt{3}}{2}$ .

6 0

3 years ago

Plss help if you I don’t how to find the answer

leonid [27]

answer

11x = 480

i dont know

5 0

2 years ago

20 is 50% of what base

Tresset [83]

The base would be 40

5 0

3 years ago

Phone Company A charges a monthly fee of $42.50, and $0.02 for each minute talk time. Phone company B charges a monthly fee of $

Y_Kistochka [10]

We need to call for x minute
+ Phone Company A charges a monthly fee of $42.50, and $0.02 for each minute talk time. So we have to spend: $42.50+ $0.02x
+ Phone company B charges a monthly fee of $25.00, and $0.09 for each minute of talk time. So we have to spend: $25.00+ $0.09x

We solve for x: $42.50+ $0.02x> $25.00+ $0.09x
or $42.50- $25.00 > $0.09x- $0.02x
and we have $0.07x<$27.50
or x< 27.50:0.07 and x< 393.86

The answer is:
If we have to call much time, at least 394 minutes, we should choose A
If not, choose B

8 0

3 years ago