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3 years ago

CAN SOMEONE PLEASE HELP!!!??

Mathematics

1 answer:

Lostsunrise [7]3 years ago

5 0

B i think because it’s the range

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How do you solve |3x|=18

Oduvanchick [21]

We can get two different equations from this:

3x = 18

and

3x = -18

So:

x = 6, x = -6

x = {6, -6}

6 0

3 years ago

Read 2 more answers

Types of bonds are divided into three categories: good risk, medium risk, and poor risk. Assume that of a total of 11, 332 bonds

Alekssandra [29.7K]

Answer:

Probabilty of not poor= 0.75

Step-by-step explanation:

total of 11332 bonds.

7311 are good risk.

1182 are medium risk.

Poor risk

= total risk-(good risk+ medium risk)

= 11332-(7311+1182)

= 11332-8493

= 2839.

Poor risk = 2839

Probabilty that the ball choosen at random is not poor= 1 - probability that the ball is poor

Probability of poor = 2839/11332

Probabilty of poor= 0.2505

Probabilty that the ball choosen at random is not poor= 1- 0.2505

= 0.7495

To two decimal place= 0.75

3 0

2 years ago

This is kinda confusing can someone help explain!!

AURORKA [14]

Answer: all i can say is this for an edu guess

2 6 and 4 would be 40 or 30

3 5 and 7 would be 110 or 120

8 would be 20 or 10

also this was the first answer

7 0

2 years ago

Read 2 more answers

Solve (1/81)^x*1/243=(1/9)^−3−1 by rewriting each side with a common base.

elena55 [62]

Answer:

$x=(243)log_{\frac{1}{81}}[(\frac{1}{81})-1]$

Step-by-step explanation:

you have the following formula:

$(\frac{1}{81})^{\frac{x}{243}}=(\frac{1}{9})^{-3}-1$

To solve this equation you use the following properties:

$log_aa^x=x$

Thne, by using this propwerty in the equation (1) you obtain for x

$log_{(\frac{1}{81})}(\frac{1}{81})^{\frac{x}{243}}=log_{\frac{1}{81}}[(\frac{1}{81})-1]\\\\\frac{x}{243}=log_{\frac{1}{81}}[(\frac{1}{81})-1]\\\\x=(243)log_{\frac{1}{81}}[(\frac{1}{81})-1]$

8 0

2 years ago

Combine and simplify the following radical expression ASAP ASAP ASAP ASAP

Montano1993 [528]

Answer:

$\sqrt[3]{3}$

Step-by-step explanation:

Our expression is: $\frac{1}{3} \sqrt[3]{81}$ .

Let's focus on the cube root of 81 first. What's the prime factorisation of 81? It's simply: 3 * 3 * 3 * 3, or $3^3*3$ . Put this in for 81:

$\sqrt[3]{81} =\sqrt[3]{3^3*3}=\sqrt[3]{3^3} *\sqrt[3]{3}$

We know that the cube root of 3 cubed will cancel out to become 3, but the cube root of 3 cannot be further simplified, so we keep that. Our outcome is then:

$\sqrt[3]{3^3} *\sqrt[3]{3}=3\sqrt[3]{3}$

Now, let's multiply this by 1/3, as shown in the original problem:

$\frac{1}{3}* 3\sqrt[3]{3}=\sqrt[3]{3}$

Thus, the answer is $\sqrt[3]{3}$ .

~ an aesthetics lover

6 0

3 years ago