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3 years ago

PLEASE ANSWER BOTH OF THEM ASAP!!!!

Mathematics

2 answers:

Ne4ueva [31]3 years ago

6 0

125 and 86.8 (If I did the math correctly). Remember that the area of right triangles are base times height times 1/2.

Annette [7]3 years ago

6 0

I know #2 is C, because lengthxwidthxheight equals volume on a regular solid object. I'm sorry I can only answer part 2.

You might be interested in

Six more than 1/5 a number is 12. What is the number

suter [353]

Answer:

Step-by-step explanation:

let x be the number

6+1/5x=12

1/5x=12-6

1/5x=6

x=5*6

x=30

check: 6+(1/5*30)=6+6=12

8 0

2 years ago

The sum of 2 times a number and 8 is 7

My name is Ann [436]

Answer:

The number is -1/2

Step-by-step explanation:

Let x be the number

2x+8 = 7

Subtract 8 from each side

2x+8-8 = 7-8

2x = -1

Divide each side by 2

2x/2 =-1/2

x = -1/2

The number is -1/2

8 0

3 years ago

Read 2 more answers

When an oil tank is 7/12 full, it contains 5 1/4 gallons. How many gallons does it hold when full

blondinia [14]

<h3>Answer:</h3>

The tank can hold up to 9 gallons.

Quick Review:

By doing the inverse of 7/12, 12/7, when you multiply it, you'll get a value of 1.

7/12 * 12/7 = 1

Doing the Problem:

By multiplying 5 1/4 by 12/7, you'll get the answer. But first, convert 5 1/4 into an improper fraction.

5 1/4 ---> 21/4

Okay, lets go!

21/4 * 12/7 = 252/28

<u>When reduced, this gives you the answer 9/1, or 9 gallons.</u>

Hope that helps, :)

6 0

3 years ago

What’s the domain and range of:<br> log(√(2x-1) + 3 )<br> Please explain how you got it too!!

Radda [10]

Two main facts are needed here:

1. The logarithm $\log x$ , regardless of the base of the logarithm, exists for $x>0$ .

2. The square root $\sqrt x$ exists for $x\ge0$ .

(in both cases we're assuming real-valued functions only)

By (2) we know that $\sqrt{2x-1}$ exists if $2x-1\ge0$ , or $x\ge\dfrac12$ .

By (1), we know that $\log(\sqrt{2x-1}+3)$ exists if $\sqrt{2x-1}+3>0$ , or $\sqrt{2x-1}>-3$ . But as long as the square root exists, it will always be positive, so this condition will always be met.

Ultimately, then, we only require $x\ge\dfrac12$ , so the function has domain $\left[\dfrac12,\infty)$ .

To determine the range, we need to know that, in their respective domains, $\sqrt x$ and $\log x$ increase monotonically without bound. We also know that $x=\dfrac12$ at minimum, at which point the square root term vanishes, so the least value the function takes on is $\log3$ . Then its range would be $[\log3,\infty)$ .