Given ABC PQR, m < B = 3v +4, and m<Q = 8v -6, find m<B and m<Q
A.22
B.11
C.10
D.25

Answer:

[tex]2.25\times 10^1[/tex] times greater is the weight of the elephant than the weight of the lion.

Step-by-step explanation:

Given : A zookeeper weighed an African elephant to be [tex]9\times 10^3[/tex] pounds and an African lion to be [tex]4\times 10^2[/tex] pounds.  

To find : How many times greater is the weight of the elephant than the weight of the lion?

Solution :  

A zookeeper weighed an African elephant to be [tex]9\times 10^3[/tex] pounds.

A zookeeper weighed an African lion to be [tex]4\times 10^2[/tex] pounds

Now, Divide the weighed an African elephant with weighed an African lion to determine times greater is the weight of the elephant than the weight of the lion,

[tex]n=\frac{9\times 10^3}{4\times 10^2}[/tex]

[tex]n=2.25\times 10^1[/tex]

Therefore, Option A is correct.

[tex]2.25\times 10^1[/tex] times greater is the weight of the elephant than the weight of the lion.

The required answer of the given expression is 8/3.

Simplification generally means finding an answer for the complex calculation that may involve numbers on division, multiplication, square roots, cube roots, plus and minus.

The given expression is,

7/3 + 3(2/3 − 1/3)²

Solving the bracket,

= 7/3 + 3{(2− 1)/3}²

solving the fraction,

= 7/3 + 3(1/3)²

solving the square,

= 7/3 + 3(1/9)

= 7/3 + 1/3

adding them we get,

= 8/3

which is the required answer of the given expression.

Thus, the required answer of the given expression is 8/3.

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Let us say that:

s = speed of Donna

s + 2 = speed of April

Then we are also given that:

t + (25/60) = time of Donna

t = time of April

The distance taken by the two should be equal. We know that distance is the product of speed and time, so:

s * (t + 25/60) = (s + 2) t

We also know that:

(s + 2) * t = 5

s = 5/t – 2

So that:

(5/t – 2) * (t + 25/60) = 5

5 + 125/60t – 2t – 50/60 = 5

25/12t – 2t – 5/6 = 0

Multiply everything by t:

- 2t^2 – 5/6 t + 25/12 = 0

t^2 + 5/12 t = 25/24

Completing the square:

(t + 5/24)^2 = 25/24 + (5/24)^2

t + 5/24 = ± 1.04

t = -1.25 h, 0.83 h

Since time cannot be negative, therefore t = 0.83 h

So,

(s + 2) * t = 5

s + 2 = 5/0.83 = 6 miles per hour

s = 4 miles per hour

Answers:

speed of Donna = 4 miles / hr

speed of April = 6 miles / hr

To solve this problem, the two consecutive odd integers are defined as x and x+2. A mathematical equation is created according to the problem conditions and solved. The consecutive odd integers are found to be 81 and 83.

Let's define the two consecutive odd integers as x (the smaller) and x+2 (the larger). Given the condition from the question that the sum of the smaller integer and three times the larger integer is 330, we can form the equation x + 3*(x+2) = 330.

We'll simplify that equation, which gives us: x + 3x + 6 = 330, and combining like terms, we get 4x + 6 = 330. We can solve for x by subtracting 6 from both sides and then dividing both sides by 4, which results in x = 81. Given that x represents the smaller odd integer, the larger integer would be x + 2 or 83.

So, the two consecutive odd integers fulfilling the conditions given are 81 and 83.

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The two consecutive odd integers such that the sum of the smaller and 3 times the larger is 330 are 81 and 83.

Denote the smaller odd integer as x and the larger consecutive odd integer as x + 2.

According to the problem, the sum of the smaller integer and 3 times the larger integer is 330. So, we have the equation:

x + 3(x + 2) = 330

Now, solve for x :

1. Distribute the 3 on the left side:

x + 3x + 6 = 330

2. Combine like terms:

4x + 6 = 330

3. Subtract 6 from both sides to isolate terms with x:

4x = 324

4. Divide both sides by 4 to solve for x:

[tex]\[ x = \frac{324}{4} \]\[ x = 81 \][/tex]

So, the smaller odd integer x is 81. Now find the larger consecutive odd integer:

x + 2 = 81 + 2 = 83

Therefore, the two consecutive odd integers are 81 and 83.

To verify:

- The sum of the smaller integer and 3 times the larger integer:

[tex]\[ 81 + 3 \cdot 83 = 81 + 249 = 330 \][/tex]

The verification confirms that the solution is correct.

So, the two consecutive odd integers such that the sum of the smaller and 3 times the larger is 330 are 81 and 83.

The number 0.912912 can be written in bar notation as 0.912, with a bar over 912 to indicate the repeating decimals. This method is commonly used in mathematics to represent repeating decimals.

To rewrite the number 0.912912 using bar notation, consider the part of the decimal that repeats. In this case, the repeating section is 912. Therefore, the number can be written as 0.912 where the 912 has a bar over it to represent the fact that it repeats. Bar notation is a way in mathematics to represent repeating decimals. The bar, or vinculum, shows which numbers repeat in the decimal. For this number, 0.912 with a bar over 912 is the way it would be written in bar notation.

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80% as a fraction in reduced form is 4/5.

To express 80% as a fraction in reduced form, you need to convert the percentage to a fraction. The term reduced form means simplifying the fraction to its lowest terms.

Here's a step-by-step explanation:

Step 1: Understand what a percentage represents

A percentage is a way to express a fraction or ratio out of 100. So, when you see 80%, it means 80 out of 100.

Step 2: Write the percentage as a fraction

To convert a percentage to a fraction, you can write it as a fraction with the percentage as the numerator and 100 as the denominator. Therefore, 80% can be written as 80/100.

Step 3: Simplify the fraction

To reduce the fraction to its lowest terms, you need to find the greatest common divisor (GCD) between the numerator and denominator. In this case, both 80 and 100 can be divided by 20. Dividing both the numerator and denominator by 20, we get:

80/100 = (80 ÷ 20) / (100 ÷ 20) = 4/5.

Step 4: Interpret the reduced fraction

The fraction 4/5 is the reduced form of 80%. It means that 80% is equivalent to four-fifths (4 parts out of 5).

Therefore, 80% as a fraction in reduced form is 4/5.

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Answer:

With 16.32 square yards, James gets 12 patches.

With 32.64 square yards, he gets 6 patches.

Step-by-step explanation:

James has 195.84 square yards of land he can use to grow vegetables.

He wants to divide this into patches of 16.32 square yards.

So, he will get [tex]\frac{195.84}{16.32}=12[/tex] equal patches.

If he wants to create garden patches with an area of 32.64 square yards, then he gets [tex]\frac{195.84}{32.64}=6[/tex] patches.

The change in the width of the beachfront from low to high tide is 102 feet, and the distance from the ocean to the homes at high tide is 248 feet.

a. To find the change in width from low to high tide, we multiply the rate of change (-17 feet/hour) by the time it takes for the ocean to move from low to high tide (6 hours).

So, -17 feet/hour * 6 hours = -102 feet. Therefore, the width of the beachfront decreases by 102 feet from low to high tide.

b. At high tide, the distance from the ocean to the homes is the sum of the original distance (350 feet) and the change in width during the tide change (-102 feet), resulting in 350 feet + (-102 feet) = 248 feet.

The function is increasing in:

                  x< -1

and it is decreasing in:

            x > -1

and nowhere constant.