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2 years ago

Evaluate 8a-1+0.5b8a−1+0.5b8, a, minus, 1, plus, 0, point, 5, b when a=\dfrac14a=

Mathematics

2 answers:

professor190 [17]2 years ago

7 0

Answer: $6$

Step-by-step explanation:

Given the following expression:

$8a-1+0.5b$

You need to substitute the given values of "a" and "b" into the expression. Notice that these values are:

$a=\frac{1}{4}\\\\b=10$

Then;

$8a-1+0.5b=8(\frac{1}{4})-1+0.5(10)$

Now you must solve the multiplications:

$=\frac{8}{4}-1+5=2-1+5$

The final step is to add the numbers. Therefore, you get the following answer:

$=6$

mart [117]2 years ago

5 0

Answer: 6

Step-by-step explanation:

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Aaron borrows $150 from his friend Austin. He promises to pay back the money in 4 monthly installments. Each month he wants to p

Maurinko [17]

Answer C. $80

80-40-20-10=150

5 0

2 years ago

The length, l cm, of a simple pendulum is directly proportional to the square of its period (time taken to complete one oscillat

Greeley [361]

Answer:

1) $L \propto T^2$

Using the condition given:

$2.205 m = K (3)^2$

$K = 0.245 \approx \frac{g}{4\pi^2}$

So then if we want to create an equation we need to do this:

$L = K T^2$

With K a constant. For this case the period of a pendulumn is given by this general expression:

$T = 2\pi \sqrt{\frac{L}{g}}$

Where L is the length in m and g the gravity $g = 9.8 \frac{m}{s^2}$ .

2) $T = 2\pi \sqrt{\frac{L}{g}}$

If we square both sides of the equation we got:

$T^2 = 4 \pi^2 \frac{L}{g}$

And solving for L we got:

$L = \frac{g T^2}{4 \pi^2}$

Replacing we got:

$L =\frac{9.8 \frac{m}{s^2} (5s)^2}{4 \pi^2} = 6.206m$

3) $T = 2\pi \sqrt{\frac{0.98m}{9.8\frac{m}{s^2}}}= 1.987 s$

Step-by-step explanation:

Part 1

For this case we know the following info: The length, l cm, of a simple pendulum is directly proportional to the square of its period (time taken to complete one oscillation), T seconds.

$L \propto T^2$

Using the condition given:

$2.205 m = K (3)^2$

$K = 0.245 \approx \frac{g}{4\pi^2}$

So then if we want to create an equation we need to do this:

$L = K T^2$

With K a constant. For this case the period of a pendulumn is given by this general expression:

$T = 2\pi \sqrt{\frac{L}{g}}$

Where L is the length in m and g the gravity $g = 9.8 \frac{m}{s^2}$ .

Part 2

For this case using the function in part a we got:

$T = 2\pi \sqrt{\frac{L}{g}}$

If we square both sides of the equation we got:

$T^2 = 4 \pi^2 \frac{L}{g}$

And solving for L we got:

$L = \frac{g T^2}{4 \pi^2}$

Replacing we got:

$L =\frac{9.8 \frac{m}{s^2} (5s)^2}{4 \pi^2} = 6.206m$

Part 3

For this case using the function in part a we got:

$T = 2\pi \sqrt{\frac{L}{g}}$

Replacing we got:

$T = 2\pi \sqrt{\frac{0.98m}{9.8\frac{m}{s^2}}}= 1.987 s$

8 0

2 years ago

You have heard that customers will try a new caterer if the price is 20% below their existing caterer. Your competitor prices a

Evgesh-ka [11]

Answer:

b) $640

Step-by-step explanation:

Competitor's price = $800

<u>Price, 20% below of competitor's:</u>

$800 - 20% =
$800*(100-20)/100 =
$800*0.8 =
$640

Correct answer choice is b) $640

3 0

2 years ago

. A density curve consists of the line segment connecting the points (0,1) and (0.5,1) and the segment connecting (0.5, 1) to th

Marat540 [252]

hey thanks for the point's i really needed them.. Cheers!

many

8 0

2 years ago

Diane reed had office supply sales of $3,285 this month. this is a 78% of last months supply sales. what was the amount of last

Hoochie [10]

To solve this problem, we can set up an equation, letting the unknown value of last month's supply sales be represented by the variable x.

First, we must convert 78% to its decimal equivalent, by dividing 78/100. This is because percentages are parts of a whole 100 percent.

78/100 = 0.78

Now, we are ready to set up our equation, using our knowledge that the word "of" refers to multiplication in mathematics.

3,285 = 0.78x

To solve our equation, we must divide both sides of the equation by 0.78 to get the variable x alone on the right side of the equation.

x = 4211.53846154

This means that last month's supply sales were about $4,211.54 (we round the last digit up because 8 is greater than 5).

Hope this helps!

8 0

2 years ago