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3 years ago

Ticket costs £14.50 add 7% increase in price

2 answers:

8 0

Solutions

We know that a ticket costs £14.50 and it is increased by 7% in price. To solve the problem lets use variables.

Let the Variable Y = 7% of 14.5

Equation: Y = P% × (X)

Solve for our equation Y

Y = P% × (X)

Y = 7% × 14.5

Now lets convert the percent to a decimal:

Percent = Out of 100

Percent = 7% / 100 = 0.07

Y = 0.07 × 14.5

Y = 1.015

SCORPION-xisa [38]3 years ago

3 0

If you would like to know the cost of the ticket after the 7% increase, you can calculate this using the following steps:

7% of £14.50 = 7% * 14.50 = 7/100 * 14.50 = £1.015

£14.50 + £1.015 = £15.515

The final cost of the ticket would be £15.515.

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THERE ARE 3 BAGS OF APPLES WEIGHING A TOTAL OF 22 1/2 POUNDS. TWO OF THE BAGS WEIGH 6 3/8 POUNDS AND 3 1/4 POUNDS. HOW MUCH DOES

Mkey [24]

I wasn't going to click on this one, but the all-caps enthralled me and hypnotized me.

first off, let's change all the mixed fractions to "improper" and proceed from there, keep in mind that if we subtract the two bags' weight from the total, what's leftover is the 3rd bag's weight.

$\bf \stackrel{mixed}{22\frac{1}{2}}\implies \cfrac{22\cdot 2+1}{2}\implies \stackrel{improper}{\cfrac{45}{2}}
\\\\\\
\stackrel{mixed}{6\frac{3}{8}}\implies \cfrac{6\cdot 8+3}{8}\implies \stackrel{improper}{\cfrac{51}{8}}
\\\\\\
\stackrel{mixed}{3\frac{1}{4}}\implies \cfrac{3\cdot 4+1}{4}\implies \stackrel{improper}{\cfrac{13}{4}}\\\\
-------------------------------\\\\$

$\bf \stackrel{\textit{sum of the two bags}}{\cfrac{51}{8}+\cfrac{13}{4}}\impliedby \textit{our LCD is 8}\implies \cfrac{(1)51+(2)13}{8}
\\\\\\
\cfrac{51+26}{8}\implies \cfrac{77}{8}\\\\
-------------------------------\\\\
\stackrel{total}{\cfrac{45}{2}}~-~\stackrel{two~bags}{\cfrac{77}{8}}\impliedby \textit{our LCD is again 8}\implies \cfrac{(4)45-(1)77}{8}
\\\\\\
\cfrac{180~~-~~77}{8}\implies \cfrac{103}{8}\implies \stackrel{third~bag}{12\frac{7}{8}}$

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3 years ago

What is the equation process to this question?

Andreyy89

The given equation is $h(t)=-16t^2+80t+4$ where h is the height, in feet, of a ball and t is the time, in seconds.

Part a: The height of the ball when t = 2 seconds:

The height of the ball above the ground 2 seconds after it is released can be determined by substituting t= 2 in the equation $h(t)=-16t^2+80t+4$ , we get;

$h(2)=-16(2)^2+80(2)+4$

Simplifying the terms, we get;

$h(2)=-64+160+4$

$h(2)=100$

Thus, the height of the ball after 2 seconds is 100 feet.

Part b: The height of the ball when t = 4 seconds:

The height of the ball above the ground 4 seconds after it is released can be determined by substituting t = 4 in the equation $h(t)=-16t^2+80t+4$ , we get;

$h(4)=-16(4)^2+80(4)+4$

Simplifying the terms, we get;

$h(4)=-256+320+4$

$h(4)=68$

Thus, the height of the ball after 4 seconds is 68 feet.

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3 years ago

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Answer:

3x^2

Step-by-step explanation:

The missing term is the product of x and 3x.

x * 3x = 3x^2

Answer: 3x^2

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3 years ago

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3 years ago

The school has a population of 500 students.

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Answer:

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2. 370

3. 470

4. 100

Step-by-step explanation:

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3 years ago