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3 years ago

Write an inequality to determine the number, t, of tickets the commitee

Mathematics

1 answer:

djyliett [7]3 years ago

5 0

Answer:

If it is the dance committee question the answer is 400<72+4t. Hope this helps!

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Absolute zero is -459.67 Fahrenheit. How many degrees above absolute zero is the freezing point of water?

grin007 [14]

Answer:

491.67 above

Step-by-step explanation:

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3 years ago

Write a definition for a unite rate

aalyn [17]

A unit rate is a special type of ratio (also called a single-unit rate). It will compare 1 unit of some quantity to a different number of units of a different quantity. Any ratio that is not a unit rate can be turned into one.

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3 years ago

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If y varies inversely as x2 and y=6 when x=30, what is y when x=6

ladessa [460]

Answer:

y = 150

Step-by-step explanation:

y varies inversely as x²

yx² = k

----------------

y = 6 when x = 30

6(30²) = k

6(900) = 5400

--------------------------

what is y when x = 6

y(6²) = 5400

y(36) = 5400

y = 5400/36

y = 150

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3 years ago

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Mr Lim read 3 books in 2 hours. He took the same amount of time to read the first two books. He took 15 minutes longer to read t

Naya [18.7K]

He took 75 minutes to read the third book.

5 0

3 years ago

Please help!!!! Will mark brainliest!! 18 points.

andreyandreev [35.5K]

Method 1:

You can use the formula of an area of equilateral triangle:

$A=\dfrac{a^2\sqrt3}{4}$

a - length of side

We have a = 10 yd. Substitute:

$A=\dfrac{10^2\sqrt3}{4}=\dfrac{100\sqrt3}{4}=25\sqrt3\ yd^2\\\\\sqrt3\approx1.7\\\\A=25(1.7)=\boxed{42.5\ yd^2}$

Method 2:

Use the Pythagorean theorem (look at he picture).

$5^2+h^2=10^2$

$25+h^2=100$ <em>subtract 25 from both sides</em>

$h^2=75\to h=\sqrt{75}\\\\h=\sqrt{25\cdot3}\\\\h=\sqrt{25}\cdot\sqrt3\\\\h=5\sqrt3\ yd$

The formula of an area of a trinagle:

$A=\dfrac{1}{2}bh$

b - base

h - height

We have b = 10 yd and h = 5√3 yd. Substitute:

$A=\dfrac{1}{2}(10)(5\sqrt3)=(5)(5\sqrt3)=25\sqrt3\ yd\\\\\sqrt3\approx1.7\\\\A=25(1.7)=\boxed{42.5\ yd^2}$

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3 years ago

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