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3 years ago

Dave was invited to a birthday party that ended at 7:40. If it lasted 3 hours and 45 minutes what time did the party start?

Mathematics

2 answers:

MrRissso [65]3 years ago

7 0

The party would have started at 3:55.

irinina [24]3 years ago

3 0

The party startef at 3.55

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a recipe requires 1/2 cup of flour for every batch of cookies how many full batches of cookies can be made with 5 1/2 cups of fl

Fantom [35]

Hello There!

You want to know how many cookies can be made using 5 1/2 cups of flour.

$Solving$

1 x 2 2
-------------- = -----
2 x 11 22

2/22 simplified is 1/11

So, You can make 11 cookies with 5 1/2 cups of flour.

Hope this helped!

4 0

3 years ago

Read 2 more answers

Find m∠TUL.

pogonyaev

Step-by-step explanation:

Actual graph for this problem is attached below

m∠TUV = 167°

m∠TUL = (x + 11)°

m∠LUV = (11x)°

m∠TUV= m∠TUL+ m∠LUV

now plug in the angles for each

m∠TUV= m∠TUL+ m∠LUV

$167= x+11 +11x$

solve the equation for x

$167= x+11 +11x\\167=11 +12x$

Subtract 11 from both sides

$156=12x$

divide both sides by 12

x=13

m∠TUL = (x + 11)°

m∠TUL = (13+ 11)° = (24)°

answer:

24°

4 0

3 years ago

i think of a number. if i divide the sum of 5 and the number by 3, the result is 2 less than one quarter of the number. find the

Marrrta [24]

Answer:

x = 4.75

Step-by-step explanation:

<u>(5 +X</u><u> )/3</u> = 2 - <u> X/4</u>

<h3> <u>(5 +X</u><u> )/3</u> = 8 - X </h3><h3>5+ X = 3 (8 - X) </h3><h3> 5+ X = 24 - 3X transposing terms</h3><h3>X+ 3X= 24 -5</h3><h3> 4X = 19</h3><h3> X = 19/4 </h3><h3>X = 4.75</h3>

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2 years ago

How to solve (2.31*10^-6)+(5.87*10^-4)

S_A_V [24]

We want to calculate the following

$2.31\cdot10^{-6}+5.87\cdot10^{-4}$

Using the properties of exponents, we have that

$10^{-6}=10^{-4}\cdot10^{-2}$

So we have

$2.31\cdot10^{-6}+5.87\cdot10^{-4}=2.31\cdot10^{-2}\cdot10^{-4}+5.87\cdot10^{-4}$

So, if factor 10^-4 on the right side, we have

$10^{-4}(2.31\cdot10^{-2}+5.87)$

Note that

$2.31\cdot10^{-2}=0.0231$

Then,

$5.87+2.31\cdot10^{-2}=5.87+0.0231=5.8931^{}$

So we have that

$2.31\cdot10^{-6}+5.87\cdot10^{-4}=5.8931\cdot10^{-4}$

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1 year ago

How do extraneous solutions arise from radical equations?

Blababa [14]

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3 years ago

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