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pantera1 [17]
3 years ago
8

What is StartFraction 5 Over 8 EndFraction divided by One-fourth?

Mathematics
2 answers:
RideAnS [48]3 years ago
8 0

Answer:

1/3

Step-by-step explanation:

Bogdan [553]3 years ago
5 0

Answer:

2 1/2

Step-by-step explanation:

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Stella wants to buy colored pencils that cost $3.50. She has 1 dollar, 5 quarters, 4 dimes, and 7 pennies. Does she have enough
Andrews [41]

Answer:

no

Step-by-step explanation:

Lets she how much money she has

1.00   dollars

5 * .25 = 1.25 quarters

4 * .10 = .40 dimes

7 * .01 = .07 pennies

Add it all together

1 + 1.25+ .40 +.07 =2.72

3.50 - 2.72 =.78

She is 78 cents short

6 0
3 years ago
A study of a new medical procedure selected 27 candidates from a pool of 583 potential candidates.Each of the candidates had an
nikklg [1K]

The answer is random sampling

7 0
3 years ago
Keisha says that all functions are relations but not all relations are functions. kevin says that all relations are functions bu
vladimir1956 [14]

Keisha is correct, because as per the definition <u>A function is a special relationship where each input has a single output</u>.

A function is a special relation. In other words, a relation if and only if it has a specific characteristic where each input has a single output, then it is called a Function.

All functions are relations but not all relations are functions.


3 0
3 years ago
Need help for 18) 21) Solve each equation and check. Show all work please
Alexxandr [17]
18) (10)7d/10=35(10)
7d/7=350/7
d=50
21) 3/4 w=27
(4/3)3/4 w=27(4/3)
w=36
Hope this helps!
5 0
3 years ago
Simplify the problem: Sin theta Sec Theta
kherson [118]

Answer:

\tan \theta

Step-by-step explanation:

We have to simplify the following expression as given by

\sin \theta \times \sec \theta

= \frac{\sin \theta}{\cos \theta}

= \tan \theta ( Answer )

Because, we know that \sec \theta =\frac{1}{\cos \theta} and \tan \theta = \frac{\sin \theta}{\cos \theta}

If we consider \sin \theta = \frac{Perpendicular}{Hypotenuse} and \sec \theta = \frac{Hypotenuse}{Base},

then \sin \theta \times \sec \theta = \frac{Perpendicular}{Hypotenuse}\times \frac{Hypotenuse}{Base} = \frac{Perpendicular}{Base}= \tan \theta

3 0
3 years ago
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