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3 years ago

5d-2/3c if d =4 and c =3 evaluate fast fast

Mathematics

1 answer:

Mademuasel [1]3 years ago

4 0

Answer:

Step-by-step explanation:

Plug 4 in for d and 3 in for c

5d-2/3c

5(4)-2/3(3)

Multiply in the numerator and the denominator

20-2/9

Subtract in the numerator

18/9

Hope this helps! :)

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Given a right triangle with legs a, b and hypotenuse c, solve for b if a= 72 and c= 75

kotykmax [81]

Answer:

The answer is a. 21.

Step-by-step explanation:

We solve this question by using Pythagoras theorem to relate the sides of a right angled triangle.

Here,

Hypotenuse of the given triangle(c)=75

Two sides of right angled triangle are:

a=72

b=?

Then,

for a given right angled triangle abc,

Using Pythagoras theorem,

$c=\sqrt{a^{2} +b^{2} }$

Squaring on both sides,

$c^{2} =a^{2}+b^{2}$

or, $75^{2}=72^{2}+b^{2}$

or, $b^{2}=75^{2}-72^{2}$

or, $b^{2}=5625- 5184$

or, $b^{2} =441$

∴ $b=21$

So, the value of b is obtained to 21 by the use of Pythagoras theorem.

7 0

3 years ago

you currently have 24 credit hours and a 2.8 gpa you need a 3.0 gpa to get into the college. if you are taking a 16 credit hours

Juliette [100K]

Answer:

$\sum_{i=1}^n w_i *X_i = 2.8*24 = 67.2$

And for this case we want a gpa of 3.0 taking in count that in this semester he/ she is going to take 16 credits so then the new mean would be given by:

$\bar X_f = \frac{\sum_{i=1}^n w_i *X_i+w_f *X_f }{24+16} = 3.0$

And we can solve for $\sum_{i=1}^n w_f *X_f$ and solving we got:

$3.0 *(24+16) =\sum_{i=1}^n w_i *X_i +\sum_{i=1}^n w_f *X_f$

And from the previous result we got:

$3.0 *(24+16) =67.2 +\sum_{i=1}^n w_f *X_f$

And solving we got:

$\sum_{i=1}^n w_f *X_f =120 -67.2= 52.8$

And then we can find the mean with this formula:

$\bar X_2 = \frac{\sum_{i=1}^n w_f *X_f}{16}= \frac{52.8}{16}=16=3.3$

So then we need a 3.3 on this semester in order to get a cumulate gpa of 3.0

Step-by-step explanation:

For this case we know that the currently mean is 2.8 and is given by:

$\bar X = \frac{\sum_{i=1}^n w_i *X_i }{24} = 2.8$

Where $w_i$ represent the number of credits and $X_i$ the grade for each subject. From this case we can find the following sum:

$\sum_{i=1}^n w_i *X_i = 2.8*24 = 67.2$

And for this case we want a gpa of 3.0 taking in count that in this semester he/ she is going to take 16 credits so then the new mean would be given by:

$\bar X_f = \frac{\sum_{i=1}^n w_i *X_i+w_f *X_f }{24+16} = 3.0$

And we can solve for $\sum_{i=1}^n w_f *X_f$ and solving we got:

$3.0 *(24+16) =\sum_{i=1}^n w_i *X_i +\sum_{i=1}^n w_f *X_f$

And from the previous result we got:

$3.0 *(24+16) =67.2 +\sum_{i=1}^n w_f *X_f$

And solving we got:

$\sum_{i=1}^n w_f *X_f =120 -67.2= 52.8$

And then we can find the mean with this formula:

$\bar X_2 = \frac{\sum_{i=1}^n w_f *X_f}{16}= \frac{52.8}{16}=16=3.3$

So then we need a 3.3 on this semester in order to get a cumulate gpa of 3.0

6 0

3 years ago

Help find the answer for problem number 4

Olin [163]

Adult ticket (a) = $5
Child ticket (c) = $2

785 tickets = $3280

a + c = 785 tickets
5a + 2c = $3280

c = 215 child tickets
a = 570 adult tickets

570 + 215 = 785 tickets
5(570) + 2(215) = $3280

There were 215 child tickets sold on Saturday

3 0

4 years ago

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Sidana [21]

He has come from the hole in the roof because the text states that he did not come from a door, a window, or a chimney, and that is the only answer that was left.

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3 years ago

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Nataly [62]

Solution: Z= -9

Reasoning: -6 - 3 = -9

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3 years ago