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2 years ago

E. Makalo can spend no more than $60

Mathematics

1 answer:

Liono4ka [1.6K]2 years ago

6 0

Answer:

Step-by-step explanation:

because add 5 to 45 and then you 10 to 50 and that equals 60 so 15

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Write the first 3 terms of the sequence an=n+2

frutty [35]

Answer:

Step-by-step explanation:

$a_n = n + 2$

$a_1 = 1 + 2 = 3\\a_2 = 2 + 2 = 4\\a_3 = 3 + 2 = 5$

8 0

3 years ago

23 over.30 ÷20 over 24

alexgriva [62]

Remark

Don't try and do this all at once. Break it down, otherwise you'll have layers and brackets all over the place.

Step One

Find 23/0.3

X = 23/0.3 = 76.7

Step Two

Now Divide by 20

x1 = 76.7 / 20

x1 = 3.83

Step Three

Take this result and put it over 24

x2 = x1/24

x2 = 3.83 / 24

x2 = 0.1597 <<<< Answer

6 0

2 years ago

Read 2 more answers

Integration of ∫(cos3x+3sinx)dx

Murljashka [212]

Answer:

$\boxed{\pink{\tt I = \dfrac{1}{3}sin(3x) - 3cos(x) + C}}$

Step-by-step explanation:

We need to integrate the given expression. Let I be the answer .

$\implies\displaystyle\sf I = \int (cos(3x) + 3sin(x) )dx \\\\\implies\displaystyle I = \int cos(3x) + \int sin(x)\ dx$

Let u = 3x , then du = 3dx . Henceforth 1/3 du = dx .
Now , Rewrite using du and u .

$\implies\displaystyle\sf I = \int cos\ u \dfrac{1}{3}du + \int 3sin \ x \ dx \\\\\implies\displaystyle \sf I = \int \dfrac{cos\ u}{3} du + \int 3sin\ x \ dx \\\\\implies\displaystyle\sf I = \dfrac{1}{3}\int \dfrac{cos(u)}{3} + \int 3sin(x) dx \\\\\implies\displaystyle\sf I = \dfrac{1}{3} sin(u) + C +\int 3sin(x) dx \\\\\implies\displaystyle \sf I = \dfrac{1}{3}sin(u) + C + 3\int sin(x) \ dx \\\\\implies\displaystyle\sf I = \dfrac{1}{3}sin(u) + C + 3(-cos(x)+C) \\\\\implies \underset{\blue{\sf Required\ Answer }}{\underbrace{\boxed{\boxed{\displaystyle\red{\sf I = \dfrac{1}{3}sin(3x) - 3cos(x) + C }}}}}$

6 0

3 years ago

What are three sets of fractions that can be multiplied to get 15/56?

vredina [299]

Much more than just 3.

3 0

3 years ago

Please helpp middle school math<br><br>suppose 4<a<7. find all the possible values of 15-2a

bixtya [17]

9514 1404 393

Answer:

1 < 15 -2a < 7

Step-by-step explanation:

There are a couple of ways you can do this.

1) Put the minimum and maximum values of a into the expression to see what its corresponding values are:

15-2a for a=4:

15-2(4) = 7

15-2a for a=7:

15-2(7) = 1

Then ...

1 < 15-2a < 7

2) Solve for a in terms of the value of 15-2a, then impose the limits on a.

x = 15 -2a

2a = 15 -x

a = (15 -x)/2

Now, impose the given limits:

4 < (15 -x)/2 < 7

8 < 15 -x < 14 . . . multiply by 2

-7 < -x < -1 . . . . . . subtract 15

7 > x > 1 . . . . . . . . multiply by -1

1 < 15-2a < 7 . . . . . use x=15-2a

_____

The vertical extent of the attached graph is the range of possible values of 15-2a. It goes from 1 to 7.

8 0

3 years ago