All categories

3 years ago

Find a pattern for the sum of 4 times a number and twice the sum of the same number and 3

Mathematics

1 answer:

Minchanka [31]3 years ago

8 0

The answer is 4x + 2(x+3)

You might be interested in

Solving equations -3x+7=28?

AURORKA [14]

Answer:

x = - 7

Step-by-step explanation:

- 3x + 7 = 28

- 3x + 7 - 7 = 28 - 7

- 3x = 21

- 3x/3 = 21/3

- x = 7

x = - 7

3 0

3 years ago

Read 2 more answers

Write an equation for an ellipse centered at the origin, which has foci at (0,\pm\sqrt{63})(0,± 63 )left parenthesis, 0, comma

steposvetlana [31]

Answer:

$\frac{x^{2} }{4312 } + \frac{y^{2} }{8281 }$

Step-by-step explanation:

Since the foci are at(0,±c) = (0,±63) and vertices (0,±a) = (0,±91), the major axis is the y- axis. So, we have the equation in the form (with center at the origin) $\frac{x^{2} }{b^{2} } + \frac{y^{2} }{a^{2} }$ .

We find the co-vertices b from b = ±√(a² - c²) where a = 91 and c = 63

b = ±√(a² - c²)

= ±√(91² - 63²)

= ±√(8281 - 3969)

= ±√4312

= ±14√22

So the equation is

$\frac{x^{2} }{(14\sqrt{22}) ^{2} } + \frac{y^{2} }{91^{2} } = \frac{x^{2} }{4312 } + \frac{y^{2} }{8281 }$

8 0

3 years ago

Help me I don't know what it is.

kotegsom [21]

Answer:

a is the correct answer

Step-by-step explanation:

-6-(-3)

-6+3

-3

8 0

3 years ago

Please help me with this question

san4es73 [151]

Step-by-step explanation:

Given: $f'(x) = x^2e^{2x^3}$ and $f(0) = 0$

We can solve for f(x) by writing

$\displaystyle f(x) = \int f'(x)dx=\int x^2e^{2x^3}dx$

Let $u = 2x^3$

$\:\:\:\:du=6x^2dx$

Then

$\displaystyle f(x) = \int x^2e^{2x^3}dx = \dfrac{1}{6}\int e^u du$

$\displaystyle \:\:\:\:\:\:\:=\frac{1}{6}e^{2x^3} + k$

We know that f(0) = 0 so we can find the value for k:

$f(0) = \frac{1}{6}(1) + k \Rightarrow k = -\frac{1}{6}$

Therefore,

$\displaystyle f(x) = \frac{1}{6} \left(e^{2x^3} - 1 \right)$

5 0

3 years ago

What is 6,010,499,999 rounded to the nearest billion?

aleksandr82 [10.1K]

Answer:

6000000000

Step-by-step explanation:

Given the following question:

$6,010,499,999$

<u>Find the billions place value:
</u>

$billions=6$

<u>Round to the nearest billion:</u>
$6,010,499,999$
$0$
$6000000000$

Your answer is "6000000000."

Hope this helps.

3 0

2 years ago