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3 years ago

Guys I need help on this please

Mathematics

2 answers:

iogann1982 [59]3 years ago

5 0

Answer:

First option. The graphs domain is -∞<x<∞ and its range is -5 ≤ y < ∞

Step-by-step explanation:

I graphed this function

Domain: x ∈ R

Range: y ∈ [-5, + ∞

Minimum (1, -5)

Verticle intercept (0, -4)

rodikova [14]3 years ago

3 0

Answer:

The correct answer should be the first option.

Step-by-step explanation:

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Factor the expression x2 + 4x − 12.

mrs_skeptik [129]

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(x-2) (x+6)

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3 years ago

Simplify the radical expression. Use absolute value symbols if needed.

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3 years ago

35% of salt is in feta cheese. What percentage of salt is in that 350 g cheese?

xxMikexx [17]

The amount of salt is in the 350 g of feta cheese, when there is 35% of salt is in cheese, is 122.5 grams.

<h3>What is percentage of a number?</h3>

Percentage of a number is the part of the whole number which is expressed in the fraction of hundredth. It is represented with "%" symbol.

35% of salt is in feta cheese. The percentage of salt is in that 350 g cheese has to be found out.

Let suppose there is x grams of slat in feta cheese of 350 g. Thus, the amount of salt (35% of 350) is,

$x=\dfrac{35}{100}\times350\\x=122.5\rm\; gm$

Thus, the amount of salt is in the 350 g of feta cheese, when there is 35% of salt is in cheese, is 122.5 grams.

Learn more about the percentage here;

brainly.com/question/2085058

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5 0

2 years ago

Solve by completing the square.<br> X^2-2x-15=0 (help)

gladu [14]

Step-by-step explanation:

x² - 2x - 15 = 0

x1 = 5

x2 = -3

8 0

3 years ago

Read 2 more answers

Please help me with the below question.

VMariaS [17]

By letting

$y = \displaystyle \sum_{n=0}^\infty c_n x^{n+r}$

we get derivatives

$y' = \displaystyle \sum_{n=0}^\infty (n+r) c_n x^{n+r-1}$

$y'' = \displaystyle \sum_{n=0}^\infty (n+r) (n+r-1) c_n x^{n+r-2}$

a) Substitute these into the differential equation. After a lot of simplification, the equation reduces to

$5r(r-1) c_0 x^{r-1} + \displaystyle \sum_{n=1}^\infty \bigg( (n+r+1) c_n + (n + r + 1) (5n + 5r + 1) c_{n+1} \bigg) x^{n+r} = 0$

Examine the lowest degree term $\left(x^{r-1}\right)$ , which gives rise to the indicial equation,

$5r (r - 1) + r = 0 \implies 5r^2 - 4r = r (5r - 4) = 0$

with roots at r = 0 and r = 4/5.

b) The recurrence for the coefficients $c_k$ is

$(k+r+1) c_k + (k + r + 1) (5k + 5r + 1) c_{k+1} = 0 \implies c_{k+1} = -\dfrac{c_k}{5k+5r+1}$

so that with r = 4/5, the coefficients are governed by

$c_{k+1} = -\dfrac{c_k}{5k+5} \implies \boxed{g(k) = -\dfrac1{5k+5}}$

c) Starting with $c_0=1$ , we find

$c_1 = -\dfrac{c_0}5 = -\dfrac15$

$c_2 = -\dfrac{c_1}{10} = \dfrac1{50}$

so that the first three terms of the solution are

$\displaystyle \sum_{n=0}^2 c_n x^{n + 4/5} = \boxed{x^{4/5} - \dfrac15 x^{9/5} + \frac1{50} x^{13/5}}$

4 0

2 years ago

Guys I need help on this please ​

Guys I need help on this please