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3 years ago

PLESEEEEEEEE SOMEONE HELP ME I DIDNT STUDY FOR THIS QUIZ

Mathematics

2 answers:

Nataly [62]3 years ago

7 0

Answer:

Ans 1 :-6

And 2:- y=x²

And 3 :- ×= -2

4vir4ik [10]3 years ago

7 0

1. d
2. b
3. c

good luck

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100-point Question!!!

FrozenT [24]

Answer:

Two or more independent functions (say f(x) and g(x)) can be combined to generate a new function (say g(x)) using any of the following approach.

h(x) = f(x) + g(x)h(x)=f(x)+g(x) h(x) = f(x) - g(x)h(x)=f(x)−g(x)

h(x) = \frac{f(x)}{g(x)}h(x)=

g(x)

f(x)

h(x) = f(g(x))h(x)=f(g(x))

And many more.

The approach or formula to use depends on the question.

In this case, the combined function is:

f(x) = 75+ 10xf(x)=75+10x

The savings function is given as

s(x) = 85s(x)=85

The allowance function is given as:

a(x) = 10(x - 1)a(x)=10(x−1)

The new function that combined his savings and his allowances is calculated as:

f(x) = s(x) + a(x)f(x)=s(x)+a(x)

Substitute values for s(x) and a(x)

f(x) = 85 + 10(x - 1)f(x)=85+10(x−1)

Open bracket

f(x) = 85 + 10x - 10f(x)=85+10x−10

Collect like terms

mark as brainiest

f(x) = 85 - 10+ 10xf(x)=85−10+10x

f(x) = 75+ 10xf(x)=75+10x

4 0

2 years ago

A costume designer bought 1.4 kg of sequince. She used 250 grams to make a crown and 400 grams to make a cape. How many grams of

Salsk061 [2.6K]

Find the total amount in grams.

1 kg = 1000 g

1.4 * 1000 = 1400

So she bought a total of 1400 grams.

Add up the grams she used:

400 + 250 = 650

Subtract it from the total:

1400 - 650 = 750

So 750 grams of sequence were left.

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3 years ago

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demarco wrote two correct variable expressions to represent the phrase “the product of a number and 7”. which variable expressio

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The question is asking to choose among the following choices that states the expressions that could be represented in the said phrase and in my further understanding, I would say that the answer would be n.7, I hope you are satisfied with my answer and feel free to ask for more

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3 years ago

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Write out the form of the partial fraction decomposition of the function. Do not determine the numerical values of the coefficie

Dvinal [7]

For part (a), you have

$\dfrac x{x^2+x-6}=\dfrac x{(x+3)(x-2)}=\dfrac a{x+3}+\dfrac b{x-2}$
$x=a(x-2)+b(x+3)$

If $x=2$ , then $2=b(2-3)\implies b=-2$ .

If $x=-3$ , then $-3=a(-3-2)\implies a=\dfrac35$ .

So,

$\dfrac x{x^2+x-6}=\dfrac 3{5(x+3)}-\dfrac 2{x-2}$

For part (b), since the degrees of the numerator and denominator are the same, you first need to find the quotient and remainder upon division.

$\dfrac{x^2}{x^2+x+2}=\dfrac{x^2+x+2-x-2}{x^2+x+2}=1-\dfrac{x+2}{x^2+x+2}$

In the remainder term, the denominator $x^2+x+2$ can't be factorized into linear components with real coefficients, since the discriminant is negative $(1-4\times1\times2=-7)$ . However, you can still factorized over the complex numbers, so a partial fraction decomposition in terms of complexes does exist.

$x^2+x+2=0\implies x=-\dfrac12\pm\dfrac{\sqrt7}2i$
$\implies x^2+x+2=\left(x-\left(-\dfrac12+\dfrac{\sqrt7}2i\right)\right)\left(x-\left(-\dfrac12-\dfrac{\sqrt7}2i\right)\right)$
$\implies x^2+x+2=\left(x+\dfrac12-\dfrac{\sqrt7}2i\right)\left(x+\dfrac12+\dfrac{\sqrt7}2i\right)$

Then you have

$\dfrac{x+2}{x^2+x+2}=\dfrac a{x+\dfrac12-\dfrac{\sqrt7}2i}+\dfrac b{x+\dfrac12+\dfrac{\sqrt7}2i}$
$x+2=a\left(x+\dfrac12+\dfrac{\sqrt7}2i\right)+b\left(x+\dfrac12-\dfrac{\sqrt7}2i\right)$

When $x=-\dfrac12-\dfrac{\sqrt7}2i$ , you have

$-\dfrac12-\dfrac{\sqrt7}2i+2=b\left(-\dfrac12-\dfrac{\sqrt7}2i+\dfrac12-\dfrac{\sqrt7}2i\right)$
$\dfrac32-\dfrac{\sqrt7}2i=-\sqrt7ib$
$b=\dfrac12+\dfrac3{2\sqrt7}i=\dfrac1{14}(7+3\sqrt7i)$

When $x=-\dfrac12+\dfrac{\sqrt7}2i$ , you have

$-\dfrac12+\dfrac{\sqrt7}2i+2=a\left(-\dfrac12+\dfrac{\sqrt7}2i+\dfrac12+\dfrac{\sqrt7}2i\right)$
$\dfrac32+\dfrac{\sqrt7}2i=\sqrt7ia$
$a=\dfrac12-\dfrac3{2\sqrt7}i=\dfrac1{14}(7-3\sqrt7i)$

So, you could write

$\dfrac{x^2}{x^2+x+2}=1-\dfrac{x+2}{x^2+x+2}=1-\dfrac {7-3\sqrt7i}{14\left(x+\dfrac12-\dfrac{\sqrt7}2i\right)}-\dfrac {7+3\sqrt7i}{14\left(x+\dfrac12+\dfrac{\sqrt7}2i\right)}$

but that may or may not be considered acceptable by that webpage.

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