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3 years ago

Pease help me ASAP....................

Mathematics

2 answers:

Rasek [7]3 years ago

5 0

A.
$-5 +5(x+4)\\ -5 +5x+20\\ 5x+15$

b.
$3m+2(5m-7)\\ 3m+10m-14\\ 13m-14$

c.
$4(6t+9)-10t\\ 24t+36-10t\\ 14t+36$

d.
$8k-6(3-2k)\\ 8k-18+12k\\ 20k-18$

AfilCa [17]3 years ago

3 0

First one, x+4

second one, 13m-7

third one, 14x+36

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An equation is shown below: 8(2x – 14) + 13 = 4x – 27 Part A: Write the steps you will use to solve the equation, and explain ea

blsea [12.9K]

1) Distribute 8(2x-14) + 13 = 4x - 27 then simplify like terms to get: 16x - 99 = 4x - 27.
2) Subtract 4x on both sides and at the same time add 99 to both sides to get: 12x=72.
3) Divide both sides by 12 to get x=6.
PART B: 6 makes the equation true because you just solved for it.

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3 years ago

In the Venn diagram, consider U = {whole numbers 1-100}

Dmitry [639]

The question is what numbers satisfy A ∩ C.

The symbol ∩ means intersection, .i.e. you need to find the numbers that belong to both sets A and C. Those numbers might belong to the set C or not, because that is not a restriction.

Then lets find the numbers that belong to both sets, A and C.

Set A: perfect squares from A to 100:

1^2 = 1
2^2 = 4
3^2 = 9
4^2 = 16
5^2 = 25
6^2 = 36
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=> A = {1, 4, 9, 16, 25, 36, 49, 64, 81, 100}

Set C: perfect fourths

1^4 = 1
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3^4 = 81

C = {1, 16, 81?

As you see, all the perfect fourths are perfect squares, so the intersection of A and C is completed included in A. this is:

A ∩ C = C or A ∩ C = 1, 16, 81

On the other hand, the perfect cubes are:

1^3 = 1
2^3 = 8
3^3 = 27
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B = {1, 8, 27, 81}

That means that the numbers 1 and 81 belong to the three sets, A, B, and C.

In the drawing you must place the number 16 inside the region that represents the intersection of A and C only, and the numbers 1 and 81 inside the intersection of the three sets A, B and C.

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3 years ago

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Find dy/dx given that y = sin x / 1 + cos x

kobusy [5.1K]

Answer:

$\frac{1}{1 + \cos(x) }$

Step-by-step explanation:

$y = \frac{ \sin(x) }{1 + \cos(x) }$

<u>differentiating numerator wrt x :-</u>

(sinx)' = cos x

<u>differentiating denominator wrt x :- </u>

(1 + cos x)' = (cosx)' = - sinx

Let's say the denominator was "v" and the numerator was "u"

$(\frac{u}{v} )' = \frac{v. \: (u)' - u.(v)' }{ {v}^{2} }$

here,

since u is the numerator u= sinx and u = cos x
v(denominator) = 1 + cos x; v' = - sinx

$= \frac{((1 + \cos \: x) \cos \: x )- (\sin \: x. ( - \sin \: x) ) }{( {1 + \cos(x)) }^{2} }$

$= \frac{ \cos(x) + \cos {}^{2} (x) + \sin {}^{2} (x) }{(1 + \cos \: x) {}^{2} }$

since cos²x + sin²x = 1

$= \frac{ \cos \: x + 1}{(1 + \cos \: x) {}^{2} }$

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$= \frac{1}{1 + \cos(x) }$

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3 years ago

A list of 5 steps Cecile used to factor 16 x squared minus 9. Analyze Cecile’s work. Is it correct? No, adding in 0x keeps an eq

suter [353]

The Cecile factored the polynomial correctly by factor 16 x squared minus 9 using the X-method of factorization.

<h3>What is factor of polynomial?</h3>

Factor of a polynomial is the terms in linear form, which are when multiplied together, give the original polynomial equation as result.

The given polynomial expression is,

$f(x)=16x^6-9$

Now, Cecile add 0x as,

$f(x)=16x^6+0x-9$

Now, here the product of 16 and 9 is 144. Therefore, the factors must have the product of 144 and sum of 0. Thus, by using the split the factor method as,

$f(x)=16x^6+12x-12x-9\\f(x)=4x^3(4x^3+3)+(-3)(4x^3+3)\\f(x)=(4x^3+3)(4x^3-3)$

Thus, the Cecile factored the polynomial correctly by factor 16 x squared minus 9 using the X-method of factorization.

Learn more about factor of polynomial here;

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2 years ago

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I need help? can anyone help

erica [24]

I think it’s 5 but don’t get mad if it’s wrong

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3 years ago