Solve for x

6x + 11= - (6x + 5)

Answer:

product of power

Step-by-step explanation:

I think this will help you

Step-by-step explanation:

\underline{\textsf{Given:}}

Given:

\mathsf{Polynomial\;is\;x^3+7x^2+7x-15}Polynomialisx

3

+7x

2

+7x−15

\underline{\textsf{To find:}}

To find:

\mathsf{Factors\;of\;x^3+7x^2+7x-15}Factorsofx

3

+7x

2

+7x−15

\underline{\textsf{Solution:}}

Solution:

\textsf{Factor theorem:}Factor theorem:

\boxed{\mathsf{(x-a)\;is\;a\;factor\;P(x)\;\iff\;P(a)=0}}

(x−a)isafactorP(x)⟺P(a)=0

\mathsf{Let\;P(x)=x^3+7x^2+7x-15}LetP(x)=x

3

+7x

2

+7x−15

\mathsf{Sum\;of\;the\;coefficients=1+7+7-15=0}Sumofthecoefficients=1+7+7−15=0

\therefore\mathsf{(x-1)\;is\;a\;factor\;of\;P(x)}∴(x−1)isafactorofP(x)

\mathsf{When\;x=-3}Whenx=−3

\mathsf{P(-3)=(-3)^3+7(-3)^2+7(-3)-15}P(−3)=(−3)

3

+7(−3)

2

+7(−3)−15

\mathsf{P(-3)=-27+63-21-15}P(−3)=−27+63−21−15

\mathsf{P(-3)=63-63}P(−3)=63−63

\mathsf{P(-3)=0}P(−3)=0

\therefore\mathsf{(x+3)\;is\;a\;factor}∴(x+3)isafactor

\mathsf{When\;x=-5}Whenx=−5

\mathsf{P(-5)=(-5)^3+7(-5)^2+7(-5)-15}P(−5)=(−5)

3

+7(−5)

2

+7(−5)−15

\mathsf{P(-5)=-125+175-35-15}P(−5)=−125+175−35−15

\mathsf{P(-5)=175-175}P(−5)=175−175

\mathsf{P(-5)=0}P(−5)=0

\therefore\mathsf{(x+5)\;is\;a\;factor}∴(x+5)isafactor

\underline{\textsf{Answer:}}

Answer:

\mathsf{x^3+7x^2+7x-15=(x-1)(x+3)(x+5)}x

3

+7x

2

+7x−15=(x−1)(x+3)(x+5)

\underline{\textsf{Find more:}}

Find more:

Answer:

A. x-2y-3z = 1/2

Step-by-step explanation:

yess.. if you take solve every equation then you will get same equation from all the equations

Answer:

"D"

Step-by-step explanation:

just add the two functions

5x^2 - 8x^2 = -3x^2 etc

Answer:

set notation _ A set is denoted or represented by a capital letter and enclosed in a curly bracket For example {A,B,P,Q}.

1. I think in Brooklyn, Manhattan, and the other different cities in New Jersey, the rents are overpriced.

2. Some factors are if it looks nice, there are no bugs or rodents, and if it's not a rip off.

**I can't answer the second part bcuz i don't see the tables.  

Answer:

it is summer how do y'all still have school,like don't they give y'all a break

Answer:

12

Step-by-step explanation:

(18+3)=21

(9-5)=4

4x21=84

84/2=42

42-30=12

Answer:

12

Step-by-step explanation:

((18+3)x(9-5))/2-30

=((21)x(4))/2-30

=(84)/2-30

=42-30

=12

Answer:

1. $11747.083

2. $29,253

3. $24222.04

4. $1542.23

Step-by-step explanation:

Message for explanation if you want.

Answer:

Because it's a negative.

Step-by-step explanation:

The value of a positive number is still a positive number.

Answer:

f = 10

g = 2 sqrt(3)

h = 20

Step-by-step explanation:

The short leg is opposite the smaller angle so it is f

The longer leg is opposite the larger angle so it is g

The hypotenuse is opposite the right angle so it is 20

We know f = x

g = x sqrt(3)

h = 2x = 20

2x = 20  so x = 10

f = 10

g = 2 sqrt(3)

h = 20

Answer:

x + 11 is an even number.

Step-by-step explanation:

Even numbers can only be obtained from the sum of two odd numbers or two even numbers. Since we know that x + 7 is even, x + 11 must be even as well.

Let the number be x

ATQ

[tex]\\ \sf\longmapsto 5x+6=2x-3[/tex]

[tex]\\ \sf\longmapsto 5x-2x=-3-6[/tex]

[tex]\\ \sf\longmapsto 3x=-9[/tex]

[tex]\\ \sf\longmapsto x=\dfrac{-9}{3}[/tex]

[tex]\\ \sf\longmapsto x=-3[/tex]

Answer:

The bias is they only picked men to test the new medication on to see if it was ready for the general public to use.

Step-by-step explanation:

Answer: length of DE = 18

Step-by-step explanation:

Since the angles of these two triangles are the same, we can assume that they are similar triangles. To solve for a missing side of a triangle that is similar, you can set up a proportion:

[tex]\frac{27}{15}=\frac{x}{10}[/tex]

To solve for x (the missing length), you would do:

27 × 10 ÷ 15 = 18

You can set up the proportion in any way between the two triangles as long as the two sides correspond to each other. For example, you could have also used the following proportion to solve for x:

[tex]\frac{x}{27} =\frac{10}{15}[/tex]

When you solve for x using this proportion, you would get the same value for x:

10 × 27 ÷ 15 = 18

12/16 = (3x-12)/(2x+8)

16(3x-12)=12(2x+8)

48x-192=24x+96

48x-24x=192+96

24x=288

X=288/24

X=12

3x-12

=(12x3)-12

=36-12

=24

C is the answer

Hope this helps!

Answer:

48

Step-by-step explanation:

2x+8=3x-12(ABCP ~FDQE)

2x-3x= -8-12

-x= -20

x=20

now,

CP=3x-12

3*20-12

48

Answer:

.4761

I have a feeling that you are not quite getting the concept of these questions

a "Z" table "maps" a value (x-mu/sigma) to a number on a z-table chart

Your calculator has the function but the chart is simple for this questions

the numbers IN the chart represent "the percent UNDER THE NORMAL CURVE"

a z =0 is the MIDDLE so it is 50%

your questions is negative so it is to the left of the "middle"

all you have to do is find the z number on the chart and that is the answer to this question

Step-by-step explanation:

Answer: 19.23 meters

Step-by-step explanation:

1/1.3 = x/25

(1) (25) / 1.3

= 19.23 meters

Answer:

No, because the ratio of pay to hours is not the same for each pair of value.

Answer:

6 percent increase

Step-by-step explanation:

2650 divided by 2500 gives you 1.06 which means if 2500 is 100 percent 2650 would be 106 percent so the answer is a 6 percent increase

Answer:

6%

Step-by-step explanation:

another way of doing this type of calculation is...

NEW-OLD  =  2650-2500  = 150     = .06 = 6%

   OLD                2500            2500

Answer:

y-axis.

Step-by-step explanation:

The x coordinate of both points is 4, so it is parallel to they-axis.

Answer:

We have 60% salt and 85% salt solutions and we need 140 ounces of a solution that is 80% salt.

We set up 2 equations:

A) x + y = 140

B) .60x + .85y = .80 * 140

Multiply equation A) by -.60

A) -.60x -.60y = -84  then we add this to B)

B) .60x + .85y = 112

.25y = 28

y = 112 ounces of 85% salt

x = 28 ounces of 60% salt.

Double Check

112 * .85 = 95.2 AND 28 * .60 = 16.80

95.2 + 16.80 = 112

and 112/140 = 80 per cent!!

Source: http://www.1728.org/mixture.htm

Step-by-step explanation:

Answer:

106 5/6

Step-by-step explanation:

55 < 66 < 74 < 89 < 106

the fractions are irrelevant because the whole numbers are all different

The greatest of the numbers will be 106 5/6 since 106 is greater than the rest of the integers present in the mixed fractions.

The greatest value is the highest value in a given set of number.

Given the following numbers

89 3/4

66 1/2

55 2/3

74 1/4

106 5/6

The greatest of the numbers will be 106 5/6 since 106 is greater than the rest of the integers present in the mixed fractions.

Learn more on mixed fractions here: https://brainly.com/question/414559

#SPJ6

Answer:

Hypotenuse=10 miles.

Short leg=6 miles.

Step-by-step explanation:

Answer:

Step-by-step explanation:

The reason you haven't gotten an answer to this is because in its current formatting, there is no answer. Here's what I mean:

The first equation is going to be concerning the NUMBER of memberships sold, where m is male and f is female. The total number of memberships was 10:

m + f = 10

Now for the money equation. The total amount of money made from that number of memberships was 3000, where male memberships cost $300 and so do the female memberships, giving us an equation of

300m + 300f = 3000

Go back to the first equation and solve for either m or f. I solved for m in terms of f:

m = 10 - f and sub that into the second equation for m to get:

300(10 - f) + 300f = 3000 and

3000 - 300f + 300f = 3000. Here is where you find the problem. The -300f and the 300f cancel each other out, leaving you with the fact that

3000 = 3000 which it does, but it doesn't give us any viable answer.

I would have to say that since we can't do math on this, that the most credible answer you'll find is that the same number of male and female memberships were sold because 5 male memberships cost $1500 (5 memberships at $300 a piece is $1500) and 5 female memberships also cost $1500.

1500 + 1500 = 3000

Answer:

CI ≈ (173.8 < μ < 196.2)

Step-by-step explanation:

We are told that laboratory tested twelve chicken eggs. Thus;

n = 12

Mean; x¯ = 185 mg

S.D; s = 17.6 mg

DF = n - 1 = 12 - 1 = 11

We have a 95% confidence level. Thus; α = 0.05

Since n < 30, we will use t-sample test.

Thus, from t-table attached at 95% Confidence level and DF = 11, we have;

t = 2.201

Thus,formula for Confidence interval is;

CI = (x¯ - t(s/√n)) < μ < (x¯ + t(s/√n))

CI = (185 - 2.201(17.6/√12)) < μ < (185 + 2.201(17.6/√12))

CI = (185 - 11.1825) < μ < (185 + 11.1825)

CI = (173.8175 < μ < 196.1825)

CI ≈ (173.8 < μ < 196.2)

Given:

The given expression is:

[tex]6x^2y-3xy-24xy^2+12y^2[/tex]

To find:

Part A: The expression by factoring out the greatest common factor.

Part B: Factor the entire expression completely.

Solution:

Part A:

We have,

[tex]6x^2y-3xy-24xy^2+12y^2[/tex]

Taking out the highest common factor 3y, we get

[tex]=3y(2x^2-x-8xy+4y)[/tex]

Therefore, the required expression is [tex]3y(2x^2-x-8xy+4y)[/tex].

Part B:

From part A, we have,

[tex]3y(2x^2-x-8xy+4y)[/tex]

By grouping method, we get

[tex]=3y(x(2x-1)-4y(2x-1))[/tex]

[tex]=3y(x-4y)(2x-1)[/tex]

Therefore, the required factored form of the given expression is [tex]3y(x-4y)(2x-1)[/tex].

Answer:

So, remember that:

cos(x) > 0   for  -pi/2 < x < pi/2

cos(x) < 0 for  pi/2 < x < (3/2)*pi

and

sin(x) > 0 for 0 < x < pi

sin(x) < 0 for -pi < x <0  or  pi < x < 2pi

Also, we have the periodicty of the sine and cocine equations, such that:

sin(x) = sin(x + 2pi)

cos(x) = cos(x + 2pi)

Now let's solve the problem:

[tex]sin(\frac{13*\pi}{36} )[/tex]

here we have:

x = (13/36)π

This is larger than zero and smaller than π:

0 < (13/36)π < π

then:

[tex]sin(\frac{13*\pi}{36} )[/tex]

Is positive.

The next one is:

[tex]cos(\frac{7*\pi}{12} )[/tex]

Here we have x = (7/12)*pi

notice that:

7/12 > 1/2

Then:

(7/12)*π > (1/2)*π

Then:

[tex]cos(\frac{7*\pi}{12} )[/tex]

is negative.

next one:

[tex]sin(\frac{47*\pi}{36} )[/tex]

here:

x = (47/36)*π

here we have (47/36) > 1

then:

(47/36)*π > π

then:

[tex]sin(\frac{47*\pi}{36} )[/tex]

is negative.

the next one is:

[tex]cos(\frac{17*\pi}{10} )[/tex]

Here we have x = (17/10)*π

if we subtract 2*π (because of the periodicity) we get:

(17/10)*π - 2*π

(17/10)*π - (20/10)*π

(-3/10)*π

this is in the range where the cosine function is positive, thus:

[tex]cos(\frac{17*\pi}{10} )[/tex]

is positive.

the next one is:

[tex]tan(\frac{41*\pi}{36} ) = \frac{sin(\frac{41*\pi}{36} )}{cos(\frac{41*\pi}{36} )}[/tex]

here we have:

x = (41/36)*π

Notice that both functions, sine and cosine are negatives for that value, then we have the quotient of two negative values, so:

[tex]tan(\frac{41*\pi}{36} ) = \frac{sin(\frac{41*\pi}{36} )}{cos(\frac{41*\pi}{36} )}[/tex]

is positive.

The final one is:

[tex]tan(\frac{5*\pi}{9} ) = \frac{sin(\frac{5*\pi}{9} )}{cos(\frac{5*\pi}{9} )}[/tex]

Here:

x = (5/9)*π

The sin function is positive with this x value, while the cosine function is negative, thus:

[tex]tan(\frac{5*\pi}{9} ) = \frac{sin(\frac{5*\pi}{9} )}{cos(\frac{5*\pi}{9} )}[/tex]

Is negative.

Hi there!

To find the Trigonometric Equation, we have to isolate sin, cos, tan, etc. We are also given the interval [0,2π).

First Question

What we have to do is to isolate cos first.

[tex] \displaystyle \large{ cos \theta = - \frac{1}{2} }[/tex]

Then find the reference angle. As we know cos(π/3) equals 1/2. Therefore π/3 is our reference angle.

Since we know that cos is negative in Q2 and Q3. We will be using π + (ref. angle) for Q3. and π - (ref. angle) for Q2.

Find Q2

[tex] \displaystyle \large{ \pi - \frac{ \pi}{3} = \frac{3 \pi}{3} - \frac{ \pi}{3} } \\ \displaystyle \large \boxed{ \frac{2 \pi}{3} }[/tex]

Find Q3

[tex] \displaystyle \large{ \pi + \frac{ \pi}{3} = \frac{3 \pi}{3} + \frac{ \pi}{3} } \\ \displaystyle \large \boxed{ \frac{4 \pi}{3} }[/tex]

Both values are apart of the interval. Hence,

[tex] \displaystyle \large \boxed{ \theta = \frac{2 \pi}{3} , \frac{4 \pi}{3} }[/tex]

Second Question

Isolate sin(4 theta).

[tex] \displaystyle \large{sin 4 \theta = - \frac{1}{ \sqrt{2} } }[/tex]

Rationalize the denominator.

[tex] \displaystyle \large{sin4 \theta = - \frac{ \sqrt{2} }{2} }[/tex]

The problem here is 4 beside theta. What we are going to do is to expand the interval.

[tex] \displaystyle \large{0 \leqslant \theta < 2 \pi}[/tex]

Multiply whole by 4.

[tex] \displaystyle \large{0 \times 4 \leqslant \theta \times 4 < 2 \pi \times 4} \\ \displaystyle \large \boxed{0 \leqslant 4 \theta < 8 \pi}[/tex]

Then find the reference angle.

We know that sin(π/4) = √2/2. Hence π/4 is our reference angle.

sin is negative in Q3 and Q4. We use π + (ref. angle) for Q3 and 2π - (ref. angle for Q4.)

Find Q3

[tex] \displaystyle \large{ \pi + \frac{ \pi}{4} = \frac{ 4 \pi}{4} + \frac{ \pi}{4} } \\ \displaystyle \large \boxed{ \frac{5 \pi}{4} }[/tex]

Find Q4

[tex] \displaystyle \large{2 \pi - \frac{ \pi}{4} = \frac{8 \pi}{4} - \frac{ \pi}{4} } \\ \displaystyle \large \boxed{ \frac{7 \pi}{4} }[/tex]

Both values are in [0,2π). However, we exceed our interval to < 8π.

We will be using these following:-

[tex] \displaystyle \large{ \theta + 2 \pi k = \theta \: \: \: \: \: \sf{(k \: \: is \: \: integer)}}[/tex]

Hence:-

For Q3

[tex] \displaystyle \large{ \frac{5 \pi}{4} + 2 \pi = \frac{13 \pi}{4} } \\ \displaystyle \large{ \frac{5 \pi}{4} + 4\pi = \frac{21 \pi}{4} } \\ \displaystyle \large{ \frac{5 \pi}{4} + 6\pi = \frac{29 \pi}{4} }[/tex]

We cannot use any further k-values (or k cannot be 4 or higher) because it'd be +8π and not in the interval.

For Q4

[tex] \displaystyle \large{ \frac{ 7 \pi}{4} + 2 \pi = \frac{15 \pi}{4} } \\ \displaystyle \large{ \frac{ 7 \pi}{4} + 4 \pi = \frac{23\pi}{4} } \\ \displaystyle \large{ \frac{ 7 \pi}{4} + 6 \pi = \frac{31 \pi}{4} }[/tex]

Therefore:-

[tex] \displaystyle \large{4 \theta = \frac{5 \pi}{4} , \frac{7 \pi}{4} , \frac{13\pi}{4} , \frac{21\pi}{4} , \frac{29\pi}{4}, \frac{15 \pi}{4} , \frac{23\pi}{4} , \frac{31\pi}{4} }[/tex]

Then we divide all these values by 4.

[tex] \displaystyle \large \boxed{\theta = \frac{5 \pi}{16} , \frac{7 \pi}{16} , \frac{13\pi}{16} , \frac{21\pi}{16} , \frac{29\pi}{16}, \frac{15 \pi}{16} , \frac{23\pi}{16} , \frac{31\pi}{16} }[/tex]

Let me know if you have any questions!