EASY POINTS. all irregular/wrong answers will be reported and removed, this is not easy points if you don't know how to factor.
I need these to be in the greatest common factor form, i am in a business calc class and understand everything but the algebra :) i hate algebra.
1) 36x^2-24x
2)12x^2-24x
3)45x^3-30x

Answer:

3 x 10^4

(3rd option)

Step-by-step explanation:

8.9 divided by 3.3 is 2.69

8.9 is roughly 9

3.3 is roughly 3

9 divided 3 is 3

10^8 divided 10^4 is 10^4

when dividing numbers with exponents you subtract the exponents

so the best estimate is

3 x 10^4

9514 1404 393

Answer:

  congruent

Step-by-step explanation:

Angles supplementary to the same angle are congruent.

__

Here, angles 1 and 3 are both supplementary to angle 2, so ...

  angle 1 and angle 3 are congruent

Answer:

x=40

Step-by-step explanation:

The angles of both expressions (4x-28) and (3x+12) are congruent (alternating exterior angles). So, you can set up the equation 4x-28 = 3x+12 because both expressions are equivalent, so it makes sense. Then solve for x.

4x - 28 = 3x + 12

x - 28 = 12

x = 40

Answer:3500

Step-by-step explanation:

Answer:

(b) NO

Step-by-step explanation:

It must have the square to to be a right angle

Answer:

A: It is obtuse

B: It is a right angle

Step-by-step explanation:

A: Angle RST equals 110*, if it is greater than 90*, it is obtuse

B: Right Angles are 90*, so this is simple, you subtract RST by UST, so 110-20=90

Was it this instead? [tex](-1-2a^{2}b)(1-2a^{2}b)[/tex]

Answer:

[tex]4a^{4b}-1[/tex]

Step-by-step explanation:

I am assuming your question is this.

[tex](-1-2a^{2b})(1-2a^{2b})[/tex]

Apply the difference of 2 squares formula.

[tex](a-b)(a+b)=a^{2} -b^{2}[/tex]

So then we have.

[tex](-1-2a^{2b})(1-2a^{2b})=(-2a^{2b})^{2}-1^{2}[/tex]

Then simplify.

[tex]4a^{4b}-1[/tex]

Answer:

x+y=12

Step-by-step explanation:

Standard form is ax+by=c

Given, y-8 = -(x-4)

y-8=-x+4

x+y=4+8

x+y=12

Answer:

x + y = 12

Step-by-step explanation:

Standard form is Ax+By=C

so the goal is to get the y and the x on the same side without and negatives or fractions on that side.

y - 8 = - (x - 4)  first, start by simplifying

y - 8 = -x +4  next, because it is negative, we are going to add x to -x on the right side to get it to cancel out and we'll add x to y on the left side

x + y - 8 = 4  now we need to get rid of that eight so we'll add 8 to -8 so it cancels out and we'll add 8 to 4

x + y = 12 now it is in standard form

Answer:

The answer is True.

Step-by-step explanation:

The earliest civilizations developed between 4000 and 3000 BCE, when the rise of agriculture and trade allowed people to have surplus food and economic stability.

Hope this helps you!

Step-by-step explanation:

320 = 80 × 4

80 = 8 × 10

4 = 2 × 2

10 = 5 × 2

bold one's are answer.

hope this helps you.

Answer:

4/5

Step-by-step explanation:

There are 10 tiles total, and E and I are vowels. This makes 2/10 of the tiles tiles. The other 8/10 are not vowels.

8/10 can be simplified down to 4/5.

There is a 4/5 probability of picking a letter that is not a vowel.

The profit that Jude made during the party was £3700.

Profit is the difference between revenue and cost. It is given by:

Profit = Revenue - Cost

From the question:

The total cost = 300 + 2500 + 500 + (1 * 400 people) = £3700

Revenue = 400 ticket * £16 = £6400

Profit = Revenue - Cost = £6400 - £3700

The profit that Jude made during the party was £3700.

Find out more on profit at: https://brainly.com/question/23103804

Answer:

[tex]x=\frac{5+\sqrt{15}}{2},\:x=\frac{5-\sqrt{15}}{2}[/tex]

Step-by-step explanation:

[tex]\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}[/tex]

[tex]x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]

[tex]\mathrm{For\:}\quad a=2,\:b=-10,\:c=5[/tex]

[tex]x_{1,\:2}=\frac{-\left(-10\right)\pm \sqrt{\left(-10\right)^2-4\cdot \:2\cdot \:5}}{2\cdot \:2}[/tex]

[tex]\sqrt{\left(-10\right)^2-4\cdot \:2\cdot \:5}[/tex]

Apply exponent rule: (-a)^n=a^n, if n is even

[tex]\left(-10\right)^2=10^2[/tex]

[tex]=\sqrt{10^2-4\cdot \:2\cdot \:5}[/tex]

[tex]\mathrm{Multiply\:the\:numbers:}\:4\cdot \:2\cdot \:5=40[/tex]

[tex]=\sqrt{10^2-40}[/tex]

[tex]10^2=100[/tex]

[tex]=\sqrt{100-40}[/tex]

[tex]\mathrm{Subtract\:the\:numbers:}\:100-40=60[/tex]

[tex]=\sqrt{60}[/tex]

[tex]60\:\mathrm{divides\:by}\:2\quad \:60=30\cdot \:2[/tex]

[tex]=2\cdot \:30[/tex]

[tex]30\:\mathrm{divides\:by}\:2\quad \:30=15\cdot \:2[/tex]

[tex]=2\cdot \:2\cdot \:15[/tex]

[tex]15\:\mathrm{divides\:by}\:3\quad \:15=5\cdot \:3[/tex]

[tex]=2\cdot \:2\cdot \:3\cdot \:5[/tex]

[tex]2,\:3,\:5\mathrm{\:are\:all\:prime\:numbers,\:therefore\:no\:further\:factorization\:is\:possible}[/tex]

[tex]=2\cdot \:2\cdot \:3\cdot \:5[/tex]

[tex]=2^2\cdot \:3\cdot \:5[/tex]

[tex]=\sqrt{2^2\cdot \:3\cdot \:5}[/tex]

Apply Radical Rule:

[tex]=\sqrt{2^2}\sqrt{3\cdot \:5}[/tex]

Apply Radical Rule:

[tex]\sqrt{2^2}=2[/tex]

[tex]=2\sqrt{3\cdot \:5}[/tex]

[tex]\mathrm{Refine}[/tex]

[tex]=2\sqrt{15}[/tex]

[tex]x_{1,\:2}=\frac{-\left(-10\right)\pm \:2\sqrt{15}}{2\cdot \:2}[/tex]

[tex]\mathrm{Separate\:the\:solutions}[/tex]

[tex]x_1=\frac{-\left(-10\right)+2\sqrt{15}}{2\cdot \:2},\:x_2=\frac{-\left(-10\right)-2\sqrt{15}}{2\cdot \:2}[/tex]

[tex]\frac{-\left(-10\right)+2\sqrt{15}}{2\cdot \:2}[/tex]

Apply Rule -(-a)=a

[tex]=\frac{10+2\sqrt{15}}{2\cdot \:2}[/tex]

[tex]\mathrm{Multiply\:the\:numbers:}\:2\cdot \:2=4[/tex]

[tex]=\frac{10+2\sqrt{15}}{4}[/tex]

[tex]=\frac{2\left(5+\sqrt{15}\right)}{4}[/tex]

[tex]\mathrm{Cancel\:the\:common\:factor:}\:2[/tex]

[tex]=\frac{5+\sqrt{15}}{2}[/tex]

[tex]\frac{-\left(-10\right)-2\sqrt{15}}{2\cdot \:2}[/tex]

[tex]=\frac{10-2\sqrt{15}}{2\cdot \:2}[/tex]

[tex]\mathrm{Multiply\:the\:numbers:}\:2\cdot \:2=4[/tex]

[tex]=\frac{10-2\sqrt{15}}{4}[/tex]

[tex]=\frac{2\left(5-\sqrt{15}\right)}{4}[/tex]

[tex]\mathrm{Cancel\:the\:common\:factor:}\:2[/tex]

[tex]=\frac{5-\sqrt{15}}{2}[/tex]

[tex]\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}[/tex]

[tex]x=\frac{5+\sqrt{15}}{2},\:x=\frac{5-\sqrt{15}}{2}[/tex]

Answer:

12

Step-by-step explanation:

Given :-

Two numbers => 48 and 60

To Find :-

HCF of 48 and 60

Solving :-

Prime factorize the numbers.

48 = 2⁴ x 3

60 = 2² x 3 x 5

Solution :-

HCF (48, 60) = 2² x 3 = 12

The expected number of red candy that you will get out of 10 bags is 3 red candies.

In each bag there are 100 candies, such that there are:

All the candies have the same probability of being randomly selected, so the probability of getting a red candy is equal to the quotient between the number of red candies in each bag and the total number of candy.

P = 30/100 = 0.3

Then if you draw one candy of each bag of 10 bags, the expected number of red candies that you will have is 10 times the above probability:

N = 10*P = 10*0.3 = 3

3 red candies.

If you want to learn more about probability, you can read:

https://brainly.com/question/251701

A function is a relation in which there is only one y-value for every x-value.

If we given an input of 3 for x, for example, we would only get one output for y, like 1, for example.

If we look at the table, we can see that the x-value of 3 gives us two outputs: 7 and 9.

Therefore, this relation is not a function.

Answer:

what are you trying to solve?

Answer:

the answer to this is

A) nonlinear

Step-by-step explanation:

this is not linear because the line is not a line that goes in 1 direction. It has points of an increase and decrease so it is non linear

Answer:

it is not linear

Step-by-step explanation:

It is not a linear line because it is not going straight

Answer:

X = 11/36

Step-by-step explanation:

First, find the common denominator of the fractions and convert them.

(4/9) x 4 = 16/36

(3/4) x 9 = 27/36

X + 16/36 = 27/36

Now you can solve for X

X = 11/36

The fraction cannot be simplified

Answer:

3

Step-by-step explanation:

Answer:

Uh how to use segmant tool

Step-by-step explanation:

not there

========================================================

Explanation:

Going from the first term to the second, we add on -2

Then going from the second term to the third, we add on +2

This inconsistency (-2 vs +2) means we don't have a common difference, and therefore the sequence is not arithmetic.

------------

Now let's divide each term by their previous term

term2/term1 = 0/2 = 0

term3/term2 = 2/0 = undefined

We can stop here and conclude that we don't have a common ratio either, so this sequence is not geometric.

------------

Side note: The subsequence 0,2,4,6,... is arithmetic with common difference d = 2 because we add 2 to each term. It's that first 2 in the original sequence that breaks everything.

Answer:

Step-by-step explanation:

First Choice

Answer:

Suppose there are x students,

a) 4 pencils  x students+20 (*=multiplication)

b) 5 pencilsx students+10

But a = b so

4x+20=5x+10

x=10

4 10+20 or 5 10+10 = 60 pencils.

Let

[tex]I(a) = \displaystyle \int_0^\infty \frac{\sqrt x \arctan(ax)}{1+x^2} \, dx[/tex]

Differentiate with respect to a :

[tex]I'(a) = \displaystyle a \int_0^\infty \frac{x^{\frac32}}{(1+x^2)(1+a^2x^2)} \, dx[/tex]

Substitute y = √x :

[tex]I'(a) = \displaystyle 2a \int_0^\infty \frac{y^4}{(1+y)(1+a^2y)} \, dy[/tex]

Polynomial division yields

[tex]\dfrac{y^4}{(1+y)(1+a^2y)} \\\\ = \dfrac1{a^2}y^2 - \left(\dfrac1{a^2} + \dfrac1{a^4}\right)y + \dfrac1{a^2} + \dfrac1{a^4} + \dfrac1{a^6} - \dfrac{\left(1+\frac1{a^2} + \frac1{a^4} + \frac1{a^6}\right)y + \frac1{a^2} + \frac1{a^4} + \frac1{a^6}}{(1+y)(1+a^2y)}[/tex]

Computing I'(a) isn't so difficult from here. You'd find (assuming a ≥ 0)

[tex]I'(a) = \displaystyle \frac\pi{\sqrt2\left(\sqrt a + a + a^{\frac32} + a^2\right)}[/tex]

Integrate both sides with respect to a. On the right side, substituting b = √a yields

[tex]\displaystyle \int \frac{da}{\sqrt a + a + a^{\frac32} + a^2} = \int \frac{2b}{b + b^2 + b^3 + b^4} \, db \\\\ = 2 \int \frac{db}{1 + b + b^2 + b^3} \\\\ = 2 \int \frac{db}{(1+b)(1+b^2)} \\\\ = \int \left(\frac1{1+b} + \frac{1-b}{(1+b^2)}\right) \, db \\\\ = \frac14 \left(2 \arctan(b) + 2 \ln(1 + b) - \ln(1 + b^2)\right) + C \\\\ = \frac14 \left(2 \arctan(\sqrt a) + 2 \ln(1 + \sqrt a) - \ln(1 + a)\right) + C[/tex]

Noting that a = 0 makes the integral I(a) vanish, we have

[tex]0 = \dfrac14 \left(2 \arctan(\sqrt0) + 2\ln(1 + \sqrt0) - \ln(1 + 0)\right) + C \implies C = 0[/tex]

and so

[tex]\displaystyle I(a) = \frac\pi{4\sqrt2} \left(2 \arctan(\sqrt a) + 2 \ln(1 + \sqrt a) - \ln(1 + a)\right)[/tex]

We recover the integral we want with a = 1, which gives a value of

[tex]\displaystyle \int_0^\infty \frac{\sqrt x \arctan(x)}{1 + x^2} \, dx = \boxed{\frac{\pi^2 + 2\pi\ln(2)}{4\sqrt2}}[/tex]

To find out the box number the boxes, extract the same number of balls as the number of the box, calculate the total weight.

Considering you can only weigh the balls extracted once and you need to find the box with balls that are 1-kilo lighter; here are the steps you can follow:

The total weight using this method would be 6,300 kg which is the result of adding 30 kg (first box) + 60 kg (second box)... However, since there is a box with lighter balls the total weight will be less.

Let's imagine the weight is 6,290, this means there are 10 fewer kilos and therefore the box with the lighter balls is box 10 since this is the box number from which 10 balls were extracted.

Learn more about weight in: https://brainly.com/question/10069252

Answer:

x less than or equal to -2

Answer:

down

Step-by-step explanation:

c) 6 and 8

a) 7 and 1

b) 5 and 4

Answer:

alternate angled

Step-by-step explanation:

add up to 180