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Trava [24]
3 years ago
8

Write twenty-six thousand expanded notation plz

Mathematics
2 answers:
muminat3 years ago
5 0
20,000+6,000- that is the answer
ale4655 [162]3 years ago
5 0

The place value chart can help us to write a number in expanded notation. When we put 26,000 into the place value chart, we can recognize that our number is equal to

2 ten thousands + 6 thousands + 0 hundreds + 0 tens + 0 units

Image is provided

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What value of x is in the solution set of 2(3x – 1) = 4x - 6?
Andrei [34K]

Answer:

x=-2

Step-by-step explanation:

2(3x – 1) = 4x - 6

Distribute

6x-2=4x-6

Moving like terms (4x)

2x-2=-6

Moving like terms (-2)

2x=-4

Divide both sides by 2 to isolate x

x=-2

Hope this helps! Please mark brainliest :)

5 0
3 years ago
Read 2 more answers
Writing an Equation in Vertex Form
pickupchik [31]

We get the value of h and k as 9 and -8

The quadratic function g(x) = a(x - h)^2 + k, if a is  not equal to 0,which means that equation has a quadratic value then it is  said to be in standard quadratic form.

If a is positive, then the graph opens upwards,

If a is negative, then it opens downwards.

The line of symmetry of the quadratic equation  is the vertical line x = h, and the co-ordinate of the point of vertex is (h,k).

We are given the co-ordinate of the vertex point in the question and hence all that is required is to equate it.

Therefore, the values of h and k are 9 and -8 in the given equation in vertex form

You can learn more about quadratic equation in standard form here:

brainly.com/question/25025116?referrer=searchResults

Disclaimer: the question may be incomplete and hence has been answered accordingly

#SPJ9

7 0
2 years ago
WILL UPVOTE!<br> Factor the expression.<br> x^2 -12x +36 =
svlad2 [7]
(x-6)(x+6) this is the answer my student thought of in class
5 0
3 years ago
An important tool in archeological research is radiocarbon dating, developed by the American chemist Willard F. Libby.3 This is
Radda [10]

Answer:

Q(t) = Q_o*e^(-0.000120968*t)

Step-by-step explanation:

Given:

- The ODE of the life of Carbon-14:

                                       Q' = -r*Q

- The initial conditions Q(0) = Q_o

- Carbon isotope reaches its half life in t = 5730 yrs

Find:

The expression for Q(t).

Solution:

- Assuming Q(t) satisfies:

                                       Q' = -r*Q

- Separate variables:

                                      dQ / Q = -r .dt

- Integrate both sides:

                                       Ln(Q) = -r*t + C

- Make the relation for Q:

                                       Q = C*e^(-r*t)

- Using initial conditions given:

                                       Q(0) = Q_o

                                       Q_o = C*e^(-r*0)

                                      C = Q_o    

- The relation is:

                                       Q(t) = Q_o*e^(-r*t)

- We are also given that the half life of carbon is t = 5730 years:

                                       Q_o / 2 = Q_o*e^(-5730*r)

                                        -Ln(0.5) = 5730*r

                                        r = -Ln(0.5)/5730

                                        r = 0.000120968          

- Hence, our expression for Q(t) would be:

                                       Q(t) = Q_o*e^(-0.000120968*t)                                    

7 0
3 years ago
PLEASE ANSWER AS SOON AS POSSIBLE WILL GIVE "Brainliest" TO THE FIRST CORRECT PERSON!!!
saw5 [17]

Answer: x=18 ; y=6\sqrt{3}

Step-by-step explanation:

You can write two expressions that are in terms of 'x' and 'y'. This is a right triangle, so Pythagorean's theorem can be used. We can also use the angle to find the sine of 30 degrees.

<u>Pythagorean's Theorem:</u>

The values 'a' and 'b' are the two shorter sides of the triangle, while the value 'c' is the longest side of the triangle; the hypotenuse.

a^2+b^2=c^2

x^2+y^2=(12\sqrt{3} )^2

x^2+y^2=12^2*\sqrt{3} ^2=432

<u>Sine of the Angle:</u>

Sine is defined to be the opposite side divided by the hypotenuse. Let's take the sine of 30 degrees:

sin(\alpha )=opposite/hypotenuse

sin(30)=y/12\sqrt{3}

\frac{1}{2} =y/12\sqrt{3}

y=6\sqrt{3}

Plug this value of 'y' into the first expression derived from Pythagorean's theorem:

x^2+y^2=432

x^2+(6\sqrt{3} )^2=432

x^2+108=432

x^2=432-108=324

x=\sqrt{324}

x=18

Use this value of 'x' to solve for 'y' in the expression from Pythagorean's theorem:

(\sqrt{324})^2+y^2=432

324+y^2=432

y^2=432-324=108

y=\sqrt{108}

y=\sqrt{36*3}

y=6\sqrt{3}

7 0
2 years ago
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