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atroni [7]
2 years ago
12

How do i solve 6x+2=-2x+22

Mathematics
1 answer:
belka [17]2 years ago
8 0

6x+2=-2x+22

Move -2x to the other side

Sign changes from -2x to +2x

6x+2x+2=-2x+2x+22

6x+2x+2= 22

8x+2= 22

Move +2 to the other side

8x+2-2= 22-2

8x= 22-2

8x= 20

Divide both sides by 8

8x/8= 20/8

x= 20/8

Reduce fraction by dividing by 4

20/4= 5

8/4= 2

Answer:

x= 5/2

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How many decimals has the product of 3.99 and 3.3<br>Show work please
mariarad [96]

Answer:

3

Step-by-step explanation:

The product of 3.99 and 3.3 will have 3 decimals.

When multiplying decimals, you add the decimals to get the total number of decimals in your product.

2+1=3 so it will have 3 decimals.

4 0
2 years ago
What is the area of the trapezoid?
Maslowich
<h3><u>Given Information :</u></h3>

  • Length of parallel sides = 60 ft and 40 ft
  • Height of the trapezoid = 30 ft

<h3><u>To calculate :</u></h3>

  • Area of the trapezoid.

<h3><u>Calculation :</u></h3>

As we know that,

\bigstar \: \boxed{\sf {Area_{(Trapezium)} = \dfrac{1}{2} \times ( a + b) \times h}} \\

  • a and b are length of parallel sides.
  • h denotes height.

<em>S</em><em>u</em><em>b</em><em>s</em><em>t</em><em>i</em><em>t</em><em>u</em><em>t</em><em>i</em><em>n</em><em>g</em><em> </em><em>valu</em><em>es</em><em>,</em><em> </em><em>we</em><em> </em><em>get</em><em> </em>:

\longmapsto Area = \sf \dfrac{1}{2} × ( 60 + 40 ) × 30 ft\sf ^2

\longmapsto Area = \sf \dfrac{1}{2} × 100 × 30 ft\sf ^2

\longmapsto Area = 1 × 100 × 15 ft\sf ^2

\longmapsto Area = 100 × 15 ft\sf ^2

\longmapsto <u>Area = 1500 ft</u>\sf ^2

Therefore,

  • Area of the trapezoid is <u>1500 ft\sf ^2</u>

6 0
2 years ago
Scientists have discovered a vaccine for a debilitating disease. This year there were 570,000 reported new cases of the disease
Ket [755]

Answer: There are approximately 853827 new cases in 6 years.

Step-by-step explanation:

Since we have given that

Initial population = 570000

Rate at which population decreases is given by

\frac{2}{3}

Now,

First year =570000

Second year is given by

570000\times (\frac{1}{3})

Third year is given by

570000(\frac{1}{3})^2

so, there is common ratio ,

it becomes geometric progression, as there is exponential decline.

so,

570000,570000\times \frac{1}{3},570000\times( \frac{1}{3})^2,......,570000\times (\frac{1}{3})^6

a=570000

common ratio is given by

r=\frac{a_2}{a_1}=\frac{1}{3}

number of terms = 6

Sum of terms will be given by

S_n=\frac{a(1-r^n)}{(1-r)}

We'll put this value in this formula,

S_6=\frac{570000(1-(\frac{1}{3})^6}{(1-\frac{1}{3})}\\\\=853827.16

So, there are approximately 853827 new cases in 6 years.

6 0
2 years ago
A rectangle is a parallelogram. What is true of the diagonals of a rectangle
maw [93]

Answer:diagonals of rectangle are equal in length while diagonals of a parallelogram can be different lengths

Step-by-step explanation:

3 0
2 years ago
If s is an increasing function, and t is a decreasing function, find Cs(X),t(Y ) in terms of CX,Y .
Sedbober [7]

Answer:

C(X,Y)(a,b)=1−C(s(X),t(Y))(a,1−b).

Step-by-step explanation:

Let's introduce the cumulative distribution of (X,Y), X and Y :

F(X,Y)(x,y)=P(X≤x,Y≤y)

  • FX(x)=P(X≤x)
  • FY(y)=P(Y≤y).

Likewise for (s(X),t(Y)), s(X) and t(Y) :

F(s(X),t(Y))(u,v)=P(s(X)≤u

  • t(Y)≤v)
  • Fs(X)(u)=P(s(X)≤u)
  • Ft(Y)(v)=P(t(Y)≤v).

Now, First establish the relationship between F(X,Y) and F(s(X),t(Y)) :

F(X,Y)(x,y)=P(X≤x,Y≤y)=P(s(X)≤s(x),t(Y)≥t(y))

The last step is obtained by applying the functions s and t since s preserves order and t reverses it.

This can be further transformed into

F(X,Y)(x,y)=1−P(s(X)≤s(x),t(Y)≤t(y))=1−F(s(X),t(Y))(s(x),t(y))

Since our random variables are continuous, we assume that the difference between t(Y)≤t(y) and t(Y)<t(y)) is just a set of zero measure.

Now, to transform this into a statement about copulas, note that

C(X,Y)(a,b)=F(X,Y)(F−1X(a), F−1Y(b))

Thus, plugging x=F−1X(a) and y=F−1Y(b) into our previous formula,

we get

F(X,Y)(F−1X(a),F−1Y(b))=1−F(s(X),t(Y))(s(F−1X(a)),t(F−1Y(b)))

The left hand side is the copula C(X,Y), the right hand side still needs some work.

Note that

Fs(X)(s(F−1X(a)))=P(s(X)≤s(F−1X(a)))=P(X≤F−1X(a))=FX(F−1X(a))=a

and likewise

Ft(Y)(s(F−1Y(b)))=P(t(Y)≤t(F−1Y(b)))=P(Y≥F−1Y(b))=1−FY(F−1Y(b))=1−b

Combining all results we obtain for the relationship between the copulas

C(X,Y)(a,b)=1−C(s(X),t(Y))(a,1−b).

7 0
2 years ago
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