All categories

2 years ago

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

You might be interested in

How many decimals has the product of 3.99 and 3.3 Show work please

mariarad [96]

Answer:

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>Given Information :</h3>

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

<h3>To calculate :</h3>

Area of the trapezoid.

<h3>Calculation :</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.

Substituting values, we get :

$\longmapsto$ 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 $\sf ^2$

$\longmapsto$ Area = 1500 ft $\sf ^2$

Therefore,

Area of the trapezoid is 1500 ft $\sf ^2$

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