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3 years ago

CAN SOMEONE PLEASE HELP ME and answer both questions please

Mathematics

1 answer:

Alexus [3.1K]3 years ago

3 0

1. The like graph is dated day by day. To make it easier you could maybe do two days at a time. For example, 24- sep- 26- sep. This will decrease less confusion. To graph it correctly just make the line for the higher numbers taller and for the smaller, shorter.

2. I gave you the answer for this one it the last sentence . :)

You might be interested in

Factor 4x^4+7x^2-2=0

natulia [17]

Answer:

(2x-1)(2x+1)(x^2+2) = 0

Step-by-step explanation:

Here's a trick: Use a temporary substitution for x^2. Let p = x^2. Then 4x^4+7x^2-2=0 becomes 4p^2 + 7p - 2 = 0.

Find p using the quadratic formula: a = 4, b = 7 and c = -2. Then the discriminant is b^2-4ac, or (7)^2-4(4)(-2), or 49+32, or 81.

Then the roots are:

-7 plus or minus √81

p= --------------------------------

p = 2/8 = 1/4 and p = -16/8 = -2.

Recalling that p = x^2, we let p = x^2 = 1/4, finding that x = plus or minus 1/2. We cannot do quite the same thing with the factor p= -2 because the roots would be complex.

If x = 1/2 is a root, then 2x - 1 is a factor. If x = -1/2 is a root, then 2x+1 is a factor.

Let's multiply these two factors, (2x-1) and (2x+1), together, obtaining 4x^2 - 1. Let's divide this 4x^2 - 1 into 4x^4+7x^2-2=0. We get x^2+2 as quotient.

Then, 4x^4+7x^2-2=0 in factored form, is (2x-1)(2x+1)(x^2+2) = 0.

6 0

2 years ago

Prove ΔPAB is isosceles.

Licemer1 [7]

Answer:

See explanation

Step-by-step explanation:

If $PX\cong PY,$ then triangle PXY is isosceles triangle. Angles adjacent to the base XY of an isosceles triangle PXY are congruent, so

$\angle 1\cong \angle 2$

and

$m\angle 1=m\angle 2$

Angles 1 and 3 are supplementary, so

$m\angle 3=180^{\circ}-m\angle 1$

Angles 2 and 4 are supplementary, so

$m\angle 4=180^{\circ}-m\angle 2$

By substitution property,

$m\angle 4=180^{\circ}-m\angle 2=180^{\circ}-m\angle 1=m\angle 3$

Hence,

$\angle 3\cong \angle 4$

Consider triangles APX and BPY. In these triangles:

$PX\cong PY$ - given;
$\angle 5\cong \angle 6$ - given;
$\angle 3\cong \angle 4$ - proven,

so $\triangle APX\cong \triangle BPY$ by ASA postulate.

Congruent triangles have congruent corresponding sides, then

$AP\cong BP$

Therefore, triangle APB is isosceles triangle (by definition).

6 0

3 years ago

Solve for X and Y: <br> Y=2+7x and -2x+6y=-28

Furkat [3]

Answer:

x=-1 and y=-5

Step-by-step explanation:

ሰላም!

Given expressions let them given as two equations

y=2+7x....... equation (1)
-2x+6y=-28..... equation (2)

Thus, substitute y=2+7x into equation (2)

$- 2x + 6(2 + 7x) = - 28 \\$

simplify it

$- 2x + 12 + 42x = - 28 \\ 40x + 12 = - 28$

subtract 12 from both sides

$40x + 12 - 12 = - 28 - 12 \\ 40x = - 40$

divide both sides by 40

$\frac{40x}{40} = \frac{ - 40}{40} \\ x = - 1$

Therefore, as we get the value of x=-1 substitute it to equation (1) and solve for variable y

$y = 2 + 7x \\ y = 2 + 7( -1 ) \\ y = 2 - 7 \\ y = - 5$

To be more sure make a cross check on both of the two expressions

Thus,

1) y=2+7x .... substitute y=-5 and x=-1 and check if it is equal

-5=2+7(-1)

-5=2-5

-5=-5

2)-2x+6y=-28... substitute and check

-2(-1)+6(-5)=-28

2-30=-28... (-a)(-b)=ab

-28=-28

Hope it helps!

6 0

1 year 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

Simplify. 2√6−4√24+√96What is the greatest common factor of 12xy^5+60x^4y^2−24x^3y^3?

Anon25 [30]

Answer:

greatest common factor is 12xy² .

Step-by-step explanation:

6 0

2 years ago