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3 years ago

I NEED THIS BY TODAY!!

Mathematics

2 answers:

Agata [3.3K]3 years ago

7 0

Step-by-step explanation:

$(1. \: x = - \frac{y}{3}$

$(2. \: x = \frac{22 + 3y}{2}$

I think this is correct and what you wanted

Paladinen [302]3 years ago

5 0

2x-3(-3x)=22
2x+9x=22
11x=22 and divide eleven on both sides leaving x=2

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(5x^2-7x+5)-(9x^2+6x-4)

Blababa [14]

The simplified form would be -4×^2-13x+9

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4 years ago

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If one root of the equation is 4x^2- 2x+p-4 be the reciprocal of other. then find the value of p

SashulF [63]

Answer:

p = 8

Step-by-step explanation:

Let one root of the eqn. be alpha . Other root is 1/alpha .

We know that product of both roots of an quadratic eqn. is c/a where "c" is the co-efficient of the constant & "a" is the co-efficient of x^2.

Here "c" is p-4 & "a" is 4. And the product of roots is 1 ( ∵ prdouct of a number and its reciprocal is 1 )

$\frac{c}{a} = 1\\=> \frac{p - 4}{4} = 1\\=> p - 4 = 4\\=> p = 4 + 4 = 8$

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4 years ago

The scientific notation 2 x 10-2 has what value?

Vesnalui [34]

Since the exponent is negative, you move the decimal (2.0) to the left two spots leaving you with .02

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4 years ago

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Given the center of the circle (-3,4) and a point on the circle (-6,2), (10,4) is on the circle

Anastasy [175]

Answer:

Part 1) False

Part 2) False

Step-by-step explanation:

we know that

The equation of the circle in standard form is equal to

$(x-h)^{2} +(y-k)^{2}=r^{2}$

where

(h,k) is the center and r is the radius

In this problem the distance between the center and a point on the circle is equal to the radius

The formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

Part 1) given the center of the circle (-3,4) and a point on the circle (-6,2), (10,4) is on the circle.

true or false

substitute the center of the circle in the equation in standard form

$(x+3)^{2} +(y-4)^{2}=r^{2}$

Find the distance (radius) between the center (-3,4) and (-6,2)

substitute in the formula of distance

$r=\sqrt{(2-4)^{2}+(-6+3)^{2}}$

$r=\sqrt{(-2)^{2}+(-3)^{2}}$

$r=\sqrt{13}\ units$

The equation of the circle is equal to

$(x+3)^{2} +(y-4)^{2}=(\sqrt{13}){2}$

$(x+3)^{2} +(y-4)^{2}=13$

Verify if the point (10,4) is on the circle

we know that

If a ordered pair is on the circle, then the ordered pair must satisfy the equation of the circle

For x=10,y=4

substitute

$(10+3)^{2} +(4-4)^{2}=13$

$(13)^{2} +(0)^{2}=13$

$169=13$ -----> is not true

therefore

The point is not on the circle

The statement is false

Part 2) given the center of the circle (1,3) and a point on the circle (2,6), (11,5) is on the circle.

true or false

substitute the center of the circle in the equation in standard form

$(x-1)^{2} +(y-3)^{2}=r^{2}$

Find the distance (radius) between the center (1,3) and (2,6)

substitute in the formula of distance

$r=\sqrt{(6-3)^{2}+(2-1)^{2}}$

$r=\sqrt{(3)^{2}+(1)^{2}}$

$r=\sqrt{10}\ units$

The equation of the circle is equal to

$(x-1)^{2} +(y-3)^{2}=(\sqrt{10}){2}$

$(x-1)^{2} +(y-3)^{2}=10$

Verify if the point (11,5) is on the circle

we know that

If a ordered pair is on the circle, then the ordered pair must satisfy the equation of the circle

For x=11,y=5

substitute

$(11-1)^{2} +(5-3)^{2}=10$

$(10)^{2} +(2)^{2}=10$

$104=10$ -----> is not true

therefore

The point is not on the circle

The statement is false

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4 years ago

Which statement describes the graph?

Leya [2.2K]

Answer:

Step-by-step explanation:

The answer is A, the graph raises, crosses at (0, 5) and then remains constant.

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3 years ago