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3 years ago

The vertex of the parabola below is at the point (-3, -5). Which of the equations below could be the one for this parabola?

1 answer:

4 0

Hello,

Trying to put x=-3 and y=-5, we must find an equality.

A) (-3+3)²+5=0+5=5 no

B) -3(-5+5)²==-3*0=0 not -3: no

C) (-3-5)²+3=64+3 not -5 :no

D) (-3+3)²-5=0-5=-5 Yes

Answer D

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QveST [7]

Answer:

Step-by-step explanation:

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2 years ago

Callie has a new kitten. They can weigh 3 pounds less than half the weight of Callie‘s cat. Together the cat and kitten weigh 18

gavmur [86]

Answer:

The weight of cat is 14 pounds and the weight of kitten is 4 pounds.

Step-by-step explanation:

Given:

Callie has a new kitten. It can weigh 3 pounds less than half the weight of Callie‘s cat. Together the cat and kitten weigh 18 pounds.

Now, to find the weight of each animal:

Let the cat's weight be $x.$

And the kitten weight = $\frac{x}{2} -3$

Total weight of cat and kitten = 18 pounds.

Now, to set an equation to get the weight of each animal:

$(x)+(\frac{x}{2}-3)=18$

$x+\frac{x}{2} -3=18$

$\frac{2x+x-6}{2} =18$

$\frac{3x-6}{2} =18$

Multiplying both sides by 2 we get:

$3x-6=36$

Adding both sides by 6 we get:

$3x=42$

Dividing both sides by 3 we get:

$x=14\ pounds.$

The weight of cat = 14 pounds.

Substituting the value of $x$ to get the kitten's weight:

$\frac{x}{2} -3\\\\=\frac{14}{2}-3\\\\=7-3\\\\=4.$

The kitten's weight = 4 pounds.

Therefore, the weight of cat is 14 pounds and the weight of kitten is 4 pounds.

7 0

3 years ago

The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2). On a coordinate plane, line A B has points (4, 1) and (n

GarryVolchara [31]

Answer:

$(-1,1),(4,-2)$

Step-by-step explanation:

Given: The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2).

To find: coordinates of vertex of the right angle

Solution:

Let C be point $(x,y)$

Distance between points $(x_1,y_1),(x_2,y_2)$ is given by $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$AC=\sqrt{(x-4)^2+(y-1)^2}\\BC=\sqrt{(x+1)^2+(y+2)^2}\\AB=\sqrt{(4+1)^2+(1+2)^2}=\sqrt{25+9}=\sqrt{34}$

ΔABC is a right angled triangle, suing Pythagoras theorem (square of hypotenuse is equal to sum of squares of base and perpendicular)

$34=\left [ (x-4)^2+(y-1)^2 \right ]+\left [ (x+1)^2+(y+2)^2 \right ]$

Put $(x,y)=(-1,1)$

$34=\left [ (-1-4)^2+(1-1)^2 \right ]+\left [ (-1+1)^2+(1+2)^2 \right ]\\34=25+9\\34=34$

which is true. So, $(-1,1)$ can be a vertex

Put $(x,y)=(4,-2)$

$34=\left [ (4-4)^2+(-2-1)^2 \right ]+\left [ (4+1)^2+(-2+2)^2 \right ]\\34=9+25\\34=34$

which is true. So, $(4,-2)$ can be a vertex

Put $(x,y)=(1,1)$

$34=\left [ (1-4)^2+(1-1)^2 \right ]+\left [ (1+1)^2+(1+2)^2 \right ]\\34=9+4+9\\34=22$

which is not true. So, $(1,1)$ cannot be a vertex

Put $(x,y)=(2,-2)$

$34=\left [ (2-4)^2+(-2-1)^2 \right ]+\left [ (2+1)^2+(-2+2)^2 \right ]\\34=4+9+9\\34=22$

which is not true. So, $(2,-2)$ cannot be a vertex

Put $(x,y)=(4,-1)$

$34=\left [ (4-4)^2+(-1-1)^2 \right ]+\left [ (4+1)^2+(-1+2)^2 \right ]\\34=4+25+1\\34=30$

which is not true. So, $(4,-1)$ cannot be a vertex

Put $(x,y)=(-1,4)$

$34=\left [ (-1-4)^2+(4-1)^2 \right ]+\left [ (-1+1)^2+(4+2)^2 \right ]\\34=25+9+36\\34=70$

which is not true. So, $(-1,4)$ cannot be a vertex

So, possible points for the vertex are $(-1,1),(4,-2)$

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3 years ago

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Step-by-step explanation:

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3 years ago

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kiruha [24]

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3 years ago

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