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3 years ago

PLEASE HELP I WILL GIVE 50 POINTS

2 answers:

7 0

48+30 is 78, so we need to find the equation that is equal to that.
The first one is not enough, the second one is too much, and the third one is too much as well. The fourth on is the answer.

julia-pushkina [17]3 years ago

4 0

Answer:

Step-by-step explanation:

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Select the expression that is equivalent to (x+5)^2

nataly862011 [7]

Answer:7500

Step-by-step explanation:

if you divide 15000 and 2 you get 7500.and if you get 7500 you have to multiply to check if its correct so 7500 times 2 equals 15000. So the answer is 7500.

4 0

3 years ago

Help me plz thanksss

natulia [17]

Answer:

0.5j+19

Step-by-step explanation:

6 0

2 years ago

raph the equation with a diameter that has endpoints at (-3, 4) and (5, -2). Label the center and at least four points on the ci

andreyandreev [35.5K]

Answer:

Equation:

${x}^{2} + {y}^{2} + 2x - 2y - 35= 0$

The point (0,-5), (0,7), (5,0) and (-7,0)also lie on this circle.

Step-by-step explanation:

We want to find the equation of a circle with a diamterhat hs endpoints at (-3, 4) and (5, -2).

The center of this circle is the midpoint of (-3, 4) and (5, -2).

We use the midpoint formula:

$( \frac{x_1+x_2}{2}, \frac{y_1+y_2,}{2} )$

Plug in the points to get:

$( \frac{ - 3+5}{2}, \frac{ - 2+4}{2} )$

$( \frac{ -2}{2}, \frac{ 2}{2} )$

$( - 1, 1)$

We find the radius of the circle using the center (-1,1) and the point (5,-2) on the circle using the distance formula:

$r = \sqrt{ {(x_2-x_1)}^{2} + {(y_2-y_1)}^{2} }$

$r = \sqrt{ {(5 - - 1)}^{2} + {( - 2- - 1)}^{2} }$

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

$r = \sqrt{ 36+ 1 } = \sqrt{37}$

The equation of the circle is given by:

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

Where (h,k)=(-1,1) and r=√37 is the radius

We plug in the values to get:

$(x- - 1)^2 + (y-1)^2 = {( \sqrt{37}) }^{2}$

$(x + 1)^2 + (y - 1)^2 = 37$

We expand to get:

${x}^{2} + 2x + 1 + {y}^{2} - 2y + 1 = 37$

${x}^{2} + {y}^{2} + 2x - 2y +2 - 37= 0$

${x}^{2} + {y}^{2} + 2x - 2y - 35= 0$

We want to find at least four points on this circle.

We can choose any point for x and solve for y or vice-versa

When y=0,

${x}^{2} + {0}^{2} + 2x - 2(0) - 35= 0$

${x}^{2} +2x - 35= 0$

$(x - 5)(x + 7) = 0$

$x = 5 \: or \: x = - 7$

The point (5,0) and (-7,0) lies on the circle.

When x=0

${0}^{2} + {y}^{2} + 2(0) - 2y - 35= 0$

${y}^{2} - 2y - 35= 0$

$(y - 7)(y + 5) = 0$

$y = 7 \: or \: y = - 5$

The point (0,-5) and (0,7) lie on this circle.

3 0

3 years ago

Help me i’m struggllinbggg with this

Olenka [21]

The opposite angles of a parallelogram are congruent, so you have to set the values of each angle equal to each other and solve for x.

(10x-19)° = (7x+23)°
-7x -7x
3x-19 = 23
+19 +19
3x = 43
÷3 ÷3
x = 43

Then, substitute the value of x back into the equations.
(10x-19)°
(10(14)-19)°
(140-19)°
121°

(7x+23)°
(7(14)+23)°
(98+23)°
121°

8 0

2 years ago

HELP PLEASEE!!!!

ss7ja [257]

Answer:

See below ~

Step-by-step explanation:

$\textsf {Each of the steps has been reasoned below :}$