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3 years ago

14

Vertical angles must: check all that apply

1 answer:

Bingel [31]3 years ago

5 0

Answer:

Have the same vertex , and be congruent.

Step-by-step explanation:

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Factor a number, variable, or expression out of 2x2 + 3x

Lilit [14]

Answer:

x(2x+3)

Step-by-step explanation:

2x^2+3x=x(2x+3)

3 0

3 years ago

Read 2 more answers

Solve for x.<br><br> y=4+bx^2

kicyunya [14]

$y=4+bx^2$

$bx^2 = y - 4$

$x^2 = \dfrac{y-4}{b}$

$x = \pm \sqrt{ \dfrac{y-4}{b} }$

3 0

4 years ago

Solve for [x]. The polygons in each<br> 2x-20<br> 16<br> 16<br> 8

Nataly_w [17]

$\quad \huge \quad \quad \boxed{ \tt \:Answer }$

$\qquad \tt \rightarrow \:x = 16$

____________________________________

$\large \tt Solution \: :$

If two polygons are similar, their corresponding sides ratios will be equal to each other.

$\qquad \tt \rightarrow \: \cfrac{8}{16} = \cfrac{2x - 20}{24}$

$\qquad \tt \rightarrow \: \cfrac{1}{2} = \cfrac{2x - 20}{24}$

$\qquad \tt \rightarrow \: 2x - 20 = \cfrac{24}{2}$

$\qquad \tt \rightarrow \: 2x = 12 + 20$

$\qquad \tt \rightarrow \: 2x = 32$

$\qquad \tt \rightarrow \: x = 16$

Answered by : ❝ AǫᴜᴀWɪᴢ ❞

7 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

Can someone help plz!

Verizon [17]

Answer:

Do you still need this one?

Step-by-step explanation:

$x\geq 1$

3 0

3 years ago