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3 years ago

The green line below

Mathematics

2 answers:

trasher [3.6K]3 years ago

7 0

A,d I'm pretty sure because of the cy axis

Serjik [45]3 years ago

7 0

Answer:

A D E are all true. The others are false and should not be checked.

Step-by-step explanation:

A crosses the x axis at (- 2.5,0)

B is false. The line does not go through (0,0) which is the origin.

C is false. The green line is not perpendicular to anything.

D is true. It crosses the y axis at (0,5)

E is true. It is on the xy plane.

F is false. It is not parallel to either axis.

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An ellipse has a center at the origin, a vertex along the major axis at (20, 0), and a focus at (16, 0). the equation of the ell

olga_2 [115]

If the center is at (0, 0) and the vertex is at (20, 0), then the distance, a, is the length from the center to the vertex, 20. The distance from the center to the focus is c. The distance from the center to the focus is 16, so c = 16. The formula we use to find the focus is $c^2=a^2-b^2$ . We have our c value and our a value, so we will sub in those to find b. $16^2=20^2-b^2$ and 256 = 400 - b^2. -b^2 = -144, so b = 12. There you go!

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3 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

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Step-by-step explanation:

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3 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.

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3 years ago

A two digit number when read from left to right to right consiststs of consecutive integers. the nmber is six more than for time

Solnce55 [7]

I suppose you want to know such number. Since we have a two digit number consisting of two consecutive integers, the only possible numbers are:

$12,23,34,45,56,67,78,89$

Since we sorted all the cases out, we simply have to check which one satisfies the requirement. For each number, we'll write four times the the sum of its digits, and add 6, hoping to get the original number.

$\text{Number: } 12,\ \ 4\times \text{sum of digits: } 12\ \ \text{plus six: } 18\neq 12$

$\text{Number: } 23,\ \ 4\times \text{sum of digits: } 24\ \ \text{plus six: } 30\neq 23$

$\text{Number: } 34,\ \ 4\times \text{sum of digits: } 28\ \ \text{plus six: } 34$

So, the answer is 34.

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3 years ago