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2 years ago

PLS HELP ASAP ILL GIVE BRAINLKEST THANKS

Mathematics

2 answers:

kompoz [17]2 years ago

4 0

Answer:

d<u>></u>-7

Step-by-step explanation:

miss Akunina [59]2 years ago

4 0

First answer
d = greater than or equal to 7

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5. What is the equation of the tangent line to the circle x^2+y^2=1 through the point (6,0)?

notka56 [123]

Answer:

There is no tangent line of the given circle at (6, 0).

Step-by-step explanation:

Given equation of the circle,

$x^2 + y^2 = 1$

∵ equation of a circle is $(x-h)^2 +(y-k)^2 = r^2$ ,

Where, (h, k) is the center of the circle and r is the radius,

By comparing,

Center of the given circle = (0, 0),

Radius of the circle = 1 unit

Now, check whether point (6, 0) lie on the circle,

if x = 6, $6^2 + y^2 = 1$

$36 + y^2 = 1$

$y^2 = 1- 36$

$y= i\sqrt{35}\neq 0$

i.e., (6, 0) does not lie on the circle,

Hence, there is no tangent line of the given circle at (6, 0).

3 0

3 years ago

Find the missing measurement for each quadrilateral 5) this is rhombus ABCD

a_sh-v [17]

To find the area of a rhombus, multiply the lengths of the two diagonals and divide by 2 (same as multiplying by 1/2): The sides and angles of a rhombus: The sides of a rhombus are all congruent (the same length.) Opposite angles of a rhombus are congruent (the same size and measure.)

A square has two perpendicular bisectors but a square is not a rhombus (because a rhombus does not have all four angles = 90. Oh, but wait, a rhombus is a square in the same way a rectangle is a square but not the other way around.

4 0

3 years ago

What is the equation of the perpendicular bisector of line AB?

Usimov [2.4K]

It should be B

Because the slope of AB is 6/5 and when you are finding it’s perpendicular bisector then the slope of that line should be the negative reciprocal, it should be -5/6.

7 0

3 years ago

Read 2 more answers

What is 0.42 in a fraction in simplest form

RideAnS [48]

0.42 equals

- 42/100

- 21/50

The simplest form for 0.42 is 21/50

3 0

2 years ago

Type the correct answer in each box. A circle is centered at the point (5, -4) and passes through the point (-3, 2). The equatio

Galina-37 [17]

Answer:

$(x+ \boxed{-5})^2+(y+\boxed4)^2=\boxed{100}$

Step-by-step explanation:

Given:

Center of circle is at (5, -4).

A point on the circle is $(x_1,y_1)=(-3, 2)$

Equation of a circle with center $(h,k)$ and radius 'r' is given as:

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

Here, $(h,k)=(5,-4)$

Radius of a circle is equal to the distance of point on the circle from the center of the circle and is given using the distance formula for square of the distance as:

$r^2=(h-x_1)^2+(k-y_1)^2$

Using distance formula for the points (5, -4) and (-3, 2), we get

$r^2=(5-(-3))^2+(-4-2)^2\\r^2=(5+3)^2+(-6)^2\\r^2=8^2+6^2\\r^2=64+36=100$

Therefore, the equation of the circle is:

$(x-5)^2+(y-(-4))^2=100\\(x-5)^2+(y+4)^2=100$

Now, rewriting it in the form asked in the question, we get

$(x+ \boxed{-5})^2+(y+\boxed4)^2=\boxed{100}$

4 0

3 years ago