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3 years ago

Which equation represents a circle with the same center as

Mathematics

2 answers:

Stolb23 [73]3 years ago

7 0

Answer:

(x - 4)² + (y - 5)² = 4

Step-by-step explanation:

The equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k) are the coordinates of the centre and r is the radius

Given

(x - 4)² + (y - 5)² = 2

with centre (4, 5) and r = $\sqrt{2}$

The comparable equation with equivalent left side and r = 2 is

(x - 4)² + (y - 5)² = 4

ale4655 [162]3 years ago

6 0

Answer:

Yeah It's B

Step-by-step explanation:

Read more about common factors at

brainly.com/question/219464

#SPJ1

3 0

1 year ago

Two trains leave Raleigh at the same time one traveling north and the other south the first train travels at 50 miles per hour a

Elis [28]

Answer:

<h2>2.5 hours</h2>

Step-by-step explanation:

Step one:

given data

for train one

speed= 50miles per hour

for train two

speed= 60miles per hour

the total distance between both trains=275miles

Step two:

the expression for speed= distance/time

so that, distance= speed*time

let the time be t

we are going to combine both distances

train one distance =50t

train two distance =60t

Hence total distance

50t+60t=275

110t=1=275

t=275/110

t=2.5 hours

<u>It will take 2.5 hours</u>

4 0

3 years ago

Please could you find the answers to the questions in the attachment.

Fudgin [204]

( $\frac{x1+x2}{2} , \frac{y1+y2}{2}$ )we need 3 equations
1. midpoint equation which is ( $\frac{x1+x2}{2} , \frac{y1+y2}{2}$ ) when you have 2 points

2. distance formula which is D= $\sqrt{(x2-x1)^{2}+(y2-y1)^{2}}$

3. area of trapezoid formula whhic is (b1+b2) times 1/2 times height

so

x is midpoint of B and C
B=11,10
c=19,6
x1=11
y1=10
x2=19
y2=6
midpoint=( $\frac{11+19}{2} , \frac{10+6}{2}$ )
midpoint=( $\frac{30}{2} , \frac{16}{2}$ )
midpoint= (15,8)

point x=(15,8)

y is midpoint of A and D
A=5,8
D=21,0
x1=5
y1=8
x2=21
y2=0
midpoint=( $\frac{5+21}{2} , \frac{8+0}{2}$ )
midpoint=( $\frac{26}{2} , \frac{8}{2}$ )
midpoint=(13,4)

Y=(13,4)

legnths of BC and XY
B=(11,10)
C=(19,6)
x1=11
y1=10
x2=19
y2=6
D= $\sqrt{(19-11)^{2}+(6-10)^{2}}$
D= $\sqrt{(8)^{2}+(-4)^{2}}$
D= $\sqrt{64+16}$
D= $\sqrt{80}$
D= $4 \sqrt{5}$
BC= $4 \sqrt{5}$

X=15,8
Y=(13,4)
x1=15
y1=8
x2=13
y2=4
D= $\sqrt{(13-15)^{2}+(4-8)^{2}}$
D= $\sqrt{(-2)^{2}+(-4)^{2}}$
D= $\sqrt{4+16}$
D= $\sqrt{20}$
D= $2 \sqrt{5}$
XY= $2 \sqrt{5}$

the thingummy is a trapezoid
we need to find AD and BC and XY
we already know that BC= $4 \sqrt{5}$ and XY= $2 \sqrt{5}$

AD distance
A=5,8
D=21,0
x1=5
y1=8
x2=21
y2=0
D= $\sqrt{(21-5)^{2}+(0-8)^{2}}$
D= $\sqrt{(16)^{2}+(-8)^{2}}$
D= $\sqrt{256+64}$
D= $\sqrt{320}$
D= $4 \sqrt{2}$
AD= $4 \sqrt{2}$

so we have
AD= $4 \sqrt{2}$
BC= $4 \sqrt{5}$
XY= $2 \sqrt{5}$

AD and BC are base1 and base 2
XY=height
so
(b1+b2) times 1/2 times height
( $4 \sqrt{2}+4 \sqrt{5}$ ) times 1/2 times $2 \sqrt{5}$ =
( $4 \sqrt{2}+4 \sqrt{5}$ ) times \sqrt{5} [/tex] =
$4 \sqrt{10}+4*5$ = $4 \sqrt{10}+20$ =80 $\sqrt{10}$ =252.982

X=(15,8)
Y=(13,4)
BC= $4 \sqrt{5}$
XY= $2 \sqrt{5}$
Area=80 $\sqrt{10}$ square unit or 252.982 square units

7 0

3 years ago

Factor 10x^3 8x^2 - 6x. (<br> a.2x(5x 4x - 3) (<br> b.2x(5x 4x^2- 3) (<br> c.2x(5x^2 4x - 3)

Semmy [17]

You have to look at the equation and find a common factor bettween all parts. SO, we see they all have an x, so that is the first thing you can factor out. They also have
2x(5x2 4x -3) C

4 0

3 years ago

Kelly can type 120 words in 3 minutes. at that ate, how many words can she type in 5 minutes?

madam [21]

$\frac{120\ \text{words}}{3\ \text{minutes}} = \frac{40\ \text{words}}{1\ \text{minute}}$

$\frac{40\ \text{words}}{1\ \text{minute}} \times \frac{5}{5} = \frac{200\ \text{words}}{5\ \text{minutes}}$

Kelly can type 200 words in five minutes.

5 0

3 years ago