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tankabanditka [31]

2 years ago

11

Circle J is congruent to circle P. Circles J and P are congruent. Chords O K and Q R are congruent and chords K M and R T are co

ngruent. If OK = 2x, QR = 12, and RT = x + 3, what is the length of chord KM? 8 9 10 12

2 answers:

aniked [119]2 years ago

6 0

Answer: 9

Step-by-step explanation:

Molodets [167]2 years ago

4 0

Answer: 9

Step-by-step explanation:

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This is a Differential Equations problem, I only need help on part 1 from question five. I need steps as well, thank you.

jolli1 [7]

Answer:

a) $y(t) = \sqrt{2t + 1 }$

b) 1.5 hours after thickness will be 2 inches.

Step-by-step explanation:

$\frac{dy}{dt} = \frac{1}{y} \\ \\ ydy = dt \\ \\ integrating \: both \: sides \\ \\ \int y \: dy = \int 1 \: dt \\ \\ \frac{ {y}^{2} }{2} = t + c \\ \\ {y}^{2} = 2t + 2c \\ \\ y (t)= \sqrt{2t + 2c} ...(1) \\ \\ a) \: \: \: plug \: t = 0 \: in \: (1) \\ \\ y (0)= \sqrt{2(0) +2 c} \\ \\ y (0)= \sqrt{0 +2 c} \\ \\ 1 = \sqrt{2c} \: \: ( \because \: y (0)=1) \\ \\ 2c = 1 \: \implies \: c = \frac{1}{2} \\ \\ plug \: c = \frac{1}{2} \: in \: (1) \\ \\ y(t) = \sqrt{2t + 2 \times \frac{1}{2} } \\ \\ \huge \: \red {y(t) = \sqrt{2t + 1 } } \\ \\b) \: \: plug \: y(t) = 2 \: in \: above \: equation \\ \\ 2 = \sqrt{2t + 1} \\ \\ 4 = 2t + 1 \: \\ (squaring \: both \: sides) \\ \\ 4 - 1 = 2t \\ \\ 2t = 3 \\ \\ t = \frac{3}{2} \\ \\ t = 1.5 \: hours \\$

6 0

2 years ago

The vertical change between 2 points.

fredd [130]

Answer:

what does it look like we need that

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

lord [1]

Answer:

y +2 = 4(x -3)

Step-by-step explanation:

Point-slope form of the equation of a line with slope m through point (h, k) is ...

y -k = m(x -h)

For m = 4 and (h, k) = (3, -2), the equation is ...

y +2 = 4(x -3) . . . . . matches the first choice

3 0

3 years ago

Drag the tiles to the correct boxes to complete the pairs not all tiles will be used match each quadratic graph to its respectiv

ch4aika [34]

Answer:

Part 1) The function of the First graph is $f(x)=(x-3)(x+1)$

Part 2) The function of the Second graph is $f(x)=-2(x-1)(x+3)$

Part 3) The function of the Third graph is $f(x)=0.5(x-6)(x+2)$

See the attached figure

Step-by-step explanation:

we know that

The quadratic equation in factored form is equal to

$f(x)=a(x-c)(x-d)$

where

a is the leading coefficient

c and d are the roots or zeros of the function

Part 1) First graph

we know that

The solutions or zeros of the first graph are

x=-1 and x=3

The parabola open up, so the leading coefficient a is positive

The function is equal to

$f(x)=a(x-3)(x+1)$

Find the value of the coefficient a

The vertex is equal to the point (1,-4)

substitute and solve for a

$-4=a(1-3)(1+1)$

$-4=a(-2)(2)$

$a=1$

therefore

The function is equal to

$f(x)=(x-3)(x+1)$

Part 2) Second graph

we know that

The solutions or zeros of the first graph are

x=-3 and x=1

The parabola open down, so the leading coefficient a is negative

The function is equal to

$f(x)=a(x-1)(x+3)$

Find the value of the coefficient a

The vertex is equal to the point (-1,8)

substitute and solve for a

$8=a(-1-1)(-1+3)$

$8=a(-2)(2)$

$a=-2$

therefore

The function is equal to

$f(x)=-2(x-1)(x+3)$

Part 3) Third graph

we know that

The solutions or zeros of the first graph are

x=-2 and x=6

The parabola open up, so the leading coefficient a is positive

The function is equal to

$f(x)=a(x-6)(x+2)$

Find the value of the coefficient a

The vertex is equal to the point (2,-8)

substitute and solve for a

$-8=a(2-6)(2+2)$

$-8=a(-4)(4)$

$a=0.5$

therefore

The function is equal to

$f(x)=0.5(x-6)(x+2)$

3 0

3 years ago

PLS HELP I NEED HELP

Vinvika [58]

Answer:

yo dont click that link hes a scammer thats an ip grabber

4 0

2 years ago