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3 years ago

Is the point (1,3) a solution to the linear equation 5x-9y=32?

Mathematics

2 answers:

baherus [9]3 years ago

8 0

No it does not (1,-3) does

dsp733 years ago

7 0

I need a second equation to solve this, something like x + y = 12 or something

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Which of these points lies on the circle with the center (2,3) and radius 2?

yan [13]

Answer:

The answer to your question is the letter A.

Step-by-step explanation:

Data

Center = (2, 3)

radius = 2

Process

1.- Find the equation of the line

(x - h)² + (y - k)² = r²

-Substitution

(x - 2)² + (y - 3)² = 2²

-Simplification

(x - 2)² + (y - 3)² = 4

-Evaluate the points in the equation

A. (4, 3)

(4 - 2)² + (3 - 3)² = 4

2² + 0 = 4

4 = 4 This point lies in the circle

B (-1, 0)

(-1 - 2)² + (0 - 3)² = 4

-3² + (-3)² = 4

9 + 9 = 4 This point is not part of the circle

C. (1, 3)

(1 - 2)² + (3 - 3)² = 4

(-1)² + (0)² = 4

1 = 4 This point is not part of the circle

D. (3, 4)

(3 - 2)² + (4 - 3)² = 4

1² + 1² = 4

2 = 4 This point is not part of the circle

5 0

3 years ago

Read 2 more answers

NASA launches a rocket at t = 0 seconds. Its height, in meters above sea-level, as a function of time is given by h ( t ) = − 4.

Oliga [24]

Answer:

$\displaystyle 1)48.2 \: \: \text{sec}$

$\rm \displaystyle 2)3021.6 \: m$

Step-by-step explanation:

<h3>Question-1:</h3>

so when flash down occurs the rocket will be in the ground in other words the elevation(height) from ground level will be 0 therefore,

to figure out the time of flash down we can set h(t) to 0 by doing so we obtain:

$\displaystyle - 4.9 {t}^{2} + 229t + 346 = 0$

to solve the equation can consider the quadratic formula given by

$\displaystyle x = \frac{ - b \pm \sqrt{ {b}^{2} - 4 ac} }{2a}$

so let our a,b and c be -4.9,229 and 346 Thus substitute:

$\rm\displaystyle t = \frac{ - (229) \pm \sqrt{ {229}^{2} - 4.( - 4.9)(346)} }{2.( - 4.9)}$

remove parentheses:

$\rm\displaystyle t = \frac{ - 229 \pm \sqrt{ {229}^{2} - 4.( - 4.9)(346)} }{2.( - 4.9)}$

simplify square:

$\rm\displaystyle t = \frac{ - 229 \pm \sqrt{ 52441- 4( - 4.9)(346)} }{2.( - 4.9)}$

simplify multiplication:

$\rm\displaystyle t = \frac{ - 229 \pm \sqrt{ 52441- 6781.6} }{ - 9.8}$

simplify Substraction:

$\rm\displaystyle t = \frac{ - 229 \pm \sqrt{ 45659.4} }{ - 9.8}$

by simplifying we acquire:

$\displaystyle t = 48.2 \: \: \: \text{and} \quad - 1.5$

since time can't be negative

$\displaystyle t = 48.2$

hence,

at 48.2 seconds splashdown occurs

<h3>Question-2:</h3>

to figure out the maximum height we have to figure out the maximum Time first in that case the following formula can be considered

$\displaystyle x _{ \text{max}} = \frac{ - b}{2a}$

let a and b be -4.9 and 229 respectively thus substitute:

$\displaystyle t _{ \text{max}} = \frac{ - 229}{2( - 4.9)}$

simplify which yields:

$\displaystyle t _{ \text{max}} = 23.4$

now plug in the maximum t to the function:

$\rm \displaystyle h(23.4)- 4.9 {(23.4)}^{2} + 229(23.4)+ 346$

simplify:

$\rm \displaystyle h(23.4) = 3021.6$

hence,

about 3021.6 meters high above sea-level the rocket gets at its peak?

5 0

3 years ago

Solve equation by factoring.

3241004551 [841]

Hi ;-)

$x^2+6x-10=30 \ \ /-30\\\\x^2+6x-40=0\\\\a=1, \ b=6, \ c=-40\\\\\Delta=b^2-4ac=6^2-4\cdot1\cdot(-40)=36+160=196\\\\\sqrt{\Delta}=\sqrt{196}=14\\\\x_1=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-6-14}{2}=\boxed{-10}\\\\x_2=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-6+14}{2}=\boxed4$

8 0

3 years ago

Complete the given table for the equation 4x+y=9

madam [21]

If x=1 and y=5 we get as 4product of 1+5=9
=4+5=9
9=9
lhs=rhs

4 0

4 years ago

What is the gcf of 27 and 36

mestny [16]

The answer would be 9. 9x3=27 and 9x4=36. Although 3 fits into both, 9 is the BIGGER number that also fits into both of the numbers shown in the problem.

5 0

3 years ago

Read 2 more answers