All categories

3 years ago

Solve The Equation 4x×9y=7 4x-9y=9

Mathematics

1 answer:

hoa [83]3 years ago

8 0

Answer:

$\large\boxed{x=\dfrac{9}{8}-\dfrac{\sqrt{109}}{8},\ y=-\dfrac{1}{2}-\dfrac{\sqrt{109}}{18}}\\or\\\boxed{x=\dfrac{9}{8}+\dfrac{\sqrt{109}}{2},\ y=-\dfrac{1}{2}+\dfrac{\sqrt{109}}{18}}$

Step-by-step explanation:

$\left\{\begin{array}{ccc}4x\times9y=7&(1)\\4x-9y=9&(2)\end{array}\right\\\\(2)\\4x-9y=9\qquad\text{subtract}\ 4x\ \text{from both sides}\\-9y=-4x+9\qquad\text{change the signs}\\9y=4x-9\qquad\text{substitute it to (1)}\\\\4x(4x-9)=7\qquad\text{use the distributive property}\ a(b+c)=ab+ac\\(4x)(4x)+(4x)(-9)=7\\(4x)^2-36x=7\\(4x)^2-2(4x)(4.5)=7\qquad\text{add}\ 4.5^2\ \text{to both sides}\\(4x)^2-2(4x)(4.5)+4.5^2=7+4.5^2\qquad\text{use}\ (a-b)^2=a^2-2ab+b^2$

$(4x-4.5)^2=7+20.25\\(4x-4.5)=27.25\to 4x-4.5=\pm\sqrt{27.25}\\\\4x-\dfrac{45}{10}=\pm\sqrt{\dfrac{2725}{100}}\\\\4x-\dfrac{45}{10}=\pm\dfrac{\sqrt{2725}}{\sqrt{100}}\\\\4x-\dfrac{45}{10}=\pm\dfrac{\sqrt{25\cdot109}}{10}\\\\4x-\dfrac{45}{10}=\pm\dfrac{\sqrt{25}\cdot\sqrt{109}}{10}\\\\4x-\dfrac{45}{10}=\pm\dfrac{5\sqrt{109}}{10}\qquad\text{add}\ \dfrac{45}{10}\ \text{to both sides}\\\\4x=\dfrac{45}{10}\pm\dfrac{5\sqrt{109}}{10}$

$4x=\dfrac{9}{2}\pm\dfrac{\sqrt{109}}{2}\qquad\text{divide both sides by 4}\\\\x=\dfrac{9}{8}\pm\dfrac{\sqrt{109}}{8}\\\\\text{Put the values of}\ x\ \text{to (2):}\\\\9y=4\left(\dfrac{9}{8}\pm\dfrac{\sqrt{109}}{8}\right)-9\\\\9y=\dfrac{9}{2}\pm\dfrac{\sqrt{109}}{2}-\dfrac{18}{2}\\\\9y=-\dfrac{9}{2}\pm\dfrac{\sqrt{109}}{2}\qquad\text{divide both sides by 9}\\\\y=-\dfrac{1}{2}\pm\dfrac{\sqrt{109}}{18}$

You might be interested in

5 gallon = 0.6685 cubic feet how many gallon is 1 cubic foot?

gladu [14]

7.48 gallons. Vote me brainliest

8 0

3 years ago

Read 2 more answers

Backgammon is a board game for two players in which the playing pieces are moved according to the roll of two dice. players win

Ivahew [28]

The rolls of the dice are independent, i.e. the outcome of the second die doesn't depend in any way on the outcome of the first die.

In cases like this, the probability of two events happening one after the other is the multiplication of the probabilities of the two events.

So, the probability of rolling two 6s is the multiplication of the probabilities of rolling a six with the first die, and another six with the second:

$P(\text{rolling two 6s}) = P(\text{rolling a 6}) \cdot P(\text{rolling a 6}) = \dfrac{1}{6} \cdot \dfrac{1}{6} = \dfrac{1}{36}$

Similarly,

$P(\text{rolling two 3s}) = P(\text{rolling a 3}) \cdot P(\text{rolling a 3}) = \dfrac{1}{6} \cdot \dfrac{1}{6} = \dfrac{1}{36}$

Actually, you can see that the probability of rolling any ordered couple is always 1/36, since the probability of rolling any number on both dice is 1/6:

$P(\text{rolling any ordered couple}) = P(\text{rolling the first number}) \cdot P(\text{rolling the second number}) = \dfrac{1}{6} \cdot \dfrac{1}{6} = \dfrac{1}{36}$

7 0

3 years ago

liubo4ka [24]

Answer:D 4 units

Enjoy

4 0

2 years ago

A boat is 60m from the base of a lighthouse. The angle of depression between the lighthouse and the boat is 37°. How tall is the

LekaFEV [45]

Answer: 34.64 m

Step-by-step explanation:

Given: A boat is 60 m from the base of a lighthouse.

The angle of depression between the lighthouse and the boat is 37°.

By using trigonometric ratios :

$\tan x=\dfrac{\text{Side opposite to }x}{\text{Side adjacent to }x}$

here x= 37°, side opposite to x = height of lighthouse (h) , side adjacent to x = 60 m

$\tan 37^{\circ}=\dfrac{h}{60}\\\\\Rightarrow\ 0.57735=\dfrac{h}{60}\\\\\Rightarrow\ h= 60\times0.57735\approx34.64$

Hence, the lighthouse is 34.64 m tall.

6 0

3 years ago

What is the horizontal asymptote for y(t) for the differential equation dy dt equals the product of 2 times y and the quantity 1

marta [7]

First, we need to solve the differential equation.
$\frac{d}{dt}\left(y\right)=2y\left(1-\frac{y}{8}\right)$
This a separable ODE. We can rewrite it like this:
$-\frac{4}{y^2-8y}{dy}=dt$
Now we integrate both sides.
$\int \:-\frac{4}{y^2-8y}dy=\int \:dt$
We get:
$\frac{1}{2}\ln \left|\frac{y-4}{4}+1\right|-\frac{1}{2}\ln \left|\frac{y-4}{4}-1\right|=t+c_1$
When we solve for y we get our solution:
$y=\frac{8e^{c_1+2t}}{e^{c_1+2t}-1}$
To find out if we have any horizontal asymptotes we must find the limits as x goes to infinity and minus infinity.
It is easy to see that when x goes to minus infinity our function goes to zero.
When x goes to plus infinity we have the following:
$$$\lim_{x\to\infty} f(x)$$=y=\frac{8e^{c_1+\infty}}{e^{c_1+\infty}-1} = 8$
When you are calculating limits like this you always look at the fastest growing function in denominator and numerator and then act like they are constants.
So our asymptote is at y=8.

3 0

3 years ago