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3 years ago

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7. log(x+3)(2x + 3) +log(x+3)(x + 5) = 2

1 answer:

iren [92.7K]3 years ago

5 0

Answer:67

Step-by-step explanation: the first one is the best. that’s why

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Noel and Casey both start at the same place. Noel walks due south and Casey walks due east. After some time has passed, Noel is

Dovator [93]

Answer:

2.04 miles per hour

Step-by-step explanation:

Given

Noel

$n_1 =6miles$

$r_1 = 2mph$

Casey

$c_1 = 8miles$

$r_2 =1mph$

Required

The rate at which the distance increases

Their movement forms a right triangle and the distance between them is the hypotenuse.

At $n_1 =6miles$ and $c_1 = 8miles$

The distance between them is:

$d_1 = \sqrt{n_1^2 + c_1^2}$

$d_1 = \sqrt{6^2 + 8^2}$

$d_1 = \sqrt{100}$

$d_1 = 10miles$

After 1 hour, their new position is:

New = Old + Rate * Time

$n_2 = n_1 + r_1 * 1$

$n_2 = 6 + 2 * 1 = 8$

And:

$c_2 = c_1 + r_2 * 1$

$c_2 = 8 + 1 * 1 = 9$

So, the distance between them is now:

$d_2 = \sqrt{n_2^2 + c_2^2}$

$d_2 = \sqrt{8^2 + 9^2}$

$d_2 = \sqrt{145}$

$d_2 = 12.04$

The rate of change is:

$\triangle d = d_2 -d_1$

$\triangle d = 12.04 -10$

$\triangle d = 2.04$

5 0

3 years ago

A light bulb has a lifetime X that is exponentially distributed with a mean 340 hours. Find the probability that the bulb lifeti

LUCKY_DIMON [66]

Answer:

0.7026

Step-by-step explanation:

Let X denote the lifetime of light bulb. Given $X \sim Exp(\lambda)$ where the mean is $E(X) = 340 =\frac{1}{\lambda} \implies \lambda = \frac{1}{340} = 0.00294$ .

Recall that,

$\displaystyle P(X>x) = 1 - \int_0^x e^{-\lambda x} = 1 - (1 - e^{-\lambda x} = e^{-\lambda x}$

$\displaystyle P(X>220 | X>100) = \frac{P(X>220,X>100)}{P(X>100)} = \frac{e^{-\lambda \times 220}}{e^{-\lambda \times 100}} = \frac{0.5236}{0.7452} = 0.7026$

7 0

3 years ago

How to solve 9x^2+9x+2

Oksana_A [137]

Answer:

factoring: ( 3 x + 1 ) ( 3 x+ 2 )

expressing/equation: 9x²= 81x

81x + 9x + 2=

9x+81x=90x

90x + 2

Step-by-step explanation:

7 0

3 years ago

Look at the figure. Classify the pair of angles: <1 and <5

zavuch27 [327]

Answer:

where is the figure. We cant answer if we dont have the figure

7 0

3 years ago

Read 2 more answers

EXAMPLE 5 Find the maximum value of the function f(x, y, z) = x + 2y + 9z on the curve of intersection of the plane x − y + z =

geniusboy [140]

The Lagrangian,

$L(x,y,z,\lambda,\mu)=x+2y+9z-\lambda(x-y+z-1)-\mu(x^2+y^2-1)$

has critical points where its partial derivatives vanish:

$L_x=1-\lambda-2\mu x=0$

$L_y=2+\lambda-2\mu y=0$

$L_z=9-\lambda=0$

$L_\lambda=x-y+z-1=0$

$L_\mu=x^2+y^2-1=0$

$L_z=0$ tells us $\lambda=9$ , so that

$L_x=0\implies-8-2\mu x=0\implies x=-\dfrac4\mu$

$L_y=0\implies11-2\mu y=0\implies y=\dfrac{11}{2\mu}$

Then with $L_\mu=0$ , we get

$x^2+y^2=\dfrac{16}{\mu^2}+\dfrac{121}{4\mu^2}=1\implies\mu=\pm\dfrac{\sqrt{185}}2$

and $L_\lambda=0$ tells us

$x-y+z=-\dfrac4\mu-\dfrac{11}{2\mu}+z=1\implies z=1+\dfrac{19}{2\mu}$

Then there are two critical points, $\left(\pm\frac8{\sqrt{185}},\mp\frac{11}{\sqrt{185}},1\pm\frac{19}{\sqrt{185}}\right)$ . The critical point with the negative $x$ -coordinates gives the maximum value, $9+\sqrt{185}$ .

8 0

3 years ago