A car garage fits 17 rows of 30 cars on each of the 3 floors. How many cars are there if the car park is full?

Answer:

Step-by-step explanation:

tan 45=1

[tex]tan 2\alpha =\frac{2tan\alpha}{1-tan^2\alpha } \\put~\alpha =22.5\\2\alpha =2 \times 22.5=45\\tan~2\alpha =tan~45=1\\1=\frac{2tan \alpha }{1-tan^2\alpha} \\1-tan^2\alpha =2 tan\alpha \\ tan^2\alpha +2tan \alpha -1=0\\tan\alpha =\frac{-2 \pm\sqrt{4+4} }{2} \\=-1\pm\sqrt{2} \\\alpha =22.5\\so ~tan~\alpha ~is~positive.\\so ~tan~\alpha =\sqrt{2} -1[/tex]

The error happens in the last line.

The relevant log rule is ln(A) - ln(B) = ln(A/B)

The issue is that the 0 isn't ln(0), so that 0 cannot be thrown into the natural log like that.

What we can do is this set of steps

ln(3x) - 0

ln(3x) - ln(1)

ln(3x/1) ... applying the log rule mentioned

ln(3x)

But these steps are a needlessly overcomplicating things. We can simply go from ln(3x)-0 to ln(3x) in one step. Subtracting off zero doesn't change ln(3x) at all.

--------------------------

Let's continue the steps to solve for x.

ln(x^2) = ln(3x)

x^2 = 3x

x^2-3x = 0

x(x-3) = 0

x = 0 or x-3 = 0

x = 0 or x = 3

Those are the two possible solutions.

However, x = 0 isn't valid because it's not in the domain of ln(x^2). In other words, ln(0) isn't defined. The domain of y = ln(x) is x > 0

This means that x = 3 is the only solution.

Answer:

Step-by-step explanation:

Answer:

-1

Step-by-step explanation:

3-1=2

2-1=1

Answer:

A = 50x^2+3

B = 75

C = 55

Step-by-step explanation:

[tex]f(x)=2x^2+3\\g(x)=5x[/tex]

A. f(g(x))

[tex]2(5x)^2+3\\2(25x^2)+3\\=50x^2+3[/tex]

B. g(g(3))

[tex]5(5(3))\\5(15)\\=75[/tex]

C. g(f(-2))

[tex]g(2(-2)^2+3)\\g(8+3)\\g(11)\\5(11)\\=55[/tex]

each number under the leaves section would be a dog

answer: 11

Answer:

11

Step-by-step explanation:

Answer:

principal = $400

Step-by-step explanation:

amount(a)=principal(p) + interest(y).........(1)

Interest(y)=principal(p) x rate(r) x time(t)....(2)

In (1), p= a - y...........(3)

putting (3) in (2),

y = (a - y) x r% x t

y = (432 - y) x 8% x 1

y = 3456%-8y%

y + 0.08y = 3456%

1.08y = 34.56

y = 32

but a - y = p

432 - 32 = p

p = 400

principal = $400

Answer:

[tex]\textsf{B.}\quad -2(\sqrt{2}-1)[/tex]

Step-by-step explanation:

Given integral:

[tex]\displaystyle \int^{\pi}_{\frac{\pi}{2}}\dfrac{\cos \theta}{\sqrt{1+ \sin \theta}}\:\:d\theta[/tex]

Solve by using Integration by Substitution

Substitute u for one of the functions of [tex]\theta[/tex] to give a function that's easier to integrate.

[tex]\textsf{Let }u=1+\sin \theta[/tex]

Find the derivative of u and rewrite it so that [tex]d \theta[/tex] is on its own:

[tex]\implies \dfrac{du}{d \theta}=\cos \theta[/tex]

[tex]\implies d \theta=\dfrac{1}{\cos \theta}\:du[/tex]

Use the substitution to change the limits of the integral from [tex]\theta[/tex]-values to u-values:

[tex]\textsf{When }\theta=\pi \implies u=1[/tex]

[tex]\textsf{When }\theta=\dfrac{\pi}{2} \implies u=2[/tex]

Substitute everything into the original integral and solve:

[tex]\begin{aligned}\displaystyle \int^{\pi}_{\frac{\pi}{2}}\dfrac{\cos \theta}{\sqrt{1+ \sin \theta}}\:\:d\theta & =\int^{1}_2}\dfrac{\cos \theta}{\sqrt{u}}\:\cdot \dfrac{1}{\cos \theta}\:\:du\\\\& =\int^{1}_{2}\dfrac{1}{\sqrt{u}} \:\:du \\\\& =\int^{1}_{2} u^{-\frac{1}{2}}\:\:du \\\\& = \left[ 2u^{\frac{1}{2}} \right]^{1}_{2}\\\\& = \left(2(1)^{\frac{1}{2}}\right)-\left(2(2)^{\frac{1}{2}}\right)\\\\& = 2-2\sqrt{2}\\\\& = -2(\sqrt{2}-1)\end{aligned}[/tex]

Reason:

sorry but here is no options so I am giving you a example so that you can understand by your own!

Answer;

5r+3

2x-6

X4+10

4y-1

Step by step answer;

So the method of identifying the Monomail is if you see a variable in a constant so it is a Monomail !

Answer:

Step-by-step explanation:

Comment

Either you know the answer to this or you don't. There is no point in eliminating the wrong answers until you get to the right one. It is better to get to the right one immediately.

A rhombus and its diagonals have many interesting properties.  This one is one of those properties. The diagonals are not the same length. The diagonals bisect each other at right angles.

Answer: B

Answer:

10

Step-by-step explanation:

use PEMDAS

PEMDAS order is Parentheses, Exponents, Multiplication and Division (from left to right), Addition and Subtraction (from left to right).

2[32 — (4 − 1)³] =

2[32 — (3)³] =

2[32 — 27] =

2[5] =

10

khanacademy

Answer: i think no?

Step-by-step explanation:

AD and BE are both parallel lines (they are parallel to eachother), so they will never intersect

CF and GF intersect at point F (i think)

Bg and Ac intersect at point B (i think)

i dont want to give a definite answer in the event im wrong bc I just learned this like a few weeks ago-

Based on the given mean absolute deviations, the sample mean absolute deviation will be 2 and the population mean absolute deviation will be 3.8.

= Sum of sample mean absolute deviation / Number

= (0 + 0.5 + 4.5 + 0.5 + 0 + 4 + 4.5 + 4 + 0) / 9

= 2

First find the population mean:

= (10 + 11 + 19) / 3

= 13.3

Population mean absolute deviation:
= 1/3 x ( (10 - 13.67) + (11 - 13.3) + (19 - 13.3))

= 3.8

We can therefore see that the sample mean deviations is a biased estimate of the population mean absolute deviation.

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The series diverges. You Consider concluding this because the series is in arithmetic progression.

An arithmetic sequence is a sequence of integers with its adjacent terms differing with one common difference.

The explicit formula for any arithmetic series is given by the formula,

[tex]a_n = a_1 + (n-1)d[/tex]

where d is the difference and a₁ is the first term of the sequence.

The given series of numbers 1/4+3/2+11/4+4+21/4+.... is an arithmetic series because the difference between any two consecutive terms is 5/4. Therefore, the blanks can be filled as,

The series diverges. You Consider conclude this because the series is in arithmetic progression.

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The expected enrolment in 7 years is  1845.

The formula for calculating future value of the number of students is an exponential equation with this form:

FV = P (1 + r) ^n

1500 x (1.03)^7 = 1845

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Answer:

1/12  mile per minute

Step-by-step explanation:

1/4 mile in 3 minutes

 then miles per minute   is   1/4 mile / 3 min = 1/12  mile/min

Answer: two congruent segments.

Step-by-step explanation:

idlk im not smart

When Ф lies in quadrant IV the value of cos Ф lies in [0, 1].

One fourth of the part of a circle is said to be the quadrant of the circle.

Given the angle Ф lies in quadrant IV.

As we know, cos Ф is positive only in first and fourth quadrant i.e when Ф lies in 0<Ф<90 and 270<Ф<360.

cosФ range from [-1,1]

So, cos Ф is positive when Ф lies in 1st and 4th quadrant.

&  cos Ф is negative when Ф lies in 2nd and 3rd quadrant.

If cos ∅ lies in the quadrant IV, anything between [0, 1] can be the value of cos ∅.

Hence, when Ф lies in quadrant IV the value of cos Ф lies in [0, 1].

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Answer:

[tex]-\dfrac{69}{14}[/tex]

Step-by-step explanation:

Substitute the given values of the variables into the given expression:

[tex]\implies \dfrac{60 \left(\dfrac{1}{3}\right)\left(-\dfrac{1}{5}\right)\left(\dfrac{1}{4}\right)\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{4}\right)}{\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\left(\dfrac{1}{3}+\dfrac{1}{4}\right)}[/tex]

When multiplying fractions, simply multiply the numerators and the denominators:

[tex]\implies \dfrac{60 \left(\dfrac{1 \cdot -1 \cdot 1}{3\cdot 5 \cdot 4}\right)\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{4}\right)}{\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\left(\dfrac{1}{3}+\dfrac{1}{4}\right)}[/tex]

[tex]\implies \dfrac{60 \left(-\dfrac{1}{60}\right)\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{4}\right)}{\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\left(\dfrac{1}{3}+\dfrac{1}{4}\right)}[/tex]

[tex]\implies \dfrac{\left(-1\right)\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{4}\right)}{\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\left(\dfrac{1}{3}+\dfrac{1}{4}\right)}[/tex]

When adding or subtracting fractions, make the denominators the same:

[tex]\implies \dfrac{\left(-1\right)\left(\dfrac{20}{60}-\dfrac{12}{60}+\dfrac{15}{60}\right)}{\left(\dfrac{5}{15}-\dfrac{3}{15}\right)\left(\dfrac{4}{12}+\dfrac{3}{12}\right)}[/tex]

Now add the numerators and put that over the denominator:

[tex]\implies \dfrac{\left(-1\right)\left(\dfrac{20-12+15}{60}\right)}{\left(\dfrac{5-3}{15}\right)\left(\dfrac{4+3}{12}\right)}[/tex]

[tex]\implies \dfrac{\left(-1\right)\left(\dfrac{23}{60}\right)}{\left(\dfrac{2}{15}\right)\left(\dfrac{7}{12}\right)}[/tex]

Again, multiply the fractions:

[tex]\implies \dfrac{\left(-1 \cdot \dfrac{23}{60}\right)}{\left(\dfrac{2 \cdot 7}{15 \cdot 12}\right)}[/tex]

[tex]\implies \dfrac{-\dfrac{23}{60}}{\dfrac{14}{180}}[/tex]

To divide the fractions, turn the second fraction upside down, then multiply it by the first fraction:

[tex]\implies -\dfrac{23}{60} \cdot \dfrac{180}{14}[/tex]

[tex]\implies -\dfrac{4140}{840}[/tex]

Finally, simplify:

[tex]\implies -\dfrac{4140 \div 60}{840 \div 60}[/tex]

[tex]\implies -\dfrac{69}{14}[/tex]

The figures show that the number of births for Florida and Michigan reduced

From the data sourced, the number of births in Florida in 2010 was 214519 while in 2011, it was 213237. For Michigan, there was decrease from 114717 to 114159.

When a person born during 2011 in the United States is chosen randomly, the person is more likely to have been born in Florida.

The probability that a randomly selected us resident in 2011 was living in Florida will be:

= 18,801,310/308,745,538

= 0.061

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Answer:

w = -1/5

OR

w = -3

Step-by-step explanation:

Given equation:

−16w-3 = 5w²

Solution:

Subtracting 5w^2 from both sides,we get

Factor the LHS of this equation using middle term factor:

Now,

Solving for equation 1:

Solving for equation 2:

[tex] - 16w - 3 = 5 {w}^{2} \\ \\ 0 = 5 {w}^{2} + 16w + 3 \\ \\ 5 {w}^{2} + 16w + 3 = 0 \\ \\ 5 {w}^{2} + w + 15w + 3 = 0 \\ \\ (5 {w}^{2} + w) + (15w + 3) = 0 \\ \\ w(5w + 1) + 3(5w + 1) = 0 \\ \\ (w + 3)(5w + 1) = 0. [/tex]

The value of w is -3 and -1/5 .

Answer:.20

Step-by-step explanation:

According to what are written in the table the probability for x=-10 is .20

The number of cubes needed to fill the rectangular prism 4608 cubes.

volume of rectangular prism = lwh

where

Therefore,

volume of the rectangular prism = 8 × 4 × 9 / 4 = 72 inches³

volume of the cube = l³

where

Therefore,

volume = (1 / 4)³

volume of the cube = 1  / 64 inches³

Therefore,

The number of cubes needed to fill the rectangular prism = 72 ÷ 1 / 64

The number of cubes needed to fill the rectangular prism = 72 × 64 = 4608

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Answer: Choice D) set of even integers

=======================================================

Explanation:

U = universal set = {all positive integers} = {1, 2, 3, 4, 5, ...}

A = {stuff in set U that is odd} = {1, 3, 5, 7, 9, ...}

[tex]A^c[/tex] = {stuff in set U but NOT from set A} = {2, 4, 6, 8, ...}

[tex]A^c[/tex] = {set of even numbers from set U}

In set theory, the complement is simply the complete opposite. If the number 7 is found in set A for instance, then 7 is not going to live in the complement [tex]A^c[/tex]

The opposite of the odd numbers is the set of even numbers.

Therefore, we go for choice D as our final answer.

Side note: if we union set A with its complement [tex]A^c[/tex] then we get the universal set as a result. [tex]A \cup A^c = \text{universal set} = \text{set of all positive integers}[/tex]

The solution is Option D.

Ac =  {x|x ∈ U and is an even positive integer} where Ac represents the complement of set A

What is union and intersection of sets?

The union of two sets A and B is the set of all those elements which are either in A or in B, i.e. A ∪ B, whereas the intersection of two sets A and B is the set of all elements which are common. The intersection of these two sets is denoted by A ∩ B

The union of two sets is a new set that contains all of the elements that are in at least one of the two sets

The intersection of two sets is a new set that contains all of the elements that are in both sets

Given data ,

Let the set U be = {all positive integers}

So , the values of set U = { 1 , 2 , 3 , 4 , 5 .. }

And , let the set A = {x|x ∈ U and x is an odd positive integer}

So , the value of set A = { 1 , 3 , 5 , 7 , ... }

And , the complement of set  A is Ac

The value of set Ac is all the even positive odd integers

So , the value of set Ac = { 2 , 4 , 6 , 8 .. }

Therefore , the value of set Ac = { 2 , 4 , 6 , 8 .. }

Hence , the set Ac = {x|x ∈ U and is an even positive integer}

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Answer:

     [tex]\frac{15*7^{x} }{7^{x} +9}[/tex]

Step-by-step explanation:

→ Substitute f(x) into g(x)

     [tex]\frac{15*7^{x} }{7^{x} +9}[/tex]

Answer:

Step-by-step explanation:

The last one: 79, 79, 79, 79, 79, 79, 79, 79, 79, 79 because there isn't any difference in the numbers.

S.D.: 0