In a cave, a stalactite gets 44 millimeters longer each year. this year it is 72 centimeters long. how many years until it is 1 meter long?

Answer:

Let the number be X

42 + 2x/4 =2X

cross multiply

42+2X=8x

42=8X-2x

42=6X

6X=42

X=7

Step-by-step explanation:

I'm just a smart guy

Answer:

x = 3
x = -6

Step-by-step explanation:

Hello!

Let's first set the equation to 0.

To factor, first think about two numbers that multiply to -18, but add up to 3. The two numbers are 6 and -3.

Now, expand 3x to 6x and -3x. Then, factor by grouping.

Now, set each factor to 0 and solve for x

x = 3, x = -6

Answer:

8.0

Step-by-step explanation:

The triangle is a right triangle, so we are able to use trigonometric functions. Relative to the angle of 37°, we have the opposite side which is 6 and the adjacent side which is x. The trig function that uses the opposite and adjacent is tangent(SohCahToa). We can set up the following equation:

[tex]tan(37) = \frac{opposite}{adjacent}[/tex]

[tex]tan(37) = \frac{6}{x}[/tex]

tan(37) evaluates to about , so we can plug it in and solve for x:

[tex]0.7536 = \frac{6}{x} \\0.7536x = 6\\x = 7.962[/tex]

To the nearest tenth, x rounds to 8.0.

Answer:

[tex]\sf 59 \frac{17}{20} \: in^3[/tex]

Step-by-step explanation:

Regular-size box of crackers

Dimensions:  [tex]\sf 2 \frac{1}{4} \: in \times 9 \frac{1}{2}\:in \times 14\: in[/tex]

Snack-size box of crackers

Volume = [tex]\sf \frac{1}{5}[/tex] of the volume of a regular-size box

To calculate the volume of the snack-size box of crackers, calculate the volume of the regular-size box then multiply it by [tex]\sf \frac{1}{5}[/tex].

[tex]\begin{aligned}\textsf{Volume of a rectangular prism} & = \textsf{width x length x height}\\\\\implies \textsf{Volume of regular-size box} & = \sf 9 \frac{1}{2} \times 14 \times 2 \frac{1}{4}\\\\& = \sf \dfrac{19}{2} \times 14 \times \dfrac{9}{4}\\\\& = \sf \dfrac{19 \times 14 \times 9}{2 \times 4}\\\\& = \sf \dfrac{1197}{4}\: in^3\end{aligned}[/tex]

[tex]\begin{aligned}\implies \textsf{Volume of snack-size box} & = \sf \dfrac{1}{5} \times \textsf{volume of regular-size box}\\\\& = \sf \dfrac{1}{5} \times \dfrac{1197}{4}\\\\& = \sf \dfrac{1 \times 1197}{5 \times 4}\\\\& = \sf \dfrac{1197}{20}\\\\& = \sf 59 \dfrac{17}{20}\: in^3\end{aligned}[/tex]

Therefore, the volume of the snack-size box is [tex]\sf 59 \frac{17}{20} \: in^3[/tex]

Answer:

Hi,

Step-by-step explanation:

[tex](fog)(x)=g(f(x))=g(3x+2)=(3x+2)^2+1=9x^2+6x+5[/tex]

Answer:

hi

Step-by-step explanation:

Answer:

Formula = 2nr

= 2(3,14 + 3 + 4)

= 2(10,14)

= 20,28

The water consumption based on the information for three Months will be 22267m³.

The total water consumption for the three months will be:

= 7219 + 7390 + 7658

= 22267m³

In conclusion, the information simply want you to add the values of the water given.

Therefore, the water consumption based on the information for three Months will be 22267m³.

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Answer:

-10 is the opposite of 10

Step-by-step explanation:

-10 is the opposite of 10 ,because The sum of a number and its opposite equals to zero.

Answer:

Square root 75

Step-by-step explanation:

You can just enter it in the calculator to see which is equal to 5 square root 3

The ratio of the wavelengths of the two notes is 2:1 option fourth 2 to 1 is correct.

The wavelength is the distance between two successive troughs or crests. The highest point on the wave is the crest, while the lowest point is the trough.

We have two notes

Number of cycle for blue = 8

Number of cycle for red = 4

Ratio = 8:4 = 2:1

Or

2 to 1

Thus, the ratio of the wavelengths of the two notes is 2:1 option fourth 2 to 1 is correct.

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Answer:

y=2/7x+34/7

Step-by-step explanation:

the slope of this line is 2/7.

y=2/7x+?

? is the y-intercept.

Looking at the line that we have graphed, we can see that it intersects at  34/7 y.

the equation for the line that goes through 4,6 and -10, 2 is

y=2/7x + 34/7

The length of the segment HI in the figure is 32.9

To do this, we make use of the following secant-tangent equation:

HI² = KI * JI

From the figure, we have:

KI = 21 + 24 = 45

JI = 24

So, we have:

HI² = 45 * 24

Evaluate the product

HI² = 1080

Take the square root of both sides

HI = 32.9

Hence, the length of the segment HI is 32.9

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Answer:

  7

Step-by-step explanation:

We want to find the number 4-digit of positive integers n such that removing the thousands digit divides the number by 9.

__

Let the thousands digit be 'd'. Then we want to find the integer solutions to ...

  n -1000d = n/9

  n -n/9 = 1000d . . . . . . add 1000d -n/9

  8n = 9000d . . . . . . . . multiply by 9

  n = 1125d . . . . . . . . . divide by 8

The values of d that will give a suitable 4-digit value of n are 1 through 7.

When d=8, n is 9000. Removing the 9 gives 0, not 1000.

When d=9, n is 10125, not a 4-digit number.

There are 7 4-digit numbers such that removing the thousands digit gives 1/9 of the number.

Total electrons

Answer:

Given:

To calculate how many electrons have the same mass as one dust particle, divide the mass of a dust particle by the mass of an electron:

[tex]\begin{aligned}\implies \dfrac{\textsf{Mass of a dust particle}}{\textsf{Mass of an electron}} & = \sf \dfrac{7.5 \times 10^{-10}}{9.1 \times 10^{-31}}\\\\& = \sf \dfrac{7.5}{9.1} \times \dfrac{10^{-10}}{10^{-31}}\\\\& = \sf 0.8241... \times 10^{(-10-(-31))}\\\\& = \sf 0.8241... \times 10^{21}\\\\& = \sf 8.2 \times 10^{20}\end{aligned}[/tex]

Therefore, approximately [tex]\sf 8.2 \times 10^{20}[/tex] electrons have the same mass as one dust particle.

Answer:   18.7346314730578

This value is approximate.

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

Explanation:

Use the law of sines to find side b

sin(A)/a = sin(B)/b

sin(112)/37 = sin(28)/b

b*sin(112) = 37*sin(28)

b = 37*sin(28)/sin(112)

b = 18.7346314730578

Round that value however you need to.

The portion on your screenshot says "round to the nearest", but it doesn't say to the nearest what. Is it nearest tenth? Hundredth? Something else? That part is cut off.  So I'll just write down all of the decimal digits my calculator is showing, and let you do the final rounding step.

Using proportions, it is found that the number of books that was left over was of 505.

A proportion is a fraction of a total amount, and the measures are related using a rule of three.

The number of books received by Marie is given by:

M = 772 - 345 = 427.

Melanie received one-sixth of Marie, hence:

Me = 427/6 = 71.

The total amount is 25 times Melanie's amount, hence:

T = 25 x 71 = 1775.

The amount left over is:

L = 1775 - (772 + 427 + 71) = 505

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Answer:

[tex]12\dfrac 12\%[/tex]

Step by step explanation:

[tex]\text{Let it be}~ x\%,\\\\~~~~~72 \cdot x \% = 9\\\\\\\implies 72\cdot \dfrac{x}{100} = 9\\\\\\\implies x = \dfrac{9 \times 100}{72}\\\\\\\implies x = \dfrac{100}{8}\\\\\\\implies x = \dfrac{25}2\\\\\\\implies x = 12 \dfrac 12\\\\\\\text{Hence 9 is}~ 12\dfrac 12\% ~ \text{of}~ 72[/tex]

[tex]\\ \rm\Rrightarrow \dfrac{IJ}{IL}=\dfrac{MN}{MP}[/tex]

[tex]\\ \rm\Rrightarrow \dfrac{24}{12}=\dfrac{MN}{3.8}[/tex]

[tex]\\ \rm\Rrightarrow 2=\dfrac{MN}{3.8}[/tex]

[tex]\\ \rm\Rrightarrow MN=3.8(2)[/tex]

[tex]\\ \rm\Rrightarrow MN=7.6[/tex]

The solution to the question is:

c is 6 = [tex]\sqrt{a^{2} + b^{2} -2abcosC }[/tex]

b is 5 = [tex]\sqrt{a^{2} + c^{2} -2accosB }[/tex]

cosB is 2 = [tex]\frac{a^{2} + c^{2} - b^{2} }{2ac}[/tex]

a is 4 = [tex]\sqrt{b^{2} + c^{2} -2bccosA }[/tex]

cosA is 3 = [tex]\frac{b^{2} + c^{2} -a^{2} }{2bc}[/tex]

cosC is 1 = [tex]\frac{b^{2} + a^{2} - c^{2} }{2ab}[/tex]

it is used to relate the three sides of a triangle with the angle facing one of its sides.

The square of the side facing the included angle is equal to the some of the squares of the other sides and the product of twice the other two sides and the cosine of the included angle.

Analysis:

If c is the side facing the included angle C, then

[tex]c^{2}[/tex] = [tex]a^{2}[/tex] + [tex]b^{2}[/tex] -2ab cos C-----------------1

then c =  [tex]\sqrt{a^{2} + b^{2} -2abcosC }[/tex]

if b is the side facing the included angle B, then

[tex]b^{2}[/tex] = [tex]a^{2}[/tex] + [tex]c^{2}[/tex] -2accosB-----------------2

b =  [tex]\sqrt{a^{2} + c^{2} -2accosB }[/tex]

from equation 2, make cosB the subject of equation

2ac cosB =  [tex]a^{2}[/tex] +  [tex]c^{2}[/tex] - [tex]b^{2}[/tex]

cosB =  [tex]\frac{a^{2} + c^{2} - b^{2} }{2ac}[/tex]

if a is the side facing the included angle A, then

[tex]a^{2}[/tex] = [tex]b^{2}[/tex] + [tex]c^{2}[/tex] -2bccosA--------------------3

a =  [tex]\sqrt{b^{2} + c^{2} -2bccosA }[/tex]

from equation 3, making cosA subject of the equation

2bcosA =  [tex]b^{2}[/tex] +  [tex]c^{2}[/tex]  - [tex]a^{2}[/tex]

cosA =  [tex]\frac{b^{2} + c^{2} -a^{2} }{2bc}[/tex]

from equation 1, making cos C the subject

2abcosC =  [tex]b^{2}[/tex] + [tex]a^{2}[/tex] -  [tex]c^{2}[/tex]

cos C =  [tex]\frac{b^{2} + a^{2} - c^{2} }{2ab}[/tex]

In conclusion,

c is 6 = [tex]\sqrt{a^{2} + b^{2} -2abcosC }[/tex]

b is 5 = [tex]\sqrt{a^{2} + c^{2} -2accosB }[/tex]

cosB is 2 = [tex]\frac{a^{2} + c^{2} - b^{2} }{2ac}[/tex]

a is 4 = [tex]\sqrt{b^{2} + c^{2} -2bccosA }[/tex]

cosA is 3 = [tex]\frac{b^{2} + c^{2} -a^{2} }{2bc}[/tex]

cosC is 1 = [tex]\frac{b^{2} + a^{2} - c^{2} }{2ab}[/tex]

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[tex]\huge\huge\bold{\green a \blue n \pink s \purple w \orange e \red r}[/tex]

volume of rectangular prism

Formula to calculate:

[tex]\large\bold{\green V\green = \green l \green * \green w \green * \green h}[/tex]

[tex]\large\bold{V= 7.5*4*2}[/tex]

[tex]\huge\boxed{\red V\red = \red 6 \red 0 \: {\red i \red n}^{\red 3}}[/tex]

length= 7.5 in

width= 2 in

height= 4 in

The volume of the rectangular prism or cuboid.

Formula: volume= length × width × height

so, V= 7.5 × 2 × 4

[tex]\bold{V= 60 \: {\: in}^{3}}[/tex]

hope this helps!

The largest interval centered about x = 0 is ([tex]-\frac{\pi}{2}, \frac{\pi}{2}[/tex]) for which the given initial-value problem has a unique solution: [tex]y'' + tan(x)y = e^x[/tex], y(0) = 1, y'(0) = 0.

An initial-value problem (IVP) is one kind of mathematical problem that contains obtaining a solution to a differential equation along with a set of initial conditions. It is commonly encountered in the field of differential equations.

Since it is mentioned in the theorem that in an initial-value problem y' + p(t)y = g(t), y([tex]t_o[/tex]) = [tex]y_o[/tex] has a unique solution if p(t) and g(t) are continuous functions on the interval (a, b) containing [tex]t_o[/tex].

Since [tex]e^x[/tex] is continuous everywhere and tanx is continuous on ([tex]-\frac{\pi}{2}, \frac{\pi}{2}[/tex]) containing 0.

Since R, a set of real numbers, and ([tex]-\frac{\pi}{2}, \frac{\pi}{2}[/tex]), both contain 0.

Thus the largest interval centered about x = 0 is ([tex]-\frac{\pi}{2}, \frac{\pi}{2}[/tex]).

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Answer: OD, 1/5

Step-by-step explanation:

Well what I did was take DE and seen how many time it could fit into BC. BC would take up a total of 5 DE's. So since we already have one which is DE then we would have 1/5.

HOPE THIS HELPS! ^_^

Hi student, let me help you out! :)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

We are asked to find the G.C.F. of 8x-6.

[tex]\triangle~\fbox{\bf{KEY:}}[/tex]

So what is the G.C.F. of 8x-6? We can list all of their common factors and then select the greatest one, like so:

[tex]\longmapsto\bf{Factors\;of\;8x:}[/tex] 1, 2, 4, 8, x

[tex]\longmapsto\bf{Factors\;of\;-6:}[/tex] -1, 1, 2, -3, 3, -6, 6

So the GCF is:

2

Now, we can also factor it out by placing it outside the parentheses ():

2(4x-3)

See, we factored it out by dividing both terms of the expression by the G.C.F.

[tex]\ddot\bigstar[/tex] Remember this...

[tex]\fbox{\bf{We\;factor\;out\;the\;G.C.F.\;by\;dividing\;all\;the\;terms\;of\;the\;expression\;by\;it}}[/tex]

Hope this helps you out! :D

Ask in comments if any queries arise.

~Just a smiley person helping fellow students :)

Answer: 2, you have to use math apps man it will save your life

Answer:

Yes they are.

Step-by-step explanation:

By simply moving 2x from the LHS to the RHS we'll get it's equivalent to be 3x.

i.e 5x - 2x = 3x

Replace x with -2 and solve:

A) -2 + -2 = -4 which is not less than or equal to -8

B) 2 + -2 = 0, this is greater than or equal to -8

C) -2 + -2 = -4, this is not less than -8

D) 2 + -2 = 0, this is not greater than 8

Answer: B is true

Answer:

y = 260

Step-by-step explanation:

locate x = 7 on the x- axis, go vertically up to meet the graph at (7, 260 )

when input is 7 then output y = 260

The binomial expression (p + q ) n (p+q)n to calculate the binomial distribution given will give a value of 10.

From the information given, the expression (p + q ) n (p+q)n is given to calculate a binomial distribution with n = 5 and p = 0.3.

p = 0.3

q = 1 - 0.3 = 0.7

n = 5

Therefore, expression ( p + q ) n (p+q)n will b:

= (0.3 + 0.7)× 5 + (0.3 + 0.7)× 5

= 5 + 5

= 10

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