Select the correct answer. what is the solution set of y = x2 2x 7 and y = x 7? a. {(0, 7), (-1, 6)} b. {(0, 7), (-7, 0)} c. {(0, 7), (1, 8)} d. {(-2, 0), (4, 0)}

[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]

Answer:

  3/4

Step-by-step explanation:

The relationship between time, speed, and distance can be used together with the travel time relations to find the desired ratio.

__

Let H represent the distance from Yan's location to home. Let S represent the distance to the stadium, and let w represent Yan's walking speed.

The time required for Yan to walk home is ...

  time = distance/speed = H/w

The time required for Yan to walk to the stadium is ...

  time = S/w

The time required for Yan to ride his bike from home to the stadium is ...

  time = (H +S)/(7w) . . . . . his riding speed is 7 times his walking speed.

In order for the travel times to be equal, we must have ...

  time to walk home + time to bike to stadium = time to walk to stadium

  H/w + (H +S)/(7w) = S/w

__

Multiplying by 7w gives ...

  7H +(H +S) = 7S

  8H = 6S . . . . . . . . . subtract S

  H/S = 6/8 = 3/4 . . . . divide by 8S

The ratio of Yan's distance from home to his distance from the stadium is 3/4.

_____

Additional comment

Another way to think about this is ...

The stadium is farther away than home by a distance that Yan can walk in the same time he can bike the entire distance from home to the stadium. That is, the difference in distances must be 1/7 of the total distance. Now, we have a "sum and difference" problem: S +H = 7, S -H = 1. The solution is S = (7+1)/2 = 4, H = (7-1)/2 = 3. H/S = 3/4.

Answer:

BC ≈ 11.8 m

Step-by-step explanation:

using the tangent ratio in the right triangle

tanΘ = [tex]\frac{opposite}{adjacent}[/tex] = [tex]\frac{AB}{BC}[/tex] = [tex]\frac{x}{BC}[/tex] , that is

tan32° = [tex]\frac{7.4}{BC}[/tex] ( multiply both sides by BC )

BC × tan32° = 7.4 ( divide both sides by tan32° )

BC = [tex]\frac{7.4}{tan32}[/tex] ≈ 11.8 m ( to 3 sf )

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:

Formula = 2nr

= 2(3,14 + 3 + 4)

= 2(10,14)

= 20,28

[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!

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! ^_^

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

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.

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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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:

[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]

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.

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

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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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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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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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

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:   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.

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:

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:

-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:

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

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

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:

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

The degree measure of angle COA = 128 degree.

It is defined by the property that the sum of the interior angles of a triangle is equal to 180 degrees.

To determine the value of angle COA ,

we have to draw a line connecting CA ,

as OC and OA are the radius of the circle and therefore the sides of the triangle formed will be equal and also the angle made by them ,

angle OAC = angle OCA

Therefore angle DCA = angle DAC

By angle sum property

2x + 52 = 180

2x = 128

x = 64 degree

angle DCO = 90 degree

Therefore angle ACO = 90-64 = 26 degree

Similarly, angle CAO = 26 degree

By angle sum property

26+26+x = 180

52+x = 180

x = 128 degree

The degree measure of angle COA = 128 degree.

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