Louis is comparing the populations of several nearby cities. The population of Liberty Crossing can be written as 49 people, and the population of Harrisburg can be written as 47 people. The population of Glenview is 43 times the population of Harrisburg. Drag the cities to put them in order from least population (top) to greatest population (bottom).

Answer:

100 percent of 63 is 63

Step-by-step explanation:

Answer:

63

Step-by-step explanation:

100 x 63 /100 = 63

Answer:

w= -2. Solve for w: (w-5) (w+2) = w^2-4. Expand out terms of the left hand side: w^2-3 w-10 = w^2-4. Subtract w^2-4 from both sides

Step-by-step explanation:

Answer:

Step-by-step explanation:

Slope: 3/1 (positive: line go uphill)

y-intercept: -5

Place your pencil on the vertical line (y) on -5. Go up 3 units and 1 to the right, repeat and trace the line (actually you only need two points to trace a line)

Answer:

13122

Step-by-step explanation:

Our equation is 2(3)⁸

once its solve we get 13122 as our answer

(1, 64), (2, 32), (4, 16), (8, 8).

Answer:

1 x 64

2 x 32

4 x 16

8 x 8

Step-by-step explanation:

Factors of 64: 1, 2, 4, 8, 16, 32 and 64.

Pairs:

1 x 64

2 x 32

4 x 16

8 x 8

Hope this helps :)

Answer:

1/5(2ln|x|+3*5^1/3)+c

Step-by-step explanation:

Answer:

75,388

Step-by-step explanation:

Long Multiplication form

[tex]\mathrm{Mutliply\:the\:bold\:numbers}[/tex]

[tex]\frac{\begin{matrix}\space\space&8&0&\textbf{2}\\ \times \:&\space\space&9&\textbf{4}\end{matrix}}{\begin{matrix}\space\space&\space\space&\space\space&8\end{matrix}}[/tex]

[tex]\frac{\begin{matrix}\space\space&8&\textbf{0}&2\\ \times \:&\space\space&9&\textbf{4}\end{matrix}}{\begin{matrix}\space\space&\space\space&0&8\end{matrix}}[/tex]

[tex]\mathrm{Multiply}\\\\8\cdot \:4=32[/tex]

[tex]\mathrm{Carry\:}3\mathrm{\:to\:the\:column\:on\:the\:left\:and\:write\:}2\mathrm{\:in\:the\:result\:line}[/tex]

[tex]\frac{\begin{matrix}\space\space&3&\space\space&\space\space&\space\space\\ \space\space&\space\space&\textbf{8}&0&2\\ \times \:&\space\space&\space\space&9&\textbf{4}\end{matrix}}{\begin{matrix}\space\space&\space\space&2&0&8\end{matrix}}[/tex]

[tex]\mathrm{Add\:the\:carried\:digit,\:}3\mathrm{,\:to\:the\:result}[/tex]

[tex]\frac{\begin{matrix}\space\space&3&\space\space&\space\space&\space\space\\ \space\space&\space\space&8&0&2\\ \times \:&\space\space&\space\space&9&4\end{matrix}}{\begin{matrix}\space\space&3&2&0&8\end{matrix}}[/tex]

[tex]\mathrm{Multiply\:the\:top\:number\:by\:the\:bolded\:digit\:of\:the\:bottom\:number}[/tex]

[tex]\frac{\begin{matrix}\space\space&\space\space&\textbf{8}&\textbf{0}&\textbf{2}\\ \space\space&\times \:&\space\space&\textbf{9}&4\end{matrix}}{\begin{matrix}\space\space&3&2&0&8\end{matrix}}[/tex]

[tex]\mathrm{Mutliply\:the\:bold\:numbers}:\quad \:2\cdot \:9=18[/tex]

[tex]\mathrm{Carry\:}1\mathrm{\:to\:the\:column\:on\:the\:left\:and\:write\:}8\mathrm{\:in\:the\:result\:line}[/tex]

[tex]\frac{\begin{matrix}\space\space&\space\space&\space\space&1&\space\space\\ \space\space&\space\space&8&0&\textbf{2}\\ \space\space&\times \:&\space\space&\textbf{9}&4\end{matrix}}{\begin{matrix}\space\space&3&2&0&8\\ \space\space&\space\space&\space\space&8&\space\space\end{matrix}}[/tex]

[tex]\mathrm{Add\:the\:carried\:number\:to\:the\:multiplication}:\quad \:1+0\cdot \:9=1[/tex]

[tex]\frac{\begin{matrix}\space\space&\space\space&\space\space&\textbf{1}&\space\space\\ \space\space&\space\space&8&\textbf{0}&2\\ \space\space&\times \:&\space\space&\textbf{9}&4\end{matrix}}{\begin{matrix}\space\space&3&2&0&8\\ \space\space&\space\space&1&8&\space\space\end{matrix}}[/tex]

[tex]\mathrm{Mutliply\:the\:bold\:numbers}:\quad \:8\cdot \:9=72[/tex]

[tex]\mathrm{Carry\:}7\mathrm{\:to\:the\:column\:on\:the\:left\:and\:write\:}2\mathrm{\:in\:the\:result\:line}[/tex]

[tex]\frac{\begin{matrix}\space\space&7&\space\space&1&\space\space\\ \space\space&\space\space&\textbf{8}&0&2\\ \times \:&\space\space&\space\space&\textbf{9}&4\end{matrix}}{\begin{matrix}\space\space&3&2&0&8\\ \space\space&2&1&8&\space\space\end{matrix}}[/tex]

[tex]\mathrm{Add\:the\:carried\:digit,\:}7\mathrm{,\:to\:the\:result}[/tex]

[tex]\frac{\begin{matrix}\space\space&7&\space\space&1&\space\space\\ \space\space&\space\space&8&0&2\\ \times \:&\space\space&\space\space&9&4\end{matrix}}{\begin{matrix}\space\space&3&2&0&8\\ 7&2&1&8&\space\space\end{matrix}}[/tex]

[tex]\mathrm{Add\:the\:rows\:to\:get\:the\:answer.\:\:\:Fill\:in\:trailing\:zeros\:\:(not\:required)}[/tex]

[tex]\frac{\begin{matrix}\space\space&\space\space&8&0&2\\ \space\space&\times \:&\space\space&9&4\end{matrix}}{\begin{matrix}0&3&2&0&8\\ 7&2&1&8&0\end{matrix}}[/tex]     [tex]=75,388[/tex]

See Image for the addition:

[tex]\tt\frac { 2 - 2 x } { 4 } = \frac { 4 x + 12 } { 3 }[/tex]

[tex]\tt3\left(2-2x\right)=4\left(4x+12\right) [/tex]

[tex]\tt6-6x=16x+48 [/tex]

[tex]\tt6-6x-16x=48 [/tex]

[tex]\tt6-22x=48 [/tex]

[tex]\tt-22x=42 [/tex]

[tex]\tt\:x=\frac{42}{-22} [/tex]

[tex]\boxed{\tt\:x = -\frac{21}{11} = -1\frac{10}{11} \approx -1.90}[/tex]

Method used :-

◇ ✖ Cross multiplication ❌

________

Hope it helps ⚜

Step-by-step explanation:

Given equation is [{(2-2x)/4} = {(4x+12)/3]

On applying cross multiplication then

Since, (a/b) = (c/d)

⇛ab = bc

Where, a = (2-2x), b = (4), c = (4x+12) and d = (3).

Now,

⇛3(2-2x) = 4(4x+12)

Multiply the number outside the bracket with numbers and variables in the bracket.

⇛6 - 6x = 16x + 48

Shift the variable value on LHS and constant on RHS.

⇛-6x - 16x = 48-6

Subtract the values on LHS and RHS.

⇛-22x = 42

Shift the value (-22) from LHS to RHS.

⇛x = 42/-22

Write the obtained answer in lowest form by cancelling method.

⇛x = (42÷2)/(-22÷2)

Therefore, x = 21/-11 = -(21/11)

Answer: Hence the value of x for the given problem is -21/11.

VERIFICATION:

If x = -21/11 then the equation is

[{(2-2x)/4} = {(4x+12)/3]

Substitute the value x = -21/11 in expression, then

⇛[{2-2(-21/11)}/4 = {4(-21/11)+12}/3]

Multiply (-2) from (-21) to get 42 on LHS.

⇛[{2(42/11)}/4 = {4(-21/11)+12}/3]

Again multiply 4 from (-21) to get 81 on RHS.

⇛[{2(42/11)}/4 = {(-81/11) + 12}/3]

Take the LCM of the denominator 1 and 11 is 11 on LHS

⇛[{(2*11 + 42*1)/11}/4 = {(-81*1 + 12*11)/11}/3]

Multiply the numerator on both LHS and RHS.

⇛[{(22+42)/11}/4 = {(-81 + 132)/11}/3]

Add the numerator numbers on both LHS and RHS

⇛[{(64/11)}/4 = [{(48/11)}/3]

Now, conver the division fraction in multiply fraction on both LHS and RHS.

⇛[{(64/11) * 4} = {(48/11) * 3}]

Now, write the numbers in Lower form by cancelling method, on both LHS and RHS.

⇛(16/11) = (16/11)

LHS = RHS

Please let me know if you have any other questions.

90 degrees - 78 degrees = 12 degrees

cofunction of sin is cos

answer : cos12

Answer:

Step-by-step explanation:

what is she going to buy? Small stones or big stones? if its small stones then she can buy 7 small stones. If its big stones then she can buy 5 big stones.

Step-by-step explanation:

Alrighty! here you go image.

Answer:x=15

Step-by-step explanation:

Answer:

— 3 (1.5n — 1)

Step-by-step explanation:

[tex] - 4.5n + 3 = - \frac{9}{2} n + 3 = 3( - \frac{3}{2} n + 1) = - 3( \frac{3}{2}n - 1) \\ = - 3(1.5n - 1)[/tex]

Answer:

vtdOveractivity of the pituitary gland is called hyperpituitarism. Several disorders related to an overactive pituitary gland can occur.

...

Symptoms may include:

Nervousness.Rapid or irregular heartbeat.Weight loss.Fatigue.Muscular weakness.

Answer:x[tex]\geq[/tex]1

Step-by-step explanation:

remove the brackets

7x+49≤56

collect like terms

7x≤56-49

7x[tex]\geq[/tex]7

divide both sides by 7

x[tex]\geq[/tex]1

The rate of the jet in still air is 740 mph and the rate of the wind is 110 mph.

The given parameters;

Let the rate of the jet in still air = R₁

Let the rate of the wind = R₂

The rate of the jet against the wind:

[tex]R_1 - R_2 = \frac{3780}{6} \\\\R_1 - R_2 = 630[/tex]

The rate of the jet with the wind:

[tex]R_1 + R_2 = \frac{5950}{7} \\\\R_1 + R_2 = 850\\\\[/tex]

The rate of the jet in still air is calculated as follows;

[tex]R_1 = \frac{630 + 850}{2} \\\\R_1 = 740 \ mph[/tex]

The rate of the wind is calculated as follows;

[tex]R_1 - R_2 = 630\\\\R_2 = R_1 - 630\\\\R_2 = 740 - 630 \\\\R_2 = 110 \ mph[/tex]

Thus, the rate of the jet in still air is 740 mph and the rate of the wind is 110 mph.

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Answer:

its the third choice

Step-by-step explanation:

Side angle side cuz there r 2 sides congruent in the picture shown above.

Step-by-step explanation:

6³×(½)³

=> 216÷8 = 27.

hope this helps you.

Answer:

1. A

2. B

Step-by-step explanation:

It would look like y = -m(x) + b.

For example, m is 1 and b is 0. (You can try any random numbers, but I tried to find the simplest one so it's easy for us to see.) The equation will be y= -x. (you can substitute random numbers in here, too)

If x = 3, y = -3 ; x = 5, y = -5  (x increases, y decreases)

If x = -1, y = 1 ; x = -5, y = 5 (x decreases, y increases)

As you can see from the example,

If the value of x increases, the value of y will decrease and vice versa

For the negative slope equation y = -x as the value of x increases y decreases.

Slope or the gradient is the number or the ratio which determines the direction or the steepness of the line.

Given that the graph of a linear function has a negative slope. Suppose the equation with a negative slope is y = -x.

It would look like y = -m(x) + b.

As an illustration, m is 1 and b is 0. (You can try any random number; however, I looked for the most straightforward one so that we could easily observe it.) It will be written as y=-x. (You can also omit this and use random numbers)

If x = 3, y = -3 ; x = 5, y = -5  (x increases, y decreases)

If x = -1, y = 1 ; x = -5, y = 5 (x decreases, y increases)

Therefore, for the negative slope equation y = -x as the value of x increases y decreases.

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[tex]A=7+3^2 -2 \cdot 5 = 7 + 9 - 10 = 9-3=6\\\\\\B = \left(\dfrac 34 + \dfrac 18\right) \div \dfrac 18 = \dfrac 78 \times 8 =7\\\\\\B>A[/tex]

The expression relation is 7+3²-2⋅5 > (3/4+1/8)÷1/8.

Mathematical expressions with inequalities are those in which the two sides are not equal. Contrary to equations, we compare two values in inequality. Less than (or less than or equal to), greater than (or greater than or equal to), or not equal to signs are used in place of the equal sign.

Given:

First, 7+3²-2⋅5

Solving the above expression

= 7 + 9 -2.5

= 16 - 2.5

= 13.5

Second,

(3/4+1/8)÷1/8

= (6/8 + 1/8) ÷ 1/8

= 7/8 ÷ 1/8

= 7/8 x 8/1

= 7

Hence, 7+3²-2⋅5 > (3/4+1/8)÷1/8.

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[tex]\stackrel{\textit{one added to the }\sqrt{\qquad }}{1+\sqrt{\quad }}~~,~~ \stackrel{\sqrt{~~}\textit{ of a number and one}}{\sqrt{x+1}}\implies 1+\sqrt{x+1}\stackrel{\textit{equals}}{=}4 \\\\\\ \sqrt{x+1}=3\implies \left( \sqrt{x+1} \right)^2=3^2\implies x+1=9\implies \boxed{x=8}[/tex]

Answer:

all are perfect cubes

Step-by-step explanation:

you can just cube root the numbers and bom if it aint decimal its a perfect cube

Using the probability concept, it is found that:

a) P{X + W = 1} = 0

b) P{X = 2, Y = W} = 0.0667

A probability is the number of desired outcomes divided by the number of total outcomes.

In this problem, the order in which the balls are chosen is not important, hence, the combination formula is used to find the number of outcomes.

Combination formula:

[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by:

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

In total, 8 balls are chosen from a set of 19, hence:

[tex]T = C_{19,8} = \frac{19!}{8!11!} = 75582[/tex]

Item a:

Not possible, hence:

P{X + W = 1} = 0

Item b:

Hence:

[tex]D = C_{6,2}C_{9,3}C_{4,3} = \frac{6!}{2!4!} \times \frac{9!}{3!6!} \times \frac{4!}{3!1!} = 5040[/tex]

Then, the probability is:

[tex]p = \frac{D}{T} = \frac{5040}{75582} = 0.0667[/tex]

P{X = 2, Y = W} = 0.0667

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Answer:

It's C.

Step-by-step explanation:

Answer:

B

Step-by-step explanation:

just put the x and y values into the equations and see which equation stays true.

for example, point (4, 68) rules out A and C even without detailed calculation, because these 2 expressions can produce for positive x only negative y (and 68 is clearly a positive y).

so, when using this point in B :

68 = 4×4² + 2×4 - 4 = 4×16 + 8 - 4 = 64 + 8 - 4 = 68

correct.

and it is correct also for the other points.

so, B is the right answer.

Answer:

D: Factor a q out of p^2 q+ 2q

Step-by-step explanation:

I took the test. Hope it was helpful! :)

The first step in solving this equation p²q+2q=11 for q will be to factor q out of p²q+2q. Option D is correct.

An equation is a statement that two expressions, which include variables and/or numbers, are equal. In essence, equations are questions, and efforts to systematically find solutions to these questions have been the driving forces behind the creation of mathematics.

It is given that, the equation is,

p²q+2q=11

Take q common from the L.H.S as

q(p²+2)=11

The value of q is obtained by rearranging the equation as,

q=11/(p²+2)

The value of q obtained after solving the equation p²q+2q=11 will be q=11/(p²+2).

Thus, the first step in solving this equation p²q+2q=11 for q will be to factor q out of p²q+2q. Option D is correct.

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Answer:

SAS theorem >This is the correct

Answer:

Step-by-step explanation:

[tex] \huge\colorbox{pink}{hope \: its \: help}[/tex]

Answer:

7) m∠BHE = 146°

8) m∠BAC = 25°

Step-by-step explanation:

Given that, [tex]\displaystyle\mathsf{\overline{CD}\:||\:\overline{EF}}[/tex], and that [tex]\displaystyle\mathsf{\overline{AB}}[/tex] is a transveral.

We are also provided with the following measures of the angles: m∠DGH = 2x, and m∠FHB = 5x - 51.

∠DGH and ∠FHB are also corresponding angles, as they have corresponding positions on the same side of the transversal, [tex]\displaystyle\mathsf{\overline{AB}}[/tex]. These two angles also have the same measure.  

In order to find the measure of ∠BHE, we could set up an equality statement on ∠DGH and ∠FHB to solve for the value of x.

m∠DGH = m∠FHB

2x = 5x - 51

Add 51 and subtract 2x from both sides of the equation:

2x -2x + 51  = 5x - 2x - 51 + 51

51 = 3x

Divide both sides by 3:

[tex]\displaystyle\mathsf{\frac{51}{3}\:=\:\frac{3x}{3}}[/tex]

x = 17

⇒  m∠DGH = 2x = 2(17) = 34°,

⇒  m∠FHB = 5x - 51 = 5(17) - 51 = 34°.

Since ∠FHB and ∠BHE are supplements (whose sum add up to 180°), we could determine the measure of ∠BHE as follows:

m∠BHE + m∠FHB  = 180°

m∠BHE + 34° = 180°

m∠BHE + 34°- 34° = 180°- 34°

m∠BHE = 146°.

Therefore, the measure of ∠BHE is 146°.

Given that ΔABC with [tex]\displaystyle\mathsf{\overline{AC}}[/tex] extended to D, and that m∠ABC = 63° and m∠BCD = 92°:

The Exterior Angle Theorem states that the measure of an exterior angle of a triangle is equal to the sum of the two nonadjacent interior angles. In other words: ∠BCD = ∠BAC + ∠ABC.  

Before we could apply the Exterior Angle Theorem, we must first find m∠BAC. Since ∠BCD and ∠BCA are supplements:

m∠BCD + m∠BCA = 180°

92° + m∠BCA = 180°

92° - 92°+ m∠BCA = 180° - 92°

m∠BCA = 88°

Now that we have the measure for ∠BCA, we can find m∠BAC by applying the Triangle Sum Theorem where it states that the sum of the interior angles of a triangle is equal to 180°.

m∠BCD + m∠ABC + m∠BAC = 180°

92° + 63° + m∠BAC  = 180°

125° + m∠BAC  = 180°

155° - 155° + m∠BAC  = 180° - 155°

m∠BAC  = 25°

Therefore, the measure of ∠BAC is 25°.