What is the value of x if a slope is 2, and a line passing through (x, 3) and (1,5).

Answer:

12.89

Step-by-step explanation:

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Answer:

it is 1

Step-by-step explanation:

Let's use PEMDAS (do the parenthesis first).

2 to the 3 = 2x2x2 = 8

2 to the 5 = 2x2x2x2x2 = 32

8 x 32 = 256 (all the solutions are factors of 2048).

2 to the 8 = 2x2x2x2x2x2x2x2 = 256 again

256 / 256 = 1

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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[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.

well, let's take 2 points from it and get it :) , say let's use (4 , 5) and (-1 , 5)

[tex](\stackrel{x_1}{4}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{-1}~,~\stackrel{y_2}{5}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{5}-\stackrel{y1}{5}}}{\underset{run} {\underset{x_2}{-1}-\underset{x_1}{(4)}}}\implies \cfrac{0}{-5}\implies 0[/tex]

bearing in mind that for any horizonal line it's always 0.

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

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:

13122

Step-by-step explanation:

Our equation is 2(3)⁸

once its solve we get 13122 as our answer

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:

[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:

157°

Step-by-step explanation:

Sum of interior angles of polygon

(n-2)*180 ---- n = total number of sides

7 angles, 7 sides

5*180 = 900

final angle will be

900-162-115-125-138-105-98

=157

Answer:

157

Step-by-step explanation:

it is obvious that it's a heptagon

so (n-2) x180

n=7

(7-2) x 180 =900

900-162-115-125-138-105-98

900-743=157

final unknown=157°

Answer:

SAS theorem >This is the correct

Answer:

Step-by-step explanation:

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

Answer:

equation= 4x=x+22kg

Step-by-step explanation:

i hope it helped :))

[tex]\text{Area of parallelogram} = \text{Base}~ \times~\text{Height} = (4a^3b^4)(6a^2 b^5)=24a^5 b^9[/tex]

Answer:

-28.50

Step-by-step explanation:

Given the algebraic expression, 2w² - 3x + y - 8z, where w = -0.5, x = 2, y= -3, and z = 2.5:

Substitute the given values for the variables to evaluate the algebraic expression.

2w² - 3x + y - 8z

= 2(-0.5)² - 3(2) + (-3) - 8(2.5)

= 2(0.25) - 6 - 3 - 20

= 0.50 - 6 - 3 - 20

= -28.50

Therefore, 2w² - 3x + y - 8z = -28.50.

Answer:

-28.5

Step-by-step explanation:

Substitute the values for their given variables and evaluate.

[tex]2w^2-3x+y-8z\\\\2(-0.5)^2-3(2)+(-3)-8(2.5)\\\\2(0.25)-6-3-20\\\\0.5-6-3-20\\\\-28.5[/tex]

Answer:

Both of those are continuous

Step-by-step explanation:

A discrete function is only defined for a specified countable set of x, neither of the two provided fit that description. On the other hand, both are defined for arbitrary positive real numbers.

The value of both functions is defined for a continuous interval of x thus both will be continuous.

A certain kind of relationship called a function binds inputs to essentially one output.

The nature of the relationship defines the function for example y = sinx and y = x +6 like that.

Given the first function →

y = 30.48x where x is feet and y is a centimeter of conversion.

Now, it is known that for every foot interval there will be a centimeter that exists therefore they will be defined for all continuous intervals of x so it will be continuous.

The second function →

y = 3x + 2 where x is minutes and y is the number of gallons in a tank.

Now since for every minute of the interval, there will be a certain amount of gallons in the tank so it will be also continuous.

Hence "The value of both functions is defined for a continuous interval of x thus both will be continuous".

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(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:

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:

you cant draw......................

Step-by-step explanation:

Answer:

its the third choice

Step-by-step explanation:

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

Answer:

Step-by-step explanation:

One must rearrange the function to decreasing value of exponents

f(x) = -7x⁴ + 8x³ - 3x² - 9x + 11

to see that it is a 4th degree polynomial with leading coefficient of -7

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)

[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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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°.

Answer:

5280 , 8720

Step-by-step explanation:

suppose the one is x, another is y

x+y=14000

y(6%)-x(4%)=312

x=5280

y=8720

net annual receipt means the income so you have to substract the lost from the gains

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:

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: