solve for x in equation !! ?

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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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:

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:

SAS theorem >This is the correct

Answer:

Step-by-step explanation:

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

(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 :)

[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:

— 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:

equation= 4x=x+22kg

Step-by-step explanation:

i hope it helped :))

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:

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:

[tex]y=-\frac{2}{5} x+\frac{36}{5}[/tex]

Step-by-step explanation:

→ Find the gradient

[tex]\frac{4-6}{8-3}= \frac{-2}{5} =-\frac{2}{5}[/tex]

→ Write into y = mx + c format

[tex]y=-\frac{2}{5} x+c[/tex]

→ Substitute in ( 3 , 6 )

[tex]6 = -\frac{6}{5} +c[/tex]  ∴ [tex]c = 6+\frac{6}{5} =\frac{36}{5}[/tex]

→ Rewrite into y = mx + c format

[tex]y=-\frac{2}{5} x+\frac{36}{5}[/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]

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:

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:

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

Step-by-step explanation:

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:

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

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:

1. 3 and 1/5 improper: 11/15

2. 3 and 3/4 improper: 15/16

3. 1 and 2/3 improper: 5/6

4. 2 and 1/2 improper: 5/6

5. 3 and 2/10 improper: 32/40

6. 3 and 3/9 improper: 30/36

7. 1 and 2/6 improper: 8/12

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

Answer:

12.89

Step-by-step explanation:

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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:

7x2 + 14x = 0

x2 + 3x -5 = 0

x2 - x = 3x + 7  

Step-by-step explanation:

A quadratic equation has the highest power of x to the second power.  It must have x to the second power

7x2 + 14x = 0  quadratic

x3 - 3x2 + 1 = 0   not quadratic  but cubic

5x - 7 = 0  not quadratic  but linear

x2 + 3x -5 = 0   quadratic

x - 5 = 9x + 7  not quadratic but linear

x2 - x = 3x + 7  quadratic

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]\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]

Step-by-step explanation:

LCD of 4/x²t and x/t²

= LCM(x²t , t²)

= x²t²

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] \large╚»★«╝ { \mathcal{Answer }}╚»★«╝[/tex]

The Correct option among them is B

_____________________________________

The other options have one variable each, but according to the given question there should be two variables in the inequality.

The only choice satisfying the conditions is Option B

having two variables in it ( that is "m" and "n" )

[tex] ꧁ \: \large \frak{Eternal \: Being } \: ꧂[/tex]