Find the length indicated. Round answer to the nearest tenth.

Please and thank you so very much!!!!! See picture for the full problem.

Answer:

[tex]b^2 = 14[/tex]

Step-by-step explanation:

log(base)(power) = (exponent).

In this case: logb(14) = 2

Note the variable/number in place of each term. Set the equation:

base^power = exponent

[tex]log_b 14 = 2\\b^2 = 14[/tex]

[tex]b^2 = 14[/tex] is your answer.

~

Answer:

60 inches

Step-by-step explanation:

Add 59+61 =120 which you divide by 2 for the mean of 60.

Answer:

well over 3 years and almost over 4 months

Answer: £198.27

Step-by-step explanation:

Answer:

3*30 = 90

Step-by-step explanation:

3*30 = 90

Answer:Eight to 12 inches is usually adequate. If drainage is a problem, or if the plants you are growing prefer drier soil, the bed could be taller and filled with a porous growing medium.

Step-by-step explanation:

Answer:

The correct polynomial is F. Calculation errors combining h terms and distributing -1.

Step-by-step explanation:

[tex]( - 4h + 3)( {h}^{2} + 9h - 1)[/tex]

[tex] - 4 {h}^{3} - 36 {h}^{2} + 4h + 3 {h}^{2} + 27h - 3[/tex]

[tex] - 4 {h}^{3} - 33 {h}^{2} + 31h - 3[/tex]

[tex]\qquad \qquad \huge \pink {\sf{☁Answer☁}} \\ \\ [/tex]

[tex] \large \purple { \rm{Let's \: solve→}} \\ [/tex]

[tex] \rm{3(2x-5)=4x+33} \\ \\ \rm {solving \: bracket} \\ \\ \rm{6x - 15= 4x + 33} \\ \\ \rm{arranging \: terms} \\ \\ \rm{6x - 4x = 33 + 15} \\ \\ \rm{2x = 48} \\ \\ \rm{x = \frac{48}{2} } \\ \\ \rm{x = \cancel \frac{48}{2} } \\ \\ \rm{x = 24}[/tex]

[tex]\purple{\rule{15mm}{2.9pt}} \red{\rule18mm{2.5pt}} \orange{ \rule18mm{2.5pt}}[/tex]

[tex]\sf{\:мѕнαcкεя\: ♪...}[/tex]

Answer:

[tex]\fbox{X = 24}[/tex]

Solution:

Given expression ➟ 3(2x-5) = 4x + 33,

using distributive property open the above bracket,

➳ 3×2x - 3×5 = 4x + 33

Using BODMAS solve the L.H.S,

➳ 6x - 15 = 4x + 33

now arrange the like terms aside,

➳ 6x - 4x = 33 + 15

➳ 2x = 48

➳ x = 48/2

➳ x = 24

the given expression is true for x = 24

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Answer:

c and d

Step-by-step explanation:

because

- In f(x),evey time you find  X you write(-6x+5) instead

4x²-----------4(-6x+5)²

3x+1------------3(-6x+5) +1

-In g(x),evey time you find  X you write(-4x² +3x+1)  instead          

Answer:

1.25 kg

Explanation:

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

======

So, 1250 gram = (1250/1000) kg

We know

Now

Mass in kg:-

Option A

Answer:

22 is th area

Step-by-step explanation:

The total number of people attending the game is 1700.

Percentage in simple terms can be defined as a part per 100.

Percentage is a ratio or number which has to be defined as a fraction of 100.It is denoted by %.

If we are asked to calculate the percent of a number simply divide it by the whole part or number and then multiply with 100.

Percent of people supporting the home team = 92%

Number of people supporting the home team = 1564

Let x be the total number of people attending the game.

⇒ [tex]\frac{1564}{x}[/tex] × 100 = 92

⇒ [tex]\frac{1564}{x}[/tex] = 0.92

⇒ x = [tex]\frac{1564}{0.92}[/tex]

⇒ x = 1700

Hence, the total number of people attending the game will be 1700.

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Answer:

I don't see any optional answers, but the denominator must be greater than 12.

Step-by-step explanation:

27,890*(12/X) < 27,890

Therefore (12/X) < 1

and X > 12

Answer:

the answer is B five

Step-by-step explanation:

three hundred and seven minus three hundred and twelve

Answer: 21/572

[tex]\frac{5.5}{10 . 22} - \frac{4}{52}[/tex]

The dots stand for multiplication

Step-by-step explanation:

It is given that, the sum of two positive integers is 31 and the sum of the squares of these numbers is 625 and we are to find the smaller of the numbers.

So, let the two positive integers be x and y.

Therefore,

[tex] \\ {\longrightarrow \pmb{\sf {\qquad x + y = 31 \: \: ........ \: (i) }}} \\ \\[/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x {}^{2} + y {}^{2} = 625 \: ...... \: (ii)}}} \\ \\[/tex]

Now, From the first equation we have,

[tex]\\ {\longrightarrow \pmb{\sf {\qquad x + y = 31 }}} \\ \\ [/tex]

[tex] {\longrightarrow \pmb{\sf {\qquad y = 31 - x \: ...... \: (iii)}}} \\ \\ [/tex]

Now, substituting the value of y in equation (ii) we get :

[tex] \\ {\longrightarrow \pmb{\sf {\qquad x {}^{2} + (31 - x) {}^{2} = 625}}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x {}^{2} + (31 {}^{2} - 2.x.31 + x{}^{2} ) = 625}}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x {}^{2} + (961 - 62x + x{}^{2} ) = 625}}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x {}^{2} + 961 - 62x + x{}^{2} - 625 = 0}}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad 2 x {}^{2} + 336 - 62x = 0}}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad 2 x {}^{2} - 62x + 336 = 0}}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad 2( x {}^{2} - 31 + 168) = 0}}} \\ \\[/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad \frac{2}{2} ( x {}^{2} - 31 + 168) = \frac{0}{2} }}} \\ \\[/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x {}^{2} - 31 + 168 = 0}}} \\ \\[/tex]

Now using the quadratic formula :

[tex]{\longrightarrow \pmb{\sf {\qquad x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a} }}} \\ \\[/tex]

Where,

[tex] \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x = \frac{ - (- 31) \pm \sqrt{ {31}^{2} - 4(1)(168)} }{2(1)} }}} \\ \\[/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x = \frac{ 31 \pm \sqrt{ 961 -672} }{2(1)} }}} \\ \\[/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x = \frac{ 31 \pm \sqrt{ 289} }{2(1)} }}} \\ \\[/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x = \frac{ 31 \pm 17 }{2(1)} }}} \\ \\[/tex]

Now, we have two equations,

[tex]{\longrightarrow \pmb{\sf {\qquad x = \frac{ 31 + 17 }{2} \: ... .....\: (iv)}}} \\ \\[/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x = \frac{ 31 - 17 }{2} ... .....\: (v)}}} \\ \\[/tex]

So, Equation (iv) :

[tex] \\ {\longrightarrow \pmb{\sf {\qquad x = \frac{ 31 + 17 }{2} }}} \\ \\ [/tex]

[tex]\\ {\longrightarrow \pmb{\sf {\qquad x = \frac{ 48 }{2} }}} \\ \\ [/tex]

[tex]\\ {\longrightarrow \pmb{\sf {\qquad x = 24 }}} \\ \\ [/tex]

Now, Equation (v) :

[tex] \\ {\longrightarrow \pmb{\sf {\qquad x = \frac{ 31 - 17 }{2}}}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x = \frac{ 14 }{2}}}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x = \: 7}}}\\ \\ [/tex]

Now, we are to find the value of y.

Substituting the value of x (24) in equation (iii) :

[tex] \\ {\longrightarrow \pmb{\sf {\qquad y = 31 - x \:}}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad y = 31 - 24 \:}}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad y = 7 \:}}} \\ \\ [/tex]

Again, Substituting the value of x (7) in equation (iii) :

[tex]\\ {\longrightarrow \pmb{\sf {\qquad y = 31 - x \:}}} \\ \\ [/tex]

[tex] {\longrightarrow \pmb{\sf {\qquad y = 31 - 7 \:}}} \\ \\ [/tex]

[tex]\\ {\longrightarrow \pmb{\sf {\qquad y = 24 \:}}} \\ \\ [/tex]

Therefore,

[tex] \\ [/tex]

So, The smaller of the numbers is 7 .

The sum of two positive integers is 31. If the sum of the squares of these numbers is 625, find the smaller of the numbers.

This question belongs to quadratic equations so we have to find the answer by making equation and solving it .

According to question , sum of two positive integers is 31 . So ,

According to question , sum of square

of these numbers is 625 . So ,

From equation 1 ( x + y = 31 ) , Value

of x :

Now , putting value of x in eq. 2 :

Solving it by using middle term

splitting :

First number ,

Second Number ,

According to question ,

Sum of numbers is equal to 31 :

Sum of squares of these numbers is equal to 625 :

Answer:

p = 4

Step-by-step explanation:

Given equation:

[tex]x^2+(p-3)y^2-4x+6y-16=0[/tex]

Standard equation of a circle:

[tex](x-a)^2+(y-b)^2=r^2[/tex]

(where [tex](a,b)[/tex] is the centre of the circle, and [tex]r[/tex] is the radius)

If you expand this equation, you will see that the coefficient of [tex]y^2[/tex] is always one.

Therefore, [tex]p-3=1[/tex]

[tex]\implies p=1+3=4[/tex]

Additional information

To rewrite the given equation in the standard form:

[tex]\implies x^2+y^2-4x+6y-16=0[/tex]

[tex]\implies x^2-4x+y^2+6y=16[/tex]

[tex]\implies (x-2)^2-4+(y+3)^2-9=16[/tex]

[tex]\implies (x-2)^2+(y+3)^2=16+4+9[/tex]

[tex]\implies (x-2)^2+(y+3)^2=29[/tex]

So this is a circle with centre (2, -3) and radius √29

Answer:

A , B And D

2/4 divide both numbers by to and you get 1/2

3/6 divide both numbers by 3 and you get 1/2

4/8 divide both numbers by 4 and you get 1/2

Answer:

This is hard

Step-by-step explanation:

I'm so sorry I can't do it

Answer:

[tex]\frac{69}{20}[/tex]

Step-by-step explanation:

[tex]\frac{3}{7} -\frac{11}{4} +\frac{16}{5} +\frac{18}{7}[/tex]

= [tex]\frac{3}{7} +\frac{18}{7}-\frac{11}{4} +\frac{16}{5}[/tex]

= [tex]\frac{21}{7} -\frac{11}{4} +\frac{16}{5}[/tex] As both denominators have 7 you can add their numerators

=[tex]\frac{21}{7} -(-\frac{9}{20})[/tex]     L.C.M

=[tex]\frac{21}{7} +\frac{9}{20}[/tex]   L.C.M

=[tex]\frac{69}{20}[/tex]

Answer:

yes

Step-by-step explanation:

your answer is correct

Answer:

115

Step-by-step explanation:

If 65 is subtracted from 180the answer is 115

So

f(2)

f(3)

f(4)

The ways that the policies of President Kennedy and President Johnson are similar and different is given below:

After the assassination of Kennedy, Johnson was known to have followed the footsteps of his predecessor's and went ahead on implementing some of his programs.  Johnson was known to have went ahead with the the anti-poverty program and the Civil Rights Bill.

The differences between the both presidents were:

Conclusively, these two president were said to be very were similar, and yet they were different.

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Check the picture below.

let's notice something, the equation of line B is in slope-intercept form, thus

[tex]y = 2x + \underset{\stackrel{\uparrow }{b}}{2}\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}[/tex]

so using that grid scale, we can just get two points from the line A to get its equation, let's use those in the picture below

[tex](\stackrel{x_1}{-6}~,~\stackrel{y_1}{-6})\qquad (\stackrel{x_2}{6}~,~\stackrel{y_2}{0}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{0}-\stackrel{y1}{(-6)}}}{\underset{run} {\underset{x_2}{6}-\underset{x_1}{(-6)}}}\implies \cfrac{0+6}{6+6}\implies \cfrac{1}{2}[/tex]

[tex]\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-6)}=\stackrel{m}{\cfrac{1}{2}}(x-\stackrel{x_1}{(-6)}) \\\\\\ y+6=\cfrac{1}{2}(x+6)\implies y+6=\cfrac{1}{2}x+3\implies y=\cfrac{1}{2}x-3[/tex]

Answer:

5. radius= TD

inscribed angle =RQS

diameter =TA

6. radius=SY

secant=TU

Tangent=S

Step-by-step explanation:

to figure this out you would have to take all the numbers and do the rise over run technique

Answer:

Area of cube = 6l²

= 6×2²

= 24 ft²

2 feet.

how many square feet of wood he used to build the box.

each side measures 2ft.

A=L²=(2 ft)²=4 ft²

A cube has 6 sides so the total quantity of wood is:

Wood=6×A

=6×4 ft²

=24 ft²

he used 24 ft² of wood to build the box.

Answer:

5/7.

Step-by-step explanation:

The rise = 5 and the corresponding run si 7.

Answer:

5/7

Step-by-step explanation:

so we start at the point (0,0). the slope is essentially

Rise/Run

so in this graph we see a point of (7,5)

the rise was 5 along the y axis and the run was 7 along the x axis.

so 5/7 is correct

Answer:

Hi! I'm not too bad at math so I hope I can help you. Except, I can't seem to find your question so please comment your question(s) to me so that I can help you. Have a great day:)