How do I find the radius/x of a shaded region with it's area and/or angle measurement?

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Answer:

Step-by-step explanation:

800 J/kgC

What is the specific heat of a material that gains 600 joules of energy when a 0.25 kg object increases in temperature by 3C?

it is 800 J/kgC

Answer: 9 cups

Step-by-step explanation: 8 multiply it by 9 .

Answer:

Answer:

Equation of the parabola:  or

Vertex form:

Intercept form:

Step-by-step explanation:

Answer:

∠ABC = 55°

Step-by-step explanation:

Alternate Interior Angles Theorem

If a line (AC) intersects a set of parallel lines (m and n) in the same plane at two distinct points, the alternate interior angles that are formed are congruent.

⇒  ∠BAC = ∠DCA

⇒  ∠BAC = 50°

The interior angles of a triangle sum to 180°

⇒ ∠BCA + ∠BAC+ ∠ABC = 180

⇒ 75 + 50 + ∠ABC = 180

⇒ ∠ABC = 180 - 75 - 50

⇒ ∠ABC = 55°

Answer: Total cost/month = 7 + 3x

Step-by-step explanation:

This is only the total cost per month that is why it is only 7 and 3x since the number of downloaded games is not given.

I'm not sure about my answer but I hope that it helps.

Answer:

[tex]\huge\boxed{\sf x(x + 4)}[/tex]

Step-by-step explanation:

Given expression:

x² + 4x

Take x common

x(x + 4)

[tex]\rule[225]{225}{2}[/tex]

Answer:

[tex] \frac{5}{12} [/tex]

Hope it helps............

Answer:

y = 10

Step-by-step explanation:

if the figure is to be a parallelogram, then

∠ ACB and ∠ DAC are alternate angles and are congruent , that is

5y - 13 = 7 + 3y ( subtract 3y from both sides )

2y - 13 = 7 ( add 13 to both sides )

2y = 20 ( divide both sides by 2 )

y = 10

The summation that represents the money in the account 10 years after the initial deposit is  P(10) = 300(1.055)^10

Exponential equations are equations that increase geometrically

The formula for calculating the compound. interest is expressed as:

P(t) = P0(1+r)^t

Given the following

P0 =. $300
r = 5.5% = 0.055

t = 10years

Substitute

P(10) = 300(1+0.055)^10

P(10) = 300(1.055)^10

Hence the summation that represents the money in the account 10 years after the initial deposit is  P(10) = 300(1.055)^10

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Answer: D

Step-by-step explanation:

Answer:

1152$

Step-by-step explanation:

b1 = 7/8 C

b2 = 5/6 b1 = 35/ 48 C

7/8 C + 35/48 C = C + 696

42/48 C + 35/48 C = 48/48 C + 696 * 48/48

(42 + 35 - 48 ) C = 696 * 48

29 C = 33408

C =1152.

£3.50 for each burger

3.50×4= 14 left over is 6 from the 20 pound note

Answer:

The number is 740

Step-by-step explanation:

Hello!

Let the number that we want to solve for be x.

100% of x is x, but a decrease of 5% if 95% of x.

The number is 740.

The solution is correct.

The students will colour the 1/8th part of the sheet given to them by their teacher.

The fraction is defined as the division of the whole part into an equal number of parts.

Given that:-

Now the sheet student have =  1 / 2

Students will colour 1/4th of the sheet they have = ( 1/2  x  1/4 )

= 1 / 8

Therefore the students will colour the 1/8th part of the sheet given to them by their teacher.

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Answer:

it's difficult , sorry

Step-by-step explanation:

ask easily one

Answer: 4 books

Step-by-step explanation:

x = minimum number of book

Leslee

= 3x (bought 3 times the minimum)

Denise

= x + 7 (bought 7 more than the minimum)

Together

= (3x) + (x + 7)

(3x) + (x + 7) = 23 (books they bought together)

3x + x + 7 = 23

3x + x = 23 - 7 (isolate the like variables together)

4x = 16 (divide both sides by 4)

x = 4

4 is the minimum number of books

To check:

3(4) + (4+7)

12 + 11 = 23

Answer:

x = 30

Step-by-step explanation:

The sum of the angles of a triangle add to 180

x+2x+3x = 180

Combine like terms

6x= 180

Divide each side by 6

6x/6 = 180/6

x = 30

Answer:

Volumer of cylinder

=(πd²/4)×h

=(36π/4)11

99π yd³

Answer: 4

Step-by-step explanation:

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[tex]\diamond\large\textsf{\textbf{\underline{Given question:-}}}[/tex]

                Solve the following equation for b.

13b=6b+14

[tex]\diamond\large\blue\textsf{\textbf{\underline{\underline{Answer and how to solve:-}}}}[/tex]

❖ First of all, we need to make sure that all the variables are on one side (the left-hand side, LHS) and numbers are on the right-hand side, RHS.

Here we have variables on the right-hand side, so we can move that term to the left-hand side, making sure to use the opposite operation:-

13b=6b+14              Subtract 6b on both sides, which results in:-

7b=14                      Divide by 7 on both sides, which results in:-

b=2                         We've found the value of b. Now, we need to check our solution by replacing b with the value of b, 2, and seeing whether or not we end up with a true statement:-

13(2)=6(2)+14                 Multiply.

26=12+14                       Now add.

26=26                          Since the left-hand side does equal the right-hand side, we conclude that our solution, b=2, is correct.

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Answer:

+ 25

Step-by-step explanation:

b² + 10b

using the method of completing the square

add ( half the coefficient of the b- term)²

b² + 2(5)b + 25

= (b + 5)² ← a perfect square

Answer:

I did the work and uploaded the answers i hope it helped!!

Answer:

9 / 4

Step-by-step explanation:

The slope of line V can be found using the formula for slope: (y2 - y1) / (x2 - x1).

(y2 - y1) / (x2 - x1)

(13 - 9)  / (1 - 10)

(4) / (-9)

Two slopes are perpendicular when the values are reciprocals. The slope of line V is 4 / (-9). The reciprocal of this is: 9 / 4.

The expressions which represents the slope of a tangent to the curve [tex]y=\frac{1}{x+1}[/tex] at any point (x, y) are:

                                ​[tex]f'(x) =limh \rightarrow 0\frac{-h}{h(x+1)(x+h+1)}\\\\f'(x) =limh \rightarrow 0\frac{-h}{x^2h+2xh+xh^2+h^2 +1}[/tex]

Mathematically, the slope of a tangent line to the curve is given by this equation:

[tex]f'(x) =limh \rightarrow 0\frac{f(x+h)-f(x)}{h}[/tex]

Given the function:

[tex]f(x)=y=\frac{1}{x+1}[/tex]

When (x + h), we have:

[tex]f(x+h)=y=\frac{1}{x+h+1}[/tex]

Next, we would find the derivative of f(x):

[tex]f'(x) =limh \rightarrow 0\frac{\frac{1}{x+h+1} -\frac{1}{x+1}}{h}\\\\f'(x) =limh \rightarrow 0\frac{x+1 -(x+h+1)}{h(x+1)(x+h+1)}\\\\f'(x) =limh \rightarrow 0\frac{x+1 -x-h-1}{h(x+1)(x+h+1)}\\\\f'(x) =limh \rightarrow 0\frac{-h}{h(x+1)(x+h+1)}\\\\f'(x) =limh \rightarrow 0\frac{-h}{x^2h+2xh+xh^2+h^2 +1}\\\\f'(x) = \frac{-1}{(x+1)(x+1)} \\\\f'(x) = \frac{-1}{(x+1)^2}[/tex]

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Answer:

B: 12.5%

Step-by-step explanation:

You take the 20.000- 8.500 the number decreased with= 11.5.

You then round of your answer to the nearest 10th, the number 11,5 will change to 12,50.

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[tex]\diamond\large\blue\textsf{\textbf{\underline{Question:-}}}[/tex]

          ✳︎ 5-3x=-10. what is x?

[tex]\diamond\large\blue\textsf{\textbf{\underline{\underline{Answer and how to solve:-}}}}[/tex]

 ➪ First, subtract 5 on both sides:-

-3x=-10-5

 ❒ Which simplifies to:-

 ✩ -3x=-15

Divide by -3 on both sides:-

   ✩  x=5

So we conclude that the value of x is 5.

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[tex]\large\displaystyle\text{$\begin{gathered}\sf 1) \ \left\{\begin{matrix} 5x-3y-z=1&\\ x+4y-6z=-1& \\ 2x+3y+4z=9& \end{matrix}\right. \end{gathered}$}[/tex]

The coefficient matrix associated with the system is constructed and the columns and rows are reduced.

                [tex]\left ( \begin{array}{ccc|c} 5 & -3 &-1 &1\\ 1 & 4 & -6 &-1\\ 2 & 3 & 4 & 9 \end{array} \right ) \begin{array}{c} \xrightarrow[\textup{F}_3- 2\textup{F}_2]{\textup{F}_1 - 5\textup{F}_2} \end{array} \left ( \begin{array}{ccc|c} 0 & -23& 29 &6\\ 1 & 4& -6 &-1\\ 0 & -5 &16 & 11 \end{array} \right )\\ \ \\ \ \\ \begin{array}{c} \xrightarrow[]{23\textup{F}_3 - 5\textup{F}_1} \end{array} \left ( \begin{array}{ccc|c} 0 & -23& 29 &6\\ 1 & 4& -6 &-1\\ 0 & 0 & 223 & 223 \end{array} \right )[/tex]

So, z=1. Translating the last matrix to the associated system of equations, we have that x=1, y=1, since:

                                       [tex]\large\displaystyle\text{$\begin{gathered}\sf \begin{align*} y&=\dfrac{6-29}{-23}=\dfrac{-23}{-23}=1\\ x = -1+6-4=1 \end{gathered}$}[/tex]

[tex]\large\displaystyle\text{$\begin{gathered}\sf \bf{2) \ \left\{\begin{matrix} x+y=7& \\ x\times y=12& \end{matrix}\right. } \end{gathered}$}[/tex]

Clear an unknown in the first equation and substitute the resulting expression in the second. Next, the quadratic equation is solved.

             [tex]\large\displaystyle\text{$\begin{gathered}\sf \bf{ y=7-x \quad \Longrightarrow \quad 12=x(7-x)=7x-x^2 } \end{gathered}$}[/tex]

                    [tex]\large\displaystyle\text{$\begin{gathered}\sf \bf{ 0=x^2-7x+12=(x-4)(x-3); } \end{gathered}$}[/tex]

                [tex]\large\displaystyle\text{$\begin{gathered}\sf \bf{ x_1=4\qquad x_2=3 \Longrightarrow y_1=3 \qquad y_2=4 } \end{gathered}$}[/tex]

[tex]\large\displaystyle\text{$\begin{gathered}\sf \boldsymbol{3) \ \left\{\begin{matrix} x^2+y^2=169& \\ x+y=17& \end{matrix}\right. } \end{gathered}$}[/tex]

Clear an unknown in the first equation and substitute the resulting expression in the second. Next, the quadratic equation is solved.

[tex]\large\displaystyle\text{$\begin{gathered}\sf \boldsymbol{ y=17-x \quad \Longrightarrow \quad 169=x^2 +(17-x)^2=x^2+x^2 -34x+289 } \end{gathered}$}[/tex]

[tex]\large\displaystyle\text{$\begin{gathered}\sf \boldsymbol{ 0=2x^2-34x+120=2(x^2-17+60)=2(x-12)(x-5); } \end{gathered}$}[/tex]

               [tex]\large\displaystyle\text{$\begin{gathered}\sf \boldsymbol{ x_1=4\qquad x_2=3 \Longrightarrow y_1=3 \qquad y_2=4 } \end{gathered}$}[/tex]

[tex]\large\displaystyle\text{$\begin{gathered}\sf \pmb{4 \ \left\{\begin{matrix} y^2-2y+1=x & \\ \sqrt{x}+y=5 & \end{matrix}\right. } \end{gathered}$}[/tex]

Substitute the expression that represents x into the second equation. Then both sides of the equation are squared and solved.

   [tex]\large\displaystyle\text{$\begin{gathered}\sf \pmb{ \sqrt{y^2-2y+1}+y=5\quad \Longrightarrow\quad \left ( \sqrt{y^2-2y+1} \right )^2=(5-y)^2 } \end{gathered}$}[/tex]

                  [tex]\large\displaystyle\text{$\begin{gathered}\sf \pmb{ y^2-2y+1=y^2-10y+25\quad \Longrightarrow\quad 8y=24; } \end{gathered}$}[/tex]

                                     [tex]\large\displaystyle\text{$\begin{gathered}\sf \pmb{ y=6 \qquad x=4 } \end{gathered}$}[/tex]