Which equation can be used to solve for xxx in the following diagram?

The simplified equation of [tex](x - 10)^2 + (y- 6)^2 = 4^2[/tex] is [tex]y = 6+ \sqrt{16 - (x - 10)^2}[/tex]

The complete question is added as an attachment

The equation is given as

[tex](x - 10)^2 + (y- 6)^2 = 4^2[/tex]

Subtract (x - 10)^2 from both sides

[tex](y- 6)^2 = 4^2 - (x - 10)^2[/tex]

Evaluate 4^2

[tex](y- 6)^2 =16 - (x - 10)^2[/tex]

Take the square root of  both sides

[tex]y- 6= \sqrt{16 - (x - 10)^2}[/tex]

Add 6 to both sides

[tex]y = 6+ \sqrt{16 - (x - 10)^2}[/tex]

Hence, the simplified equation of [tex](x - 10)^2 + (y- 6)^2 = 4^2[/tex] is [tex]y = 6+ \sqrt{16 - (x - 10)^2}[/tex]

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The solution to the system of equations is (-1, 3) and (3, -1)

The system of equation is given as:

y = -x^2 + x + 5

y = -x + 2

Next, we plot the graph of the system using a graphing tool

From the graph, both inequalities intersect at

(-1, 3) and (3, -1)

Hence, the solution to the system of equations is (-1, 3) and (3, -1)

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Answer:

graph 1,

3,-1 and -1,3

Step-by-step explanation:

See screenshot

The value of f[ -4 ] and g°f[-2] are [tex]\frac{14}{3}[/tex] and 13 respectively.

Given the function;

For f[ -4 ], we substitute -4 for every variable x in the function.

[tex]f(x) = \frac{3x-2}{x+1}\\\\f(-4) = \frac{3(-4)-2}{(-4)+1}\\\\f(-4) = \frac{-12-2}{-4+1}\\\\f(-4) = \frac{-14}{-3}\\\\f(-4) = \frac{14}{3}[/tex]

For g°f[-2]

g°f[-2] is expressed as g(f(-2))

[tex]g(\frac{3x-2}{x+1}) = (\frac{3x-2}{x+1}) + 5\\\\g(\frac{3x-2}{x+1}) = \frac{3x-2}{x+1} + \frac{5(x+1)}{x+1}\\\\g(\frac{3x-2}{x+1}) = \frac{3x-2+5(x+1)}{x+1}\\\\g(\frac{3x-2}{x+1}) = \frac{8x+3}{x+1}\\\\We\ substitute \ in \ [-2] \\\\g(\frac{3x-2}{x+1}) = \frac{8(-2)+3}{(-2)+1}\\\\g(\frac{3x-2}{x+1}) = \frac{-16+3}{-2+1}\\\\g(\frac{3x-2}{x+1}) = \frac{-13}{-1}\\\\g(\frac{3x-2}{x+1}) = 13[/tex]

Therefore, the value of f[ -4 ] and g°f[-2] are [tex]\frac{14}{3}[/tex] and 13 respectively.

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128000.

Let the three frithe ends x, y, z x is the first friend y is secmy ond friend

z is third friend

X+y+z=128000  

x=4z    

z=3y    

x=4(3y)=12y   put values of x and z in the first equation we got

12y+y+3y=128000

16y=128000

= 128000 - = 8000 16

Put the y value in equation 3 wee got

z=3*8000=24000

Put the z value in equation 2 we got

x=4*24000=96000

8000+24000+96000=128000.

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The equivalent expressions of T=A^1.5 are: T= (A)^3/2 and T= ((A)^1/2)^3

According to the statement

we have to show that the 1.5 is same as 3/2 are equivalent to T.

So,

Equivalent expressions are expressions with the same value.

The Given expression is

T=A^1.5

we can write it as a

T= (A)^3/2

Because we know that the

1.5 = 3/2

So, it become T= (A)^3/2.

And From

T=A^1.5

we can write it as a

T= ((A)^1/2)^3

because we know that the

3/2 = (1/2)^3.

So, The equivalent expressions of T=A^1.5 are: T= (A)^3/2 and T= ((A)^1/2)^3.

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BC is 3 across & CA is 3 up, but this creates a right-angle triangle. So, Pythagoras Theorem is used to find AB

AB = root 3^2 + 3^2

AB = 3 root 2

or 4.242640687

Thus, AB is 4.24 to 2d.p

(root means square root)

Hope this helps!

Answer:

7.8g/cm

Step-by-step explanation:

Answer:

C . Sixteen and four eighths.

Step-by-step explanation:

2 6/8 * 6

= 22/8 * 6

= 6*22 / 8

= 132/8

= 16 4/8.

Answer:

Step-by-step explanation:

1/2 57 * 19.5 =

28.5 * 19.5 =

555.75m²

Answer:

  r = 5

Step-by-step explanation:

Solving radical equations often results in extraneous solutions. These can be avoided by using a graphing calculator for the solution.

The attached shows that the value of r that makes both sides of the equation have the same value is r = 5.

Check:

  √(5·5 -9) -3 = √(5 +4) -2   ⇒   4 -3 = 3 -2 . . . true

Solution of an equation like this is typically done by isolating the radical expressions, then squaring the equation.

We can add 3, square the equation, then isolate the radical and square again:

  [tex]\displaysyle \sqrt{5r-9}-3=\sqrt{r+4}-2\\\\\sqrt{5r-9}=\sqrt{r+4}+1\qquad\text{add 3}\\\\5r-9=(r+4)+2\sqrt{r+4}+1\qquad\text{square both sides}\\\\(4r-14)^2=4(r+4)\qquad\text{subtract $(r+5)$ and square again}\\\\16r^2-116r+180=0\qquad\text{rewrite in standard form}\\\\4(4r -9)(r -5) = 0\qquad\text{factor}\\\\r=\{\dfrac{9}{4},\ 5\}\qquad\text{solutions to the factored form}[/tex]

Trying these values in the original equation, we get ...

  √((5(9/4) -9) -3 = √(9/4 +4) -2   ⇒   1.5 -3 = 2.5 -2 . . . . . false

  √(5(5) -9) -3 = √(5 +4) -2   ⇒   4 -3 = 3 -2 . . . . . true

The solution is r = 5.

Answer:

y^2 + 5y = 0

y(y + 5) = 0

y = 0, -5

The answer is y = 0 and y = -5.

The given question :

Answer:

department

Step-by-step explanation:

A "task force" group is another name for a command group.

We have,

A task force is a temporary grouping of individuals or units with a specific objective or mission.

It is typically formed to address a particular task or problem and is led by a designated commander or leader.

Task forces can be assembled from different departments or organizations to collaborate and work together towards a common goal.

Thus,

A "task force" group is another name for a command group.

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Answer:

Step-by-step explanation:

25.5625

Answer:

25.5625

Step-by-step explanation:

The solution is in the image

The next step Juan will take is to multiply the second fraction by (x-2)/(x-2).

Given that Juan needs to rewrite this difference as one expression (3x/(x²-7x+10))-(2x/(3x-15)) and factor the denominator as (3x/(x-2)(x-5))-(2x/3(x-5)).

An algebraic expression in mathematics is an expression composed of variables and constants and algebraic operations (addition, subtraction, etc.). Expressions are made up of concepts.

The given expression is (3x/(x²-7x+10))-(2x/(3x-15))

Firstly, we will factored the denominator as

(3x/(x-2)(x-5))-(2x/3(x-5)).

Now, we will multiply and divide the second fraction by (x-2), we get

(3x/(x-2)(x-5))-((2x(x-2))/3(x-5)(x-2)).

Hence, the next step when subtracting these expressions (3x/(x-2)(x-5))-(2x/3(x-5)) is multiply the second fraction by (x-2)/(x-2).

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The best point estimate for the proportion of all the residents opposed to the photo cop use is 60%  which is the fourth option.

Given that in a survey of 456 residents , 279 were opposed to the use of the photo cop for issuing traffic tickets.

We have to find the best point estimate for the proportion of all residents opposed to the photo cop use.

Total residents=456

Residents who opposed to the use of the photo cop for issuing traffic tickets=279

Proportion=number of residents who opposed/ total residents

=279/456

=0.611

=61.1%

The nearest option is fourth which is 60%.

Hence the best point estimate for the proportion for the proportion of all the residents opposed to the photo cop use is 60% which is option 4.

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Question is incomplete as it should include options which are :

1) 500

2)50%

3)300

4)60%

5)40%

The quantity of water left in the tank as described in the task content is; 111 l 220ml.

The tank described in the task content initially contains 200 liters, in which case, 1 liter corresponds to 1000ml.

Hence, when 88l 580ml was used, the remaining quantity of water is; 200l - (88l 580ml) = 111 l 420ml.

And finally, since 200 ml leaked, the quantity of water left is; 111l 220ml.

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Using the binomial distribution, the mean number of balls you would expect him to catch is of 7.2.

It is the probability of exactly x successes on n repeated trials, with p probability of a success on each trial.

The expected value of the binomial distribution is:

E(X) = np

For this problem, the parameters are:

n = 8, p = 0.9.

Hence the mean is:

E(X) = np = 8 x 0.9 = 7.2.

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Answer:

patient A has the greater weight

Answer:

(f - g)(300) = $1,300 is the profit made from 300 tablets

Step-by-step explanation:

(f - g)(300) is the profit made from 300 tablets

(f - g)(x) = (3 x + 6,000) - (20 x - 400)

- Simplify it

(f - g)(x) = 3 x + 6,000 - 20 x + 400

- Add like terms

(f - g)(x) = -17 x + 6,400

Substitute x by 300

(f - g)(300) = -17(300) + 6,400

(f - g)(300) = -5,100 + 6,400

(f - g)(300) = $1,300

The profit is $1,300

Minimum: 202

Quartile (Q1): 208

Median: 218

Quartile Q3: 229

Maximum: 237

The max and min value, the lower and upper quartiles, and the median of a data comprise the five-number summary.

Given the data: 236, 202, 218, 208, 225, 229, 211, 237, 223, 206, 217, 216, 232, 225, 208, we would have the following:

Minimum: 202 (least value)

Quartile (Q1): 208 (center of the first half of the data)

Median: 218 (center of the data when ordered)

Quartile Q3: 229 (center of the second half of the data)

Maximum: 237 (highest value).

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The angle of the given ray is of x = 150º.

The angle of the ray is given by the sum of the angles that compose the ray.

In this problem, the angles are of 65º, 40º and 45º, hence:

x = 65 + 40 + 45 = 150º

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The mean of the data given is 13.4364 and.the standard deviation is 1.788.

It should be noted that the mean is the average of the giving set of numbers. In this situation, the mean is 3.1998.

E(X²) will be:

= (0² × 0.0408) + (1² × 0.1304) .... + (12² × 0.0001)

= 13.4364

Var(X) = 13.4364 - 3.1998²

= 3.198

The standard deviation will be:

= ✓3.198

= 1.788

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By definite integrals and area formula of a rectangle, we find that the area of the region in the first quadrant is 147 / 4 square units.

Integrals can be used to determine the area of regions bounded by curves set on Cartesian plane. The upper limit of the integral of the question is initially found:

(2 / 3) · x = 7

x = 21 / 2

The area can be defined as an rectangle in the first quadrant minus the area below the linear equation:

[tex]A = \left(\frac{21}{2} \right) \cdot (7) - \frac{2}{3} \int\limits^{\frac{21}{2} }_{0} {x} \, dx[/tex]

[tex]A = \frac{147}{2} - \frac{1}{3} \cdot \left[\left(\frac{21}{2} \right)^{2}-0^{2}\right][/tex]

A = 147/4

By definite integrals and area formula of a rectangle, we find that the area of the region in the first quadrant is 147 / 4 square units.

The statement is poorly formatted and incomplete. Correct form is shown below:

A region in the first quadrant is bounded by the line y = (2/3) · x, the y-axis and the horizontal line y = 7. Determine the area of the region.

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Her annual salary after 8 years will be $57,600.

To find what is her annual salary, to the nearest $100, after 8 years of service:

Find 4% of 45,000

[tex]\frac{x}{45000} *100=4\\\frac{x}{450} =4\\x =1800[/tex]

So, 4% of $45,000 of $1800.

1st year = $45000

2nd year = 45000 + 1800 = $46800

3rd year = 46800 + 1800 = $48600

4th year = 48600 + 1800 = $50400

5th year = 50400 + 1800 = $52200

6th year = 52200 + 1800 = $54000

7th year = 54000 + 1800 = $55800

8th year = 55800 + 1800 = $57600

Therefore, her annual salary after 8 years will be $57,600.

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The salary of Sonya after 8 years will be $61,600 (nearest $100).

The interest on the a loan or deposit that is calculated using the both initial principal as well as the accrued interest from prior periods is known as compound interest (sometimes known as compounding interest).

Some features of compounding are-

Calculation for the amount after 8 years-

The principal amount is $45,000

Rate of interest is 4% per year.

Time duration is 8 years.

The formula for total amount after compounding is -

[tex]A=P\left(1+\frac{r}{n}\right)^{n t}[/tex]

A  = final amount

P  = initial principal balance

r  = interest rate

n  = number of times interest applied per time period.

t  = number of time periods elapsed.

As the rate of interest is applied only one in a year ; n = 1.

Substitute the values in the formula to get the value.

[tex]A=45,000\left(1+\frac{.04}{1}\right)^{8}\\A=45,000\left(1.04}\right)^{8}\\\\A=61585.60[/tex]

Therefore, the salary of Sonya will be $61600 (nearest $100) after 8 years.

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The equation of a circle whose centre ([tex]-5,-2[/tex]) and radius [tex]2[/tex] is [tex]x^{2}+y^{2}+10x+4y+25=0[/tex]

What is equation of a circle?

A circle is a closed curve drawn from a fixed point called centre of the circle. The distance between the centre of the circle and the arc of the circle is called the radius of the circle.

The equation of a circle with centre [tex](h, k)[/tex] radius [tex]r[/tex] is

[tex](x-h)^{2} +(y-k)^{2}=r^{2}[/tex]

Given center = [tex](-5,-2)[/tex]  and Radius = [tex]2[/tex]

Put the Given value in the equation:

= [tex](x-(-5))^{2} +(y-(-2))^{2} = 2^{2}[/tex]

= [tex](x+5)^{2} +(y+2)^{2}=2^{2}[/tex]

= [tex](x^{2} +10x+25)+(y^{2}+4y+4)= 4[/tex]

= [tex]x^{2} +y^{2}+10x+ 4y+ 29 = 4[/tex]

= [tex]x^{2} +y^{2}+10x +4y+ 29-4=0[/tex]

= [tex]x^{2} +y^{2}+10x+4y+25=0[/tex]

So, the equation of the circle whose centre is [tex](-5,-2)[/tex] and Radius [tex]2[/tex] is

[tex]x^{2}+y^{2}+10x+4y+25=0[/tex]

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If ages are normally distributed then using a significance level of 0.05 , we can claim that the population mean is greater than 24.

Given sample mean of 25 years , sample size of 16 ,standard deviation of 2 years.

We have to find whether the population average is greater than 24.

We have to use t test because n is less than 30. It is right tailed.

We have to first form Hypothesis.

[tex]H_{0}[/tex]:μ>24,

[tex]H_{1}[/tex]:μ<24.

t=X-μ/S/[tex]\sqrt{n}[/tex]

t critical at 5% significance level and degree of freedom=15    (16-1)=1.7531

t=24-25/2/[tex]\sqrt{16}[/tex]

=-1/0.5

=-2

Because 1.7531 is greater than -2 so we will accept the null hypothesis.

Hence we can say that the average of population is greater than 24.

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The illustration to determine whether side-side-angle (SSA) is a valid means for establishing triangle congruence is given below.

Congruence of triangles means that two triangles are said to be congruent if all three corresponding sides are equal and all the three corresponding angles are equal in measure.

In this case, these triangles can be slides, rotated, flipped and turned to be looked identical. If repositioned, they coincide with each other.

To illustrate the question, this will be:

Step 1: Draw  a line segment AB

Step 2:Draw an acute angle (let's say 45° ) at A  as AX

Step 3: From B draw an altitude on AX     BY ⊥ AX

Step 4  Using suitable compass width and taking  Y as center cut AX  on both sides  at C and D

Now BC = BD   as  BY is perpendicular bisector

Now compare ΔABC and ΔABD. Using suitable compass width and taking B as center cut AX at 2 points C & D  directly  so BC = BD.

This can illustrate that SSA is not a valid  means for establishing triangle congruence.

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Answer:

Below is an example to show whether side-side-angle (SSA) is a legitimate method for proving triangle congruence. How should the information be illustrated? Triangle congruence refers to the property that two triangles are congruent if their three corresponding sides and their three corresponding angles are both equal in size. In this instance, the triangles can be moved around, rotated, flipped, and turned to have a same appearance. They are in alignment with one another when moved. This can serve as an example of why SSA cannot be used to prove triangle congruence.

Step-by-step explanation:

The answer is (f . g)(x) = - (34k + 17k⁷ + 17k⁵ + 40k³ + 20k² + 60).

Remember that :

(f . g)(x) = f(x) × g(x)

Hence :

ANSWER: The third statement appears to be incorrect.