Determine the points of intersection of the equation circumference x² + (y-3) ² = 25 with the coordinate axes.

Answer:

The awnser is 64

Step-by-step explanation:

It is 64 because you devide the perimiter (32) by 4, to get all sides, and then multiply 1 side by the other to get 64

THIS METHOD ONLY WORKS FOR A SQUARE

Answer:

Range of Function :  { - 9, - 5, - 1,  4 }

Step-by-step explanation:

We know that y = 2x - 5 provided the domain ( x - values ) { - 2, 0, 2, 4 }. Let us substitute each element in this set of domain as x in the equation "y = 2x - 5" as to solve for the y - values, otherwise known as the range of the function.

{ - 2, 0, 2, 4 }

y = 2( - 2 ) - 5 = - 9,

y = 2( 0 ) - 5 = - 5,

y = 2( 2 ) - 5 = - 1,

y = 2( 4 ) - 5 = 4

We have the set of y - values as { - 9, - 5, - 1, 4 }. This is the range of our function.

Answer:

The answer is option B

Step-by-step explanation:

2ln 2 - 2ln3

= ln ( 2²/3²)

= ln (4/9)

Hope this helps you

Answer:

B) 2ln 2 – 2ln 3

Step-by-step explanation:

Answer:

3 dolla

Step-by-step explanation:

Part A

House 1: we make exponential can be used to describe the

value of house 1. After 1 year, 253480 = 249000 x 1.02

After 2 years, 259059.6 = 25398×1.02. After 3 years 264240.79 =259059.60x1.02.

House 1: we make linear. 256000=249000 + 1000.263000 = 256000 +7000.270000=263000 + 1000.

Part B.

House 1: f(x)=249000 1.02x [genometic sequence]

House 2: f(X) = 24900 + 1000x [arithmetic progression]

Part C

House 1: f(45)=249000×1.02 45 = 607025.697

House 2: f(45) = 249000 +7000x45 = 564000

"House 1 is an arithmetic sequence that increases in price each year while House 2 is an arithmetic sequence that only increases by 1000 per year.

In computational mathematics, a word problem is a problem of deciding whether two given expressions are equivalent with respect to a set of rewriting identities. A prototypical example is the word problem for groups, but there are many other instances as well.

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Answer: B.

k=3

Step-by-step explanation:

g(x) is 3 units higher than f(x)

so k must be equal to 3

Answer:

it is b i think

k=3

Step-by-step explanation:

Step-by-step explanation:

the inverse of the inverse of a function is the function itself

It's like going and coming back

Answer:

9

Step-by-step explanation:

The inverse of the inverse is the input

f^-1 (f^-1(9))  = 9

As a proof

f(x) = 4x+5

y = 4x+5

Exchange x and y

x = 4y+5

Solve for y

(x+5) = 4y

(x+5)/4 = y

f^-1 (x) =  ( x-5)/4

Then find the inverse of this function

y = ( x-5) /4

Exchange x and y

x = (y-5)/4

4x = y-5

4x+5 = y

Which gives us the original function back

f^-1( input) = 4x+5

Using the value given

f^-1 (9) =  ( 9-5)/4=4/4 =1

Using the 1 from the previous part as the input

f^-1( 1) = 4*1+5  = 4+5 = 9

Answer:

HIG=28°

Step-by-step explanation:

HI is a diameter therefore it's measure is 180°

HI-IG=HG

180-128=HG

52°=HG

HIG=1/2HG

HIG=1/2(56°)

HIG=28°

Answer:

[tex]$\frac{3\pi }{2}$[/tex]

Step-by-step explanation:

Solve for the interval: [tex][0, 2\pi)[/tex]

[tex]4 \csc(\theta)+1=-3[/tex]

[tex]4 \csc(\theta)=-4[/tex]

[tex]\csc(\theta)=-1[/tex]

[tex]$\frac{1}{\sin(\theta)} =-1 \Rightarrow \sin(\theta)=-1$[/tex]

We have [tex]\sin(\theta)=-1[/tex] for coterminal angles [tex]$\frac{3\pi }{2}+2\pi n$[/tex]

Once we just want the interval [tex][0, 2\pi)[/tex]

The solution is [tex]$\frac{3\pi }{2}$[/tex]

Answer:

hope it helps

this is my best

please tell me if I'm wrong

Answer:

mark my answer as the brainliest

Step-by-step explanation:

x = 53/4

Answer:

4×+1-9 (2^)+2)=0

4×+1-9 (4+2)=0

4×+1(-36-18)=0

4×+1 (-54)=0

4×-54=0

4×=54

×=54÷4

×=13.5

Answer:

y = 3x

Step-by-step explanation:

Assume a slope of 3

The y intercept is 0

The slope intercept form of the equation is

y= mx+b where m is the slope and b is the y intercept

y =3x+0

y = 3x

Answer:

7 < (third side length) < 37

Step-by-step explanation:

Add these two side lengths together.  We get 37.  Unless the 3rd side is LESS than 37, we cannot form a triangle.  

Next, subtract 15 from 22, obtaining 7.  The 3rd side has to be GREATER than 7 in order for the sides 7+, 15 and 22 to form a triangle.

7 < (third side length) < 37

Answer:

26.63 ??

Step-by-step explanation:

I think you use pythagorean theorum to solve.

do 22^2 and 15^2 and get 709.1569

sqrt of that is 26.63

Answer:

A and C

Step-by-step explanation:

The roots of the quadratic function f(q) = q² - 125 are q = 5 and q = -5.

A function has an input and an output.

A function can be one-to-one or onto one.

It simply indicated the relationships between the input and the output.

Example:

f(x) = 2x + 1

f(1) = 2 + 1 = 3

f(2) = 2 x 2 + 1 = 4 + 1 = 5

The outputs of the functions are 3 and 5

The inputs of the function are 1 and 2.

We have,

The quadratic function f(q) = q² - 125 can be factored as (q - 5)(q + 5).

Now,

To find the roots, we set the function equal to zero and solve for q:

So,

q² - 125 = 0

(q - 5)(q + 5) = 0

q - 5 = 0 or q + 5 = 0

q = 5 or q = -5

Therefore,

The roots of the quadratic function f(q) = q² - 125 are q = 5 and q = -5.

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Answer:

62.5°

Hope this helps :)

Answer:

The rhombus ABCD has an area of 22 square units.

Step-by-step explanation:

The coordinates of rhombus ABCD are shown in the image attached below. The area of the rhombus can be found in terms of their diagonals, which are now calculated by Pythagorean Theorem:

[tex]AC = \sqrt{[6-(-4)]^{2}+[8-(-4)]^{2}}[/tex]

[tex]AC = 15.620[/tex]

[tex]BD = \sqrt{(4-6)^{2}+[0-(-2)]^{2}}[/tex]

[tex]BD \approx 2.828[/tex]

The area of the rhombus is: ([tex]AC = 15.620[/tex] and [tex]BD \approx 2.828[/tex])

[tex]A = \frac{AC\cdot BD}{2}[/tex]

[tex]A = \frac{(15.620)\cdot (2.828)}{2}[/tex]

[tex]A = 22.087[/tex]

The rhombus ABCD has an area of 22 square units.

Answer:

22 units

Step-by-step explanation:

Answer:

According to the graph, f(−3)=3 and f(9)=-7

-7-3/9-(-3)= −10/12

Therefore the answer is -5/6

From Khan

The average rate of change over the interval [-3,9] will be 63 units per one unit change in y - value.

        [tex]b^x = \underbrace{b \times \dots \times b}_{x \text{ times}}[/tex]

        where : [m] → is slope of line and [c] → is the y - intercept

We have the following function -

f(x) = x³ - 9

The average rate of change would be -

r = {720 - (- 36)}/{9 - (- 3)}

r = 756/12

r = 63

Therefore, the average rate of change over the interval [-3,9] will be 63 units per one unit change in y - value.

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[ Complete question -

what is the average rate of change of f(x) = x³ - 9 over the interval [-3,9] ]

Answer:

C. [tex](5-\frac{1}{2})^6[/tex]

Step-by-step explanation:

Given

[tex]15(5)^2(-\frac{1}{2})^4[/tex]

Required

Determine which binomial expansion it came from

The first step is to add the powers of he expression in brackets;

[tex]Sum = 2 + 4[/tex]

[tex]Sum = 6[/tex]

Each term of a binomial expansion are always of the form:

[tex](a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......[/tex]

Where n = the sum above

[tex]n = 6[/tex]

Compare [tex]15(5)^2(-\frac{1}{2})^4[/tex] to the above general form of binomial expansion

[tex](a+b)^n = ......+15(5)^2(-\frac{1}{2})^4+.......[/tex]

Substitute 6 for n

[tex](a+b)^6 = ......+15(5)^2(-\frac{1}{2})^4+.......[/tex]

[Next is to solve for a and b]

From the above expression, the power of (5) is 2

Express 2 as 6 - 4

[tex](a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......[/tex]

By direct comparison of

[tex](a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......[/tex]

and

[tex](a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......[/tex]

We have;

[tex]^nC_ra^{n-r}b^r= 15(5)^{6-4}(-\frac{1}{2})^4[/tex]

Further comparison gives

[tex]^nC_r = 15[/tex]

[tex]a^{n-r} =(5)^{6-4}[/tex]

[tex]b^r= (-\frac{1}{2})^4[/tex]

[Solving for a]

By direct comparison of [tex]a^{n-r} =(5)^{6-4}[/tex]

[tex]a = 5[/tex]

[tex]n = 6[/tex]

[tex]r = 4[/tex]

[Solving for b]

By direct comparison of [tex]b^r= (-\frac{1}{2})^4[/tex]

[tex]r = 4[/tex]

[tex]b = \frac{-1}{2}[/tex]

Substitute values for a, b, n and r in

[tex](a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......[/tex]

[tex](5+\frac{-1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......[/tex]

[tex](5-\frac{1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......[/tex]

Solve for [tex]^6C_4[/tex]

[tex](5-\frac{1}{2})^6 = ......+ \frac{6!}{(6-4)!4!)}*(5)^{6-4}(\frac{-1}{2})^4+.......[/tex]

[tex](5-\frac{1}{2})^6 = ......+ \frac{6!}{2!!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......[/tex]

[tex](5-\frac{1}{2})^6 = ......+ \frac{6*5*4!}{2*1*!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......[/tex]

[tex](5-\frac{1}{2})^6 = ......+ \frac{6*5}{2*1}*(5)^{6-4}(\frac{-1}{2})^4+.......[/tex]

[tex](5-\frac{1}{2})^6 = ......+ \frac{30}{2}*(5)^{6-4}(\frac{-1}{2})^4+.......[/tex]

[tex](5-\frac{1}{2})^6 = ......+15*(5)^{6-4}(\frac{-1}{2})^4+.......[/tex]

[tex](5-\frac{1}{2})^6 = ......+15(5)^{6-4}(\frac{-1}{2})^4+.......[/tex]

[tex](5-\frac{1}{2})^6 = ......+15(5)^2(\frac{-1}{2})^4+.......[/tex]

Check the list of options for the expression on the left hand side

The correct answer is [tex](5-\frac{1}{2})^6[/tex]

Answer:

pen: $1.50 notebook: $2.00

Step-by-step explanation:

4×1.50 =6

5×2 = 10

6+10= 16

3×1.50= 4.50

8×2= 16

4.50 + 16 = 20.50

Answer:

Tina

Step-by-step explanation:

Since Amy owns the dog, she doesn't have the ferret and we know that Jewel doesn't have the ferret so the answer is Tina.

Answer: D

Step-by-step explanation:

In every other solution, x gets canceled out, so it has no impact on the equation.  Thus, A, B, and C all have either infinite or no solution, because they have 6x on both sides of the equation.

Further proof:

6x-6=15x+15

Add(6)

6x=15x+21

Subtract(15x)

-9x=21

Divide(-9)

x = -2 3/9

Hope it helps <3

Answer:

d) 6x - 6 = 15x + 15

Step-by-step explanation:

Well let’s solve.

a)

6x - 15 = 6x + 15

Lets use the communicative property which is the moving of numbers.

We can start with -15, so we do +15,

6x = 6x + 30

Then we can do -6x

0 = 30

This is a false statement meaning there is no solution.

b)

6x - 6 = 6x + 15

We can start by doing +6

6x = 6x + 21

-6x

0 = 21

Again, this is a false statement meaning this has no solutions.

c)

6x + 15 = 6x + 15

So we do -15

6x = 6x

If this is the case then there is infinitely many solutions.

d)

6x - 6 = 15x + 15

So we can start by doing +5p6

6x = 15x + 21

Then we can do -15x

-9x = 21

then we can do 21 / -9 which is -2 1/3.

x = -2 1/3

So this has one solution.

Answer:

  8 servings

Step-by-step explanation:

At the ratio of 15:1, the 75 grams of rice in one serving will require 75/15 = 5 g of spice. David's inventory of 40 g of spice is enough for ...

  40 g/(5 g/serving) = 8 servings

Only 8 servings can be made with the available spice.

Rice to spice ratio = 15 : 1

Number of grams per serving = 75 grams

Grams of spice available = 40 grams

Grams of rice, r required for available spice :

15 grams of rice = 1 gram of spice

r grams of rice = 40 grams of spice

Cross multiply :

r = (40 × 15)

r = 600 grams

The grams of rice which can go with the available spice :

Number of servings = grams of rice / grams per serving

Number of serving = 600 grams / 75 grams

Number of grams per serving = 8

Hence, The number of servings which can be made with the available spice is 8 servings

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Answer:

3/10

Step-by-step explanation:

1/5 of total income - spent on car

remainder income = 4/5

3/8 of remainder = 3/8 ×4/5 = 3/10(of total income spent on rent)

the word "of" means you can find it by multiplication

e.g. 2 of 1/2 means 2 × 1/2 which is 1

Answer:

8mins

Step-by-step explanation:

(2x60)/15=8 I hope this helps, hope it's right

Answer:

8 minutes

Step-by-step explanation:

You will multiply 60 times 2 and then divide by 15 leaving you with 8

Answer:

c. (x-3) is not a factor.

Step-by-step explanation:

[tex]\text{Molly used the addition property of equality}\\\\\text{In the equation shown in the question, you can see that Molly is trying}\\\text{to find the value of "d"}\\\\\text{In order to do this, she must get the "d" variable by itself}\\\\\text{To do this, she adds "3d" to both sides to get the "d" variable on one side}\\\\[/tex]

Answer:

Total mass of materials = 4.8kg

For gravel

55% of the materials are gravel

So we have

55/100 × 4.8 kg

= 2.64kg

For sand

30% of the materials are sand

We have

30/100 × 4.8

= 1.44kg

For cement

The remaining materials were cement so we add the masses of both the sand and gravel and subtract it from the total mass

That's

4.8 - (2.64 + 1.44)

4.8 - 4.08

= 0.72 kg

She earns £ 800 per week

She receives a raise of 7.5% which is

100 + 7.5 = 107.5%

So we have

107.5 / 100 × £ 800

= £ 860

She now earns £ 860 per week

Hope this helps you

Step-by-step explanation:

1)of 4.8kg concrete 55% is gravel,30% is sand and 15% is cement.

percent=Rate*Base

Pgravel=55%*4.8kg

Pgravel=2.64kg

Psand=30%*4.8kg

Psand=1.44kg

Pcement=15%*4.8kg

Pcement=0.72kg

2)percent=Rate*Base

percent=7.5%*£800

percent=60£

total earn=60£ + £800

total earn=£860

Answer:

a + b = 12

Step-by-step explanation:

Given

Quadrilateral;

Vertices of (0,1), (3,4) (4,3) and (3,0)

[tex]Perimeter = a\sqrt{2} + b\sqrt{10}[/tex]

Required

[tex]a + b[/tex]

Let the vertices be represented with A,B,C,D such as

A = (0,1); B = (3,4); C = (4,3) and D = (3,0)

To calculate the actual perimeter, we need to first calculate the distance between the points;

Such that:

AB represents distance between point A and B

BC represents distance between point B and C

CD represents distance between point C and D

DA represents distance between point D and A

Calculating AB

Here, we consider A = (0,1); B = (3,4);

Distance is calculated as;

[tex]Distance = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}[/tex]

[tex](x_1,y_1) = A(0,1)[/tex]

[tex](x_2,y_2) = B(3,4)[/tex]

Substitute these values in the formula above

[tex]Distance = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}[/tex]

[tex]AB = \sqrt{(0 - 3)^2 + (1 - 4)^2}[/tex]

[tex]AB = \sqrt{( - 3)^2 + (-3)^2}[/tex]

[tex]AB = \sqrt{9+ 9}[/tex]

[tex]AB = \sqrt{18}[/tex]

[tex]AB = \sqrt{9*2}[/tex]

[tex]AB = \sqrt{9}*\sqrt{2}[/tex]

[tex]AB = 3\sqrt{2}[/tex]

Calculating BC

Here, we consider B = (3,4); C = (4,3)

Here,

[tex](x_1,y_1) = B (3,4)[/tex]

[tex](x_2,y_2) = C(4,3)[/tex]

Substitute these values in the formula above

[tex]Distance = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}[/tex]

[tex]BC = \sqrt{(3 - 4)^2 + (4 - 3)^2}[/tex]

[tex]BC = \sqrt{(-1)^2 + (1)^2}[/tex]

[tex]BC = \sqrt{1 + 1}[/tex]

[tex]BC = \sqrt{2}[/tex]

Calculating CD

Here, we consider C = (4,3); D = (3,0)

Here,

[tex](x_1,y_1) = C(4,3)[/tex]

[tex](x_2,y_2) = D (3,0)[/tex]

Substitute these values in the formula above

[tex]Distance = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}[/tex]

[tex]CD = \sqrt{(4 - 3)^2 + (3 - 0)^2}[/tex]

[tex]CD = \sqrt{(1)^2 + (3)^2}[/tex]

[tex]CD = \sqrt{1 + 9}[/tex]

[tex]CD = \sqrt{10}[/tex]

Lastly;

Calculating DA

Here, we consider C = (4,3); D = (3,0)

Here,

[tex](x_1,y_1) = D (3,0)[/tex]

[tex](x_2,y_2) = A (0,1)[/tex]

Substitute these values in the formula above

[tex]Distance = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}[/tex]

[tex]DA = \sqrt{(3 - 0)^2 + (0 - 1)^2}[/tex]

[tex]DA = \sqrt{(3)^2 + (- 1)^2}[/tex]

[tex]DA = \sqrt{9 + 1}[/tex]

[tex]DA = \sqrt{10}[/tex]

The addition of the values of distances AB, BC, CD and DA gives the perimeter of the quadrilateral

[tex]Perimeter = 3\sqrt{2} + \sqrt{2} + \sqrt{10} + \sqrt{10}[/tex]

[tex]Perimeter = 4\sqrt{2} + 2\sqrt{10}[/tex]

Recall that

[tex]Perimeter = a\sqrt{2} + b\sqrt{10}[/tex]

This implies that

[tex]a\sqrt{2} + b\sqrt{10} = 4\sqrt{2} + 2\sqrt{10}[/tex]

By comparison

[tex]a\sqrt{2} = 4\sqrt{2}[/tex]

Divide both sides by [tex]\sqrt{2}[/tex]

[tex]a = 4[/tex]

By comparison

[tex]b\sqrt{10} = 2\sqrt{10}[/tex]

Divide both sides by [tex]\sqrt{10}[/tex]

[tex]b = 2[/tex]

Hence,

a + b = 2 + 10

a + b = 12

Answer:

a+b=6

Step-by-step explanation:

The tutor verified answer is mostly correct however, if you look under both by comperision sections you will see that it is:

[tex]4\sqrt{2}[/tex] and [tex]2\sqrt{10}[/tex] thus the answer is 4+2=6

3x+4y=12 is Linear.

y-1/2x+6 is Linear.

y=4x^2 is nonlinear.

y+2=|4| is linear.

y+6=4x is linear.

In order to tell whether an equation is linear or not, you have to first see if there is any variable^2 or more than ^1. If that is the case, it isn't linear. For example the equation x^2=4 or y^5=100 isn't linear. It also isn't linear if the equation is divided by X or Y.For example the equation 1/x=1 isn't linear or 1/y^2=5 isn't linear.

A question is also linear if you can write it in the form y=mx+b , 'm' meaning the slope of the equation and 'b' meaning the y-intercept of the equation.