Question is in the photo!

For function 1

p=r+7

On comparing to y=mx+b

For function 2

y intercept is 8

Function 2 has greater intercept

Answer:

  (c)  $63.75

Step-by-step explanation:

The discounted price will be 100% -25% = 75% of the marked price. Mike's final price will be ...

  ($80 +5)×0.75 = $63.75

Answer:

Therefore the equation is y=-x+4 (correct optlon: C)

Step-by-step explanation:

In order to find the equation that passes through both points, we can use the slope-intercept form of the linear equation:

y=mx+b

Where m is the slope and b is the y-intercept.

Using the given points on this equation, we have:

(-4,8):

(-4,8):8=m*(-4)+b

(-4,8):8=m*(-4)+bb=8+4m

(-4,8):8=m*(-4)+bb=8+4m(1,3):

(-4,8):8=m*(-4)+bb=8+4m(1,3):3=m+b

(-4,8):8=m*(-4)+bb=8+4m(1,3):3=m+b3=m+8+4m

(-4,8):8=m*(-4)+bb=8+4m(1,3):3=m+b3=m+8+4m5m=3-8

(-4,8):8=m*(-4)+bb=8+4m(1,3):3=m+b3=m+8+4m5m=3-85m=-5

(-4,8):8=m*(-4)+bb=8+4m(1,3):3=m+b3=m+8+4m5m=3-85m=-5m=-1

(-4,8):8=m*(-4)+bb=8+4m(1,3):3=m+b3=m+8+4m5m=3-85m=-5m=-1b=8+4*(-1)=8-4=4

Therefore the equation is y=-x+4 (correct optlon: C)

[tex]\qquad \qquad\huge \underline{\boxed{\sf Answer}}[/tex]

For the first figure ~

The diagonals of a kite intersect each other at 90°

So, we can apply Pythagoras theorem here :

[tex]\qquad \sf  \dashrightarrow \: CD² = OC² + OD²[/tex]

[tex]\qquad \sf  \dashrightarrow \: CD² = {7}^{2} + {9}^{2} [/tex]

[tex]\qquad \sf  \dashrightarrow \: CD² = 49 + 81[/tex]

[tex]\qquad \sf  \dashrightarrow \: CD² = 130[/tex]

[tex]\qquad \sf  \dashrightarrow \: CD=x = \sqrt{ 130}[/tex]

For the second figure ;

we have same concept of kite, and use of Pythagoras theorem !

Also, the diagonal QS bisects diagonal PR

Hence,

[tex]\qquad \sf  \dashrightarrow \: PR = 2 \times OR [/tex]

[tex]\qquad \sf  \dashrightarrow \: 10 = 2 \times OR [/tex]

[tex]\qquad \sf  \dashrightarrow \: OR = 10 \div 2[/tex]

[tex]\qquad \sf  \dashrightarrow \: OR = 5 \: mm[/tex]

now, apply pythagoras theorem ~

[tex]\qquad \sf  \dashrightarrow \: QR² = OR² + OQ²[/tex]

[tex]\qquad \sf  \dashrightarrow \: QR² = {5}^{2} + {6}^{2} [/tex]

[tex]\qquad \sf  \dashrightarrow \: QR² = 25 + 36[/tex]

[tex]\qquad \sf  \dashrightarrow \: QR² = 61[/tex]

[tex]\qquad \sf  \dashrightarrow \: QR=x = \sqrt{61} \: mm[/tex]

here, 2 OR = 2 OP = PR

so, similarly OP = 5 mm

Applying pythagoras theorem again ;

[tex]\qquad \sf  \dashrightarrow \: SP² = OS² + OP²[/tex]

[tex]\qquad \sf  \dashrightarrow \: SP² = {10}^{2} + {5}^{2} [/tex]

[tex]\qquad \sf  \dashrightarrow \: SP² = {100}^{} + 25[/tex]

[tex]\qquad \sf  \dashrightarrow \: SP² = 125[/tex]

[tex]\qquad \sf  \dashrightarrow \: SP = \sqrt{125}[/tex]

[tex]\qquad \sf  \dashrightarrow \: SP = y = 5\sqrt{5} \: mm[/tex]

Answer:

length= 12cm

perimeter= 40cm

Step-by-step explanation:

[tex]area = {96 \: cm}^{2} \\ breadth = \: 8cm \\ therefore = \: length \: \\ = 96 \div 8 \: = 12cm[/tex]

perimeter= (length+breadth)2

length= 12cm

breadth= 8cm

perimeter = (12+8)2

20×2= 40

[tex]\qquad \qquad\huge \underline{\boxed{\sf Answer}}[/tex]

In the given question, the two angles are complementary Angle pair, so the sum of their values will add up to 90°

that is ~

[tex]\qquad \sf  \dashrightarrow \: (3x + 30) \degree + x \degree = 90 \degree[/tex]

[tex]\qquad \sf  \dashrightarrow \: 3x + x + 30 \degree = 90 \degree[/tex]

[tex]\qquad \sf  \dashrightarrow \: 4x = 90 \degree - 30 \degree[/tex]

[tex]\qquad \sf  \dashrightarrow \: 4x = 60 \degree[/tex]

[tex]\qquad \sf  \dashrightarrow \: x = 60 \div 4[/tex]

[tex]\qquad \sf  \dashrightarrow \: x = 15 \degree[/tex]

Answer:

x = 15 (third option)

Step-by-step explanation:

Complementary angles sum up to 90°.

∠1 + ∠2 = 90°

x° + (3x + 30)° = 90°

x + 3x + 30 = 90

4x = 90 - 30

4x = 60

x = 60/4

x = 15

Hope it helps ⚜

Answer:

[tex] a=\frac{9c}{2}-b [/tex]

Step-by-step explanation:

Isolate the variable by dividing each side by factors that don't contain the variable.

Answer:

[tex]a=\frac{9c}{2}-b[/tex]

Step-by-step explanation:

Isoate the variable to solve

answer:

$47.92

explanation:

$3.88 × 16=$62.08

$110-$62.08=$47.92

the 16 is the days

110is the budget

3.88 is the cost

hope that helps

The total money Helen have left after 16 days will be $47.92

The unitary method is a method for solving a problem by the first value of a single unit and then finding the value by multiplying the single value. If an event can occur in m different ways and if following it, a second event can occur in n different ways, then the two events in succession can occur in m × n different ways.

Given that Helen budgets for coffee from the coffee bar located in the lobby of her apartment building was  $110. And she stops by the bar every morning and her average order costs $3.88.

The budget = 110 dollars

The average cost = 3.88 dollars

Then the average order costs for 16 days;

$3.88 × 16 = $62.08

Therefore, the money she have left after 16 days;

$110-$62.08=$47.92

Hence, The total money she have left after 16 days will be $47.92

Learn more about the unitary method, please visit the link given below;

https://brainly.com/question/23423168

#SPJ2

Answer:

Andy basically has two stickers

Answer:

The cost of one ticket will be £8.20

Answer:

C+3

Step-by-step explanation:

Answer:

b

Step-by-step explanation:

Answer:

√12

Step-by-step explanation:

Use the Phythagorean Theorem.

2² + b² = 4²

= 4 + b² = 16

 -4            -4

= b² = 12

√12

Answer:

[tex]\sqrt{12}[/tex] or [tex]2\sqrt{3}[/tex] (simplified)

Step-by-step explanation:

Pythagorean theorem is

α²+b²=c²

a is the leg, which is 2 in this triangle

b is the base, which x in this triangle

c is the hypotenuse, which is 4 in this triangle

So, we can plug in the values:

2²+x²=4²

4+x²=16

-4       -4

x²=12

x=[tex]\sqrt{12}[/tex], or if simplified: [tex]2\sqrt{3}[/tex]

Hope this helps!

Answer:

2

Step-by-step explanation:

Degree of the polynomial is the highest power of the term.

As degree of the polynomial is 3, m value should be 2

Answer: its b.) 2

Step-by-step explanation:

I got it right om test

Answer:

V =2860 in ^3

Step-by-step explanation:

The volume of a rectangular prism is given by

V = l*w*h  where l is the length, w is the width and h is the height.

V = 20 * 11 * 13

V =2860 in ^3

Answer:

1.)  12

2.)  36

3.)  36

4.)  36

5.)  72

6.)  72

7.)  120

8.)  30

Step-by-step explanation:

You're welcome make me brainliest :)

Answer: LCM=least common multiple so

Step-by-step explanation: 1)12

2)18 3)36 4)36     5)72    6)72    7)120    8)30

Answer is 0.42. Hope that helps.

Can you please answer comment i place on your question? I will give you a full description and answer for the work if you do.

Answer:

10.5

Step-by-step explanation:

8.4^2 = 70.56

6.3^2 = 39.69

110.25 = c^2

c = 10.5

Answer:

from the figure find the length unknown side

Answer:

Yes, it could be a probability

Step-by-step explanation:

The probability of an event HAS to be between the numbers 0 and 1. Not less than 0, not greater than 1. The number 0.12 is between 0 and 1, thus, making it a valid probability.

Answer:

Dilate, Reflect

Not congruent, dilations are used

Step-by-step explanation:

Dilate the figure by a scale factor of 1/3

Reflecting the resulting shape will map the figure

Step 1: Dilate by 1/3

Step 2: Reflect over the y-axis

Since dilations are used, the figures are not the same, as it changes area and side lengths.

-Chetan K

It looks like the integral might be

[tex]\displaystyle \int (9 - x^2)^{3/2} \, dx[/tex]

or perhaps

[tex]\displaystyle \int \frac1{(9 - x^2)^{3/2}} \, dx[/tex]

Take note of the fact that both integrands are defined only over the interval -3 < x < 3.

For either integral, we substitute x = 3 sin(θ) and dx = 3 cos(θ) dθ.

Note that we want this substitution to be reversible, so we must restrict -π/2 ≤ θ ≤ π/2, an interval over which sine has an inverse. Then θ = arcsin(x/3).

The first case then reduces to

[tex]\displaystyle \int (9 - (3\sin(\theta))^2)^{3/2} (3 \cos(\theta) \, d\theta) = 3 \times 9^{3/2} \int (1 - \sin^2(\theta))^{3/2} \cos(\theta) \, d\theta \\\\ = 81 \int (\cos^2(\theta))^{3/2} \cos(\theta) \, d\theta \\\\ = 81 \int |\cos^3(\theta)| \cos(\theta) \, d\theta[/tex]

By definition of absolute value,

[tex]\displaystyle 81 \int |\cos^3(\theta)| \cos(\theta) \, d\theta = \begin{cases}\displaystyle 81 \int \cos^4(\theta) \, d\theta & \text{if }\cos(\theta) \ge 0 \\ \displaystyle -81 \int \cos^4(\theta) \, d\theta & \text{if }\cos(\theta) < 0\end{cases}[/tex]

and these cases correspond to 0 ≤ θ < π/2 and π/2 < θ ≤ π, respectively. But we are assuming -π/2 ≤ θ ≤ π/2, so the negative case doesn't matter to us.

You can compute the remaining antiderivative by exploiting the half-angle identity for cosine,

[tex]\cos^2(\theta) = \dfrac{1 + \cos(2\theta)}2[/tex]

Then

[tex]\cos^4(\theta) = \left(\cos^2(\theta)\right)^2 = \dfrac{1 + 2\cos(2\theta) + \cos^2(2\theta)}4 = \dfrac{3 + 4\cos(2\theta) + \cos(4\theta)}8[/tex]

and so

[tex]\displaystyle \int \cos^4(\theta) \, d\theta = \dfrac{12\theta + 8\sin(2\theta) + \sin(4\theta)}{32} + C[/tex]

We can simplify this using the double angle identity for (co)sine,

sin(2θ) = 2 sin(θ) cos(θ)

cos(2θ) = 1 - 2 sin²(θ)

as well as the relations,

sin(arcsin(x/3)) = x/3

cos(arcsin(x/3)) = √(9 - x²)/3

which gives us

[tex]\displaystyle \int \cos^4(\theta) \, d\theta = \dfrac{12\theta + 16 \sin(\theta) \cos(\theta) + 4 \sin(\theta) \cos(\theta) (1 - 2\sin^2(\theta))}{32} + C[/tex]

Putting this in terms of x, we get

[tex]\displaystyle \int (9 - x^2)^{3/2} \, dx \\ = 81 \times \dfrac{12\arcsin\left(\frac x3\right) + 16 \times \frac x3 \times \frac{\sqrt{9-x^2}}3 + 4\times\frac x3\times\frac{\sqrt{9-x^2}}3 \left(1 - 2\left(\frac x3\right)^2\right)}{32} + C[/tex]

[tex]\displaystyle \int (9 - x^2)^{3/2} \, dx = 81 \times \dfrac{12\arcsin\left(\frac x3\right) + \frac{16x\sqrt{9-x^2}}9 + \frac{4x(9-2x^2)\sqrt{9-x^2}}{81}}{32} + C[/tex]

[tex]\boxed{\displaystyle \int (9 - x^2)^{3/2} \, dx = \dfrac{12\arcsin\left(\frac x3\right) + (180x-8x^3)\sqrt{9-x^2}}{32} + C}[/tex]

If you were asking about the other integral, the first few steps are similar and you end up with the far more trivial integral and antiderivative

[tex]\displaystyle \frac19 \int \frac{d\theta}{\cos^2(\theta)} = \frac19 \int \sec^2(\theta) \, d\theta = \frac19 \tan(\theta) + C[/tex]

Putting it back in terms of x, we get

[tex]\displaystyle \int \frac1{(9 - x^2)^{3/2}} \, dx = \frac19 \tan\left(\arcsin\left(\frac x3\right)\right) + C[/tex]

Recall that tan(θ) = sin(θ)/cos(θ), so

[tex]\displaystyle \int \frac1{(9 - x^2)^{3/2}} \, dx = \frac19 \times \frac{\frac x3}{\frac{\sqrt{9-x^2}}3} + C[/tex]

[tex]\boxed{\displaystyle \int \frac1{(9 - x^2)^{3/2}} \, dx = \frac{x}{9\sqrt{9-x^2}} + C}[/tex]

The equation a/b x b/a = 1 is a product equation

It is true that a/b x b/a = 1

The product expression is given as:

a/b x b/a = 1

Rewrite the product expression properly as follows:

[tex]\frac ab * \frac ba = 1[/tex]

Multiply the numerator of the fractions

[tex]\frac {ab}b * \frac 1a = 1[/tex]

Multiply the denominator of the fractions

[tex]\frac {ab}{ab} = 1[/tex]

Evaluate the quotient

[tex]1= 1[/tex]

Hence, it is true that a/b x b/a = 1

Read more about mathematical proofs at:

https://brainly.com/question/1788884

SOLUTION:

=) 49/64 is the answer I think