what is the length of the diagonal of a rectangle with width 3,5 cm and length 4,5 cm give your answer in simplest surd form​

Answer:

Counterclockwise rotation 90°

Step-by-step explanation:

first quadrant graph -> 2nd quadrant

Counterclockwise rotation 90° : (x,y) -> (-y , x)

clapped

Step-by-step explanation:

Angle 8 and angle 5 are vertical angles

Answer:

log

2

(

y

)

=

5

→

2

5

=

y

⇒

y

=

32

Step-by-step explanation:

Answer:

6

Step-by-step explanation:

Following PEMDAS ( parentheses, exponents, multiplication, division, addition, subtraction)

| 7 - 9 | + ( 7 - 9 ) ^ 2

| 7 - 9 | + -2 ^ 2

| 7 - 9 | + ( -2 x -2 )

| 7 - 9 | + 4

| -2 | + 4

2 + 4

6

Answer:

1/4 cup of milk

Step-by-step explanation:

= 1/2 ÷ 2

= 1/2 × 1/2

= 1/4

Answer:

3rd option

Step-by-step explanation:

a point (0, y ) lies on the y- axis

a point (x, 0 ) lies on the x- axis

a point (x, y ) lies in I quadrant

a point (- x, y ) lies in II quadrant

a point (- x, - y ) lies in III quadrant

a point (x, - y ) lies in IV quadrant

Then

point (2, - 5 ) lies in IV quadrant

-----------------------------------------------------------------

We require 2 equations with the repeating digit 7 placed after the decimal point.

let x = 0.427777... ( multiply both sides by 100 and 1000 )

100x = 42.777..... (1)

1000x = 427.777.... (2)

Subtract (1) from (2), thus eliminating the repeating digits

900x = 385 ( divide both sides by 900 )

x = [tex]\frac{385}{900}[/tex] = [tex]\frac{77}{180}[/tex]

2090

Step-by-step explanation:

220 divided by 2 hours is 110 per hour

19 times 110 equals 2090

Answer:

x = [tex]\frac{5}{4} + \frac{ \sqrt{5} }{4}[/tex]

x = [tex]\frac{5}{4} - \frac{ \sqrt{5} }{4}[/tex]

Step-by-step explanation:

Quadratic equation:

[tex]x = \frac{-b +- \sqrt{b^2 - 4ac} }{2a}[/tex]

The equation is:

[tex]4x^2 - 10x + 5 = 0\\[/tex]

So:

a = 4

b = -10

c = 5

Plug the variables into the quadratic equation:

[tex]x = \frac{-(-10) +- \sqrt{(-10)^2 - (4 * 4 * 5)} }{(2 * 4)}[/tex]

[tex]x = \frac{\sqrt{10 +- \sqrt{100 - 80} } }{8}[/tex]

[tex]x = \frac{10 + \sqrt{20} }{8}[/tex]

Divide both sides by 8:

10 ÷ 8 = [tex]\frac{5}{4}[/tex]

[tex]\sqrt{20}[/tex] ÷ 8 = [tex]\frac{\sqrt{5}}{4}[/tex]

[tex]\frac{5}{4}[/tex] +- [tex]\frac{\sqrt{5}}{4}[/tex]

This means that the solutions that satisfy the equation is:

x = [tex]\frac{5}{4} + \frac{ \sqrt{5} }{4}[/tex]

x = [tex]\frac{5}{4} - \frac{ \sqrt{5} }{4}[/tex]

Hope this helps!

Answer:

z. sjshshdbd.d.difffjfjfjf

The slope is [tex]-\frac{3}{7}[/tex]

The lowest point of the cables above the bridge is 48 feet above the bridge.

The statement give us a parabola equation in standard form, the vertical distance of the lowest point above the bridge is determined by the y-coordinate of the vertex. Hence, the lowest point of the cables above the bridge is 48 feet above the bridge.

We kindly invite to check this question on parabolae: https://brainly.com/question/8495869

Answer:

-90

Step-by-step explanation: I hope its the right answer. If not I'm sorry.... I'm just a kid...

Answer:

5.73

Step-by-step explanation:

The difference is 5.73.

Answer:

here's your answer

hope it helps

(look at attachment)

Answer:

Hi There,

Your answer is -926.318

Answer:

Solve by multiplying a scale factor of 4 to each coordinate

Step-by-step explanation:

If the coordinate of J is dilated by a scale factor of 4, the resulting coordinate of the figure J' will be at (-48, 64)

Dilation is a transformation technique used to change the size and shape of an object on a xy-plane.

Given the following coordinate point of a figure JKLM:

J (− 12 , 16 ),

K ( − 4 , − 3 )

L ( − 2 , 16 )

M ( 1 , 0 )

If the coordinate of J is dilated by a scale factor of 4, the resulting coordinate of the figure J' will be at (-48, 64).

Learn more on dilation here: https://brainly.com/question/10253650

Step-by-step explanation:

there seems to be something missing.

I assume the slope of the line is -2/3.

the slope of a line is always the factor of x in the equation.

the slope is the y/x ratio indicating how many units y changes, when x changes by a certain amount of units (e.g. when going from one point on the line to another).

the equation for the line looks like this then :

y = (-2/3)×x + b

to get b we use the provided point (-6, -5) and it's x and y values.

-5 = (-2/3)×-6 + b = 4 + b

b = -9

so, the full equation of the line is

y = (-2/3)×x - 9

Answer:

= 30,550 rupee owed

Step-by-step explanation:

yr1 = 50000 x 0.10^1 = 5000

5000 - 10000 = 5000 credit from debt.

yr 2 = 50000 x 1.10^2 = 500 debit from debt = 5000-15000= 10,000 credit from debt

yr 3 = 50000 x 1.10^3 = 50 debit from debt

5000 + 500 = 15000 where 50,000 owed initially and yr1 and yr 2 paid in full interest of 5500 and we deduct this from his payment = 25000-5500 =

19500 paid of initial and nothing owed on interest.

50000-19500= 30,500 rupees owed + 50 rupee interest.

= 30,550 rupee owed

The amount he should pay at the end of the three years to settle the loan is; 41,550 Rupees

We are given;

Investment = 50000 rupees

Annual Interest Rate = 10% = 0.1

Since the interest is compounded annually, then number of times of compounding = 3

Thus;

Amount to pay after 3 years is;

A = 50000(1 + 0.1)³

A = 50000 × 1.1³

A = 66500 Rupees

Now, we are told that he has already paid 10000 rupees after one year and 15000 rupees after two years. Thus, amount he now owes to settle the loan = 66500 - (10000 + 15000) = 41,550 Rupees

Read more on Compound Interest at; https://brainly.com/question/2455673

Answer: 3 + 7i

Step-by-step explanation:

Equation expanded: 12 + 5i - 9 + 2i // now simplfy

12 - 9 = 3

5i + 2i = 7i

// together, we get

3 + 7i

Answer:

10i

Step-by-step explanation:

(12+5i) - (9-2i)

(17i) - (7i)

10i

Answers:

T1 = 59.5677879151077 N

T2 = 45.7583538175564  N

The values are approximate. Round the values however you need to.

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

Explanation:

Refer to the diagram below. I have added red and blue lines to break up each tension vector into the x and y components. For tension T1 in red, we have x1 and y1 as the horizontal and vertical components respectively. Note that x1 points to the left and y1 points upward.

For T2 in blue, we have x2 point to the right and y2 point upward. The horizontal components x1 and x2 must be equal, and point in opposite directions, so that we have horizontal balance. Otherwise, the object would accelerate to one side rather than be stationary.

Applying a bit of trig, we can say

cos(angle) = adjacent/hypotenuse

cos(45) = (x1)/(T1)

x1 = T1*cos(45)

and also

sin(angle) = opposite/hypotenuse

sin(67) = (x2)/(T2)

x2 = T2*sin(67)

----------------------

Again, the x1 and x2 vectors must be equal in magnitude, allowing us to say this

x1 = x2

T1*cos(45) = T2*sin(67)

T1 = T2*sin(67)/cos(45)

which we'll use later.

----------------------

Doing a bit more trig set up, we know that

sin(45) = (y1)/(T1)

y1 = T1*sin(45)

and

cos(67) = (y2)/(T2)

y2 = T2*cos(67)

Adding those two y components together will yield the total upward force that the ropes apply

y1+y2 = T1*sin(45)+T2*cos(67)

y1+y2 = (T2*sin(67)/cos(45))*sin(45)+T2*cos(67)

y1+y2 = T2*( sin(67)*sin(45)/cos(45) + cos(67) )

y1+y2 = T2*( sin(67)*tan(45) + cos(67) )

y1+y2 = T2*( sin(67)*1 + cos(67) )

y1+y2 = T2*( sin(67) + cos(67) )

This combined upward pull must be equal to 60 newtons or else we wouldn't have vertical equilibrium. If the upward pull was smaller than the downward pull, then the object would be falling. If the other way around, then the object would be pulled upward at some accelerated speed (think of a spring action).

So we'll set y1+y2 equal to 60 and solve for T2

y1+y2 = 60

T2*( sin(67) + cos(67) ) = 60

T2 = 60/( sin(67) + cos(67) )

T2 = 45.7583538175564

That value is approximate. The units for T2 are in newtons. This is the amount of pull or force along the T2 line in the upper right direction.

Side note: make sure your calculator is in degree mode.

----------------------

The last thing to do is to compute T1

T1 = T2*sin(67)/cos(45)

T1 = 45.7583538175564*sin(67)/cos(45)

T1 = 59.5677879151077

That value is approximate as well. The units are also newtons here. This represents the amount of pull in the upper left direction as the diagram shows.

----------------------

Finding such values for T1 and T2 is useful to determine if a rope can handle a certain stress load. Let's say we had a rope that was rated for only 40 newtons. Anything beyond 40 N for this hypothetical rope would mean the rope breaks. We found T1 and T2 beyond this threshold which would indicate that such a rope isn't up to the task of holding this object up (at least not with the angles shown in the diagram).

The value of the stamp at the end of the third year is £976.31.

From the information, Sam buys a rare stamp for his stamp collection at an auction. he buys the stamp for 920£ the stamp increased in value by 7% in each of the first 2 years.

First year value:

= 920 + (2% × 920)

= 938.4

Second year value:

= 938.4 + (2% × 938.4)

= 957.168

Third year value:

= 957.168 + (2% × 957.168)

= £976.31

Learn more about percentages on:

brainly.com/question/24877689

#SPJ1

Complete question

sam buys a rare stamp for his stamp collection at an auction. he buys the stamp for 920£ the stamp increased in value by 7% in each of the first 2 years decreased in value by 4% in the third year. What is the cost by the end of. the third year?

Answer:

42:30

Step-by-step explanation:

10 x 3 = 30

14 x 3 = 42

Answer:

SSS~

RQP

Step-by-step explanation:

36/15 = 2.4

48/20 = 2.4

72/30 = 2.4

Triangles are similar by SSS~.

triangle JKL is similar to triangle RQP

Answer:

Step-by-step explanation:

Slope of perpendicular line = [tex]\frac{-1}{m}[/tex]

       = -1 ÷ [tex]\frac{-5}{3}[/tex]

        [tex]= -1 *\dfrac{-3}{5}\\\\=\dfrac{3}{5}[/tex]

[tex]{ \qquad{ \sf \green{Answer \downarrow}}}[/tex]

Subtract 16  from both sides of the equation.

[tex]{ \qquad{\longmapsto \sf \red{2x + 16 - 16 = 8 - 16}}}[/tex]

Subtract the numbers

[tex]{ \qquad{\longmapsto \sf \red{2x = 8}}}[/tex]

Divide both sides of the equation by the same term.

[tex]{ \qquad{\longmapsto \sf \green{ \frac{2x}{2} = \frac{ 8}{2} }}}[/tex]

Cancel terms that are in both the numerator and denominator

[tex]{ \qquad{\longmapsto \sf{ \qquad\longrightarrow\quad\sf \dfrac{\cancel{2x}}{\cancel{2} } = }}}\qquad\quad\sf\dfrac{\cancel{ 8}}{\cancel{2}}[/tex]

[tex]{ \qquad{\longmapsto \sf{x = {\qquad{ \sf \pink{ 4}}}}}}[/tex]

[tex]{ \red{ \rule{9cm}{0.1cm}}}[/tex]

Answer:

No

Step-by-step explanation:

7 + 3 = 10, and in order for this to make a triangle two sides must be greater than one side

Answer:

No

Step-by-step explanation:

two sides, say a and b, with a greater than or equal to b, ,put limits on the length of the third side, c: a-b < c < a+b.

a-b < c < a+b.

7-3 < 10 < 7+3

4 < 10 < 10

False,

Answer:

AAS

Step-by-step explanation:

I'm pretty sure its this, but im sorry if im wrong!