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2 years ago

B + 1.5 = 6 solve the equation and find the value of the variable

Mathematics

2 answers:

pickupchik [31]2 years ago

5 0

The answer is 4.5. Because 6 minus 1.5 is equal to 4.5

guapka [62]2 years ago

5 0

Answer:

$\boxed{\sf{b=4.5}}$

Step-by-step explanation:

To find the value of variable, you have to isolate the term of b from one side of the equation.

First, you subtract by 1.5 from both sides.

→ b+1.5-1.5=6-1.5

Solve.

Subtract the numbers from left to right.

⇒ 6-1.5=4.5

⇒ b=4.5

Therefore, the correct answer is b=4.5.

I hope this helps you! Let me know if my answer is wrong or not.

You might be interested in

What is a equivalent expression for 8-2y+5x-13+6y-2x+9

poizon [28]

Answer:

4y+3x+17

Step-by-step explanation:

take the similar numbers and add them together.

1. -2y+6y = 4y

2. 5x-2x = 3x

3. 8+9 = 17

Put the numbers back together into the new and simplified equation: 4y + 3x + 17

5 0

4 years ago

The water from a fire hose follows a path described by y equals 2.0 plus 0.9 x minus 0.10 x squared (units are in meters). If

Alecsey [184]

Answer:

The resultant velocity is 12.21 m/s.

Step-by-step explanation:

We are given that the water from a fire hose follows a path described by y equals 2.0 plus 0.9 x minus 0.10 x squared (units are in meters).

Also, v Subscript x is constant at 10.0 m/s.

The water from a fire hose follows a path described by the following equation below;

$y=2.0 + 0.9x-0.10x^{2}$

The velocity of the $x$ component is constant at = $v_x=10.0 \text{ m/s}$

and the point at which resultant velocity has to be calculated is (9.0,2.0).

Let the velocity of x and y component be represented as;

$v_x=\frac{dx}{dt} \text{ and } v_y=\frac{dy}{dt}$

Now, differentiating the above equation with respect to t, we get;

$y=2.0 + 0.9x-0.10x^{2}$

$\frac{dy}{dt} =0 + 0.9\frac{dx}{dt} -(0.10\times 2)\frac{dx}{dt}$

$\frac{dy}{dt} = 0.9\frac{dx}{dt} -0.2\frac{dx}{dt}$

$v_y = 0.9v_x -0.2v_x$

$v_y = 0.7v_x$

Now, putting $v_x=10.0 \text{ m/s}$ in the above equation;

$v_y = 0.7 \times 10.0$ = 7 m/s

Now, the resultant velocity is given by = $v=\sqrt{v_x^{2}+v_y^{2} }$

$v=\sqrt{10^{2}+7^{2} }$

= $\sqrt{149}$ = 12.21 m/s

5 0

3 years ago

43 over 50 0.91 7 over 8 84% least to greatest

Gre4nikov [31]

Make them all decimals or all fractions with the same denominator and it will be easy to tell by inspection which is least.

43/50= .86
.91=.91
7/8= .875
84%=.84

84%. 43/50. 7/8. .91

3 0

3 years ago

A ship sails 250km due North qnd then 150km on a bearing of 075°.1)How far North is the ship now? 2)How far East is the ship now

olga_2 [115]

Answer:

1) 288.8 km due North

2) 144.9 km due East

3) 323.1 km

4) 207°

Step-by-step explanation:

Bearing: The angle (in degrees) measured clockwise from north.

Trigonometric ratios

$\sf \sin(\theta)=\dfrac{O}{H}\quad\cos(\theta)=\dfrac{A}{H}\quad\tan(\theta)=\dfrac{O}{A}$

where:

$\theta$ is the angle
O is the side opposite the angle
A is the side adjacent the angle
H is the hypotenuse (the side opposite the right angle)

Cosine rule

$c^2=a^2+b^2-2ab \cos C$

where a, b and c are the sides and C is the angle opposite side c

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

Draw a diagram using the given information (see attached).

Create a right triangle (blue on attached diagram).

This right triangle can be used to calculate the additional vertical and horizontal distance the ship sailed after sailing north for 250 km.

Question 1

To find how far North the ship is now, find the measure of the short leg of the right triangle (labelled y on the attached diagram):

$\implies \sf \cos(75^{\circ})=\dfrac{y}{150}$

$\implies \sf y=150\cos(75^{\circ})$

$\implies \sf y=38.92285677$

Then add it to the first portion of the journey:

⇒ 250 + 38.92285677... = 288.8 km

Therefore, the ship is now 288.8 km due North.

Question 2

To find how far East the ship is now, find the measure of the long leg of the right triangle (labelled x on the attached diagram):

$\implies \sf \sin(75^{\circ})=\dfrac{x}{150}$

$\implies \sf x=150\sin(75^{\circ})$

$\implies \sf x=144.8888739$

Therefore, the ship is now 144.9 km due East.

Question 3

To find how far the ship is from its starting point (labelled in red as d on the attached diagram), use the cosine rule:

$\sf \implies d^2=250^2+150^2-2(250)(150) \cos (180-75)$

$\implies \sf d=\sqrt{250^2+150^2-2(250)(150) \cos (180-75)}$

$\implies \sf d=323.1275729$

Therefore, the ship is 323.1 km from its starting point.

Question 4

To find the bearing that the ship is now from its original position, find the angle labelled green on the attached diagram.

Use the answers from part 1 and 2 to find the angle that needs to be added to 180°:

$\implies \sf Bearing=180^{\circ}+\tan^{-1}\left(\dfrac{Total\:Eastern\:distance}{Total\:Northern\:distance}\right)$

$\implies \sf Bearing=180^{\circ}+\tan^{-1}\left(\dfrac{150\sin(75^{\circ})}{250+150\cos(75^{\circ})}\right)$

$\implies \sf Bearing=180^{\circ}+26.64077...^{\circ}$

$\implies \sf Bearing=207^{\circ}$

Therefore, as bearings are usually given as a three-figure bearings, the bearing of the ship from its original position is 207°

8 0

2 years ago

Read 2 more answers

A baseball team won 8 games and lost 6. If the team continues at this ratio, how many wins will the team have after playing 21 g

DIA [1.3K]

Answer:

12 wins and 9 losses.

Step-by-step explanation:

Because they are winning 57.14% of there games so if 8 out of 14=57.14% then that means if they would have still one at the exact pace then.

12 wins out of 21 would mean they would have one 57.14% of there games.

Hope this helps have a great afternoon:)

4 0

3 years ago