All categories

2 years ago

6= 5r - 14 What are the steps you would use to solve the equation?

Mathematics

1 answer:

gregori [183]2 years ago

7 0

6 = 5r - 14
6 + 14 = 5r
20 = 5r
r = 4

You might be interested in

Mary bought 4 small cookies for 40 cents each and 2 large cookies for 90 cents each how much did she spend all together

babunello [35]

Answer:

$3.40

Step-by-step explanation:

4x.40 is 1.60

2x.90 is 1.80

Then you take 1.80+1.60 to get $3.40 Hope this helps!

4 0

3 years ago

1) A fast food restaurant had 13 pounds of flour. If they split the flour evenly among 3

Ganezh [65]

Answer:

Each chicken would use 4.33 pounds of flour. Two whole numbers the answers relies is 4 and 5.

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Given O below, If AB and BC are congruent, what is the measure of chord BC?

zhuklara [117]

Since, AB and BC are the same length, the measure of the chord for BC would be 12.5 units

3 0

3 years ago

Read 2 more answers

What is the value of x?<br><br><br><br> Enter your answer in the box.

hoa [83]

The answer is x = 9, hope this helps- mark brainliest if you can please thanks

4 0

3 years ago

Read 2 more answers

Q3: Identify the graph of the equation and write and equation of the translated or rotated graph in general form. (Picture Provi

natta225 [31]

Answer:

b. circle; $2(x')^2+2(y')^2-5x'-5\sqrt{3}y'-6 =0$

Step-by-step explanation:

The given conic has equation;

$x^2-5x+y^2=3$

We complete the square to obtain;

$(x-\frac{5}{2})^2+(y-0)^2=\frac{37}{4}$

This is a circle with center;

$(\frac{5}{2},0)$

This implies that;

$x=\frac{5}{2},y=0$

When the circle is rotated through an angle of $\theta=\frac{\pi}{3}$ ,

The new center is obtained using;

$x'=x\cos(\theta)+y\sin(\theta)$ and $y'=-x\sin(\theta)+y\cos(\theta)$

We plug in the given angle with x and y values to get;

$x'=(\frac{5}{2})\cos(\frac{\pi}{3})+(0)\sin(\frac{\pi}{3})$ and $y'=--(\frac{5}{2})\sin(\frac{\pi}{3})+(0)\cos(\frac{\pi}{3})$

This gives us;

$x'=\frac{5}{4} ,y'=\frac{5\sqrt{3} }{4}$

The equation of the rotated circle is;

$(x'-\frac{5}{4})^2+(y'-\frac{5\sqrt{3} }{4})^2=\frac{37}{4}$

Expand;

$(x')^2+(y')^2-\frac{5\sqrt{3} }{2}y'-\frac{5}{2}x'+\frac{25}{4} =\frac{37}{4}$

Multiply through by 4; to get

$4(x')^2+4(y')^2-10\sqrt{3}y'-10x'+25 =37$

Write in general form;

$4(x')^2+4(y')^2-10x'-10\sqrt{3}y'-12 =0$

Divide through by 2.

$2(x')^2+2(y')^2-5x'-5\sqrt{3}y'-6 =0$

8 0

3 years ago