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3 years ago

Ting

Mathematics

2 answers:

Eddi Din [679]3 years ago

8 0

Answer:

Difference of Expression

5a+ 3b -6c -( 20 -4b +13c)

= 5a +3b -6c -20 +4b -13c

= 5a +3b +4b -6c -13c -20

= 5a +7 b - 19 c -20

TiliK225 [7]3 years ago

7 0

In this question, you would be solving the expression by combining like terms.

Solve:

(5a + 3b - 6c) - (20 - 4b + 13c)

Distribute the negative to the variables inside the parenthesis.

5a + 3b - 6c - 20 + 4b - 13c

Combine like terms.

5a + 7b - 6c - 20 - 13c

5a + 7b - 19c - 20

You can't combine anymore like terms now, so your answer would be 5a + 7b - 19c - 20

Answer:

5a + 7b - 19c - 20

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Can someone plz help me on this thank you sooo much

Stells [14]

Answer:

the last answer

Step-by-step explanation:

6 0

2 years ago

Maria practice the piano 5/6 of an hour everyday. How many hours does she practice in 4 days

S_A_V [24]

Answer:
10/3 hours

Explanation:
If Maria practices 5/6 hours a day, then she practices (5/6)(4) hours in 4 days.

(5/6)(4)=20/6

20/6 can be simplified to 10/3 hours.

I hope this helps!

3 0

2 years ago

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

When converted to speeds, which list is in order from slowest to fastest?

s2008m [1.1K]

Answer:

25 =84b h +43

7 0

3 years ago

The birthday candle has the form of a cylinder and cone together, if its has a common radius of 5 mm. The height of cylinder is

alexandr402 [8]

Answer:

1779.33 mm³

Step-by-step explanation:

We are given a candle in a form of a cylinder and a cone

Dimensions;

Common radius is 5 mm

Height of the cylinder is 20 mm

Height of the cone is 8 mm

We are required to determine the amount that it can hold.

To answer the question we are going to determine its volume

Cylinder;

Volume of a cylinder is given by;

V = πr²H

Taking π to be 3.14

Then ;

Volume of cylinder = 3.14 × 5mm² × 20 mm

= 1570 mm³

Cone

Volume of a cone is given by;

V = 1/3πr²H

That is;

Volume =1/3 × 3.14 × 5mm² × 8 mm

= 209.33 mm³

Therefore;

Volume of the solid = 1570 mm³ + 209.33 mm³

= 1779.33 mm³

Thus, the volume of the solid is 1779.33 mm³

7 0

3 years ago