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3 years ago

Lilly has read seven-eighths of her book and is on page 147. How many pages does lilly have left to read?

Mathematics

1 answer:

lara [203]3 years ago

4 0

Answer:

Step-by-step explanation:

Since Lilly has read 7/8 of her book (aka 147 pages), divide 147 by 7 and you get 21.

This means 1/8=21

Since she has read 7/8 of the book, and the full book is 8/8, 8/8-7/8= 1/8

Meaning she has 1/8 of the book left to read.

As said before, 1/8 =21

Thus, she has 21 pages left to read.

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3) A particular motorcycle is 9 ft long. A

Shalnov [3]

The model is 3 inches, since 3 times 3 is 9 and 1 times 3 is 3.

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3 years ago

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Please help me solve step by step it's urgent

NNADVOKAT [17]

Answer:

2i: 169.71

2ii: 0.17L

3a: 4×10⁻⁵

3b: 110011

Step-by-step explanation:

2i. The surface of the top and bottom of the tin is two times (top and bottom) π·r² = 2·π·3² = 18π cm².

The circumference of the circle is 2·π·r = 6π cm².

The area of the material connecting top and bottom is a rectangle of the tin height times the circumference: 6·6π = 36π cm².

This gives a total of 18π + 36π = 54π cm².

With π approximated by 22/7 the total surface area is 54*22/7 ≈ 169.71.

Notice how the calculation is simple by waiting until the very last moment to substitute π.

2ii. The volume is the area π·r² of the circle times the height of the tin: 9π*6 = 54π cm³ ≈ 169.71 cm³.

Since 1L = 1000 cm³ the volume is 0.16971 litres, which should be rounded to 0.17 L.

3a: If we rewrite P as 36 x 10⁻⁴ and realize that 36/2.25 = 16, then the fraction can be written as

16 x 10⁻⁴⁻⁶ = 16 x 10⁻¹⁰.

The square root of that is taking it to the power of 1/2, so (16x10⁻¹⁰)^0.5 = 4x10⁻⁵ = 0.00004

3b: 1111 1111 is 255 in decimal. 101 is 5 in decimal. 255/5 is 51 in decimal. 51 in binary is 110011.

6 0

3 years ago

Match the circle equations in general form with their corresponding equations in standard form. Not all will be used.

Xelga [282]

The standard form of the equation of a circumference is given by the following expression:

 $(x-h)^{2}+(y-k)^{2}=r^{2} \\ \\ where \ (h, k) \ is \ the \ center \ of \ the \ circumference \ and \ r \ the \ radius$

On the other hand, the general form is given as follows:

 $x^{2}+y^{2}+Dx+Ey+F=0 \\ \\ where: \\ D=-2h, \ E=-2k, \ F=h^{2}+k^{2}-r^{2}$ 

In this way, we can order the mentioned equations as follows:

Equations in Standard Form:

 $\bold{a)} \ (x-6)^{2}+(y-4)^{2}=56 \\ \bold{b)} \ (x-2)^{2} + (y+6)^{2}=60 \\ \bold{c)} \ (x+2)^{2}+(y+3)^{2}=18 \\ \bold{d)} \ (x+1)^{2}+(y-6)^{2}=46$

Equations in General Form:

$\bold{1)} \ x^{2}+y^{2}-4x+12y-20=0 \\ \bold{2)} \ x^{2}+y^{2}+6x-8y-10=0 \\ \bold{3)} \ 3x^{2}+3y^{2}+12x+18y-15=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 3, \ the \ equation \ becomes: \\ x^{2}+y^{2}+4x+6y-5=0 \\ \\ \bold{4)} \ 5x^{2}+5y^{2}-10x+20y-30=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 5, \ the \ equation \ becomes: \\ x^{2}+y^{2}-2x+4y-6=0 \\ \\ \bold{5)} \ 2x^{2}+2y^{2}-24x-16y-8=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 2, \ the \ equation \ becomes: \\ x^{2}+y^{2}-12x-8y-4=0$

$\bold{6)} \ x^{2}+y^{2}+2x-12y$

So let's match each equation:

$\bold{From \ a)} \\ \\ (h,k)=(6,4),\ r=2\sqrt{14} \\ D=-12, \ E=-8 \\ F=-4$

Then, its general form is:

$x^{2}+y^{2}-12x-8y-4=0$

First. a) matches 5) 

$\bold{From \ b)} \\ \\ (h,k)=(2,-6),\ r=2\sqrt{15} \\ D=-4, \ E=12 \\ F=-20$

Then, its general form is:

$x^{2}+y^{2}-4x+12y-20=0$

Second. b) matches 1) 

$\bold{From \ c)} \\ \\ (h,k)=(-2,-3),\ r=3\sqrt{2} \\ D=4, \ E=6 \\ F=-5$

Then, its general form is:

$x^{2}+y^{2}+4x+6y-5=0$

Third. c) matches 3)

$\bold{From \ d)} \\ \\ (h,k)=(-1,6),\ r=\sqrt{46} \\ D=2, \ E=-12, \ F=-9$

Then, its general form is: $x^{2}+y^{2}+2x-12y-9=0$

Fourth. d) matches 6)

6 0

3 years ago

Read 2 more answers

Please really need help

zavuch27 [327]

Answer:

IDK

Step-by-step explanation:

IDK

6 0

3 years ago

Which equation shows y=34x−52y=34x−52 in standard form?

ladessa [460]

It would be A, most definitely.

5 0

3 years ago