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3 years ago

For quiz pls let me know

Mathematics

1 answer:

Phantasy [73]3 years ago

7 0

Answer:

Step-by-step explanation:

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20 equals 5 times negative 3 plus x

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Answer:

20=5*-3+x or 20=-15+x or 35=x

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How to factorise 15x-10x^3

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The answer is 5x(3-2x^2)

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Two circles have the same radius. Complete the description for whether the combined area of the two circles is the same as the a

Elodia [21]

Answer:

see below

Step-by-step explanation:

A = pi r^2

If they have the same radius they have the same area

A two circles = pi r^2 +pi r^2

= 2 pi r^2

If we double the radius

A = pi (2r)^2

= pi 4r^2

The combined area of two circles is 1/2 the area as the area of a circle with twice the radius.

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3 years ago

Read 2 more answers

PLEASE HELP ASAP

maria [59]

Answer:

First option: $3x^2-5x=-8$

Second option: $2x^2=6x-5$

Fourth option: $-x^2-10x=34$

Step-by-step explanation:

Rewrite each equation in the form $ax^2+bx+c=0$ and then use the Discriminant formula for each equation. This is:

$D=b^2-4ac$

1) For $3x^2-5x=-8$ :

$3x^2-5x+8=0$

Then:

$D=(-5)^2-4(3)(8)=-71$

Since $D$ this equation has no real solutions, but has two complex solutions.

2) For $2x^2=6x-5$ :

$2x^2-6x+5=0$

Then:

$D=(-6)^2-4(2)(5)=-4$

Since $D$ this equation has no real solutions, but has two complex solutions.

3) For $12x=9x^2+4$ :

$9x^2-12x+4=0$

Then:

$D=(-12)^2-4(9)(4)=0$

Since $D=0$ this equation has one real solution.

4) For $-x^2-10x=34$ :

$-x^2-10x-34=0$

Then:

$D=(-10)^2-4(-1)(-34)=-36$

Since $D$ , this equation has no real solutions, but has two complex solutions.

3 0

4 years ago

1 Point

I am Lyosha [343]

Option C

The ratio for the volumes of two similar cylinders is 8 : 27

<h3><u>Solution:</u></h3>

Let there are two cylinder of heights "h" and "H"

Also radius to be "r" and "R"

$\text { Volume of a cylinder }=\pi r^{2} h$

Where π = 3.14 , r is the radius and h is the height

Now the ratio of their heights and radii is 2:3 .i.e

$\frac{\mathrm{r}}{R}=\frac{\mathrm{h}}{H}=\frac{2}{3}$

<em><u>Ratio for the volumes of two cylinders</u></em>

$\frac{\text {Volume of cylinder } 1}{\text {Volume of cylinder } 2}=\frac{\pi r^{2} h}{\pi R^{2} H}$

Cancelling the common terms, we get

$\frac{\text {Volume of cylinder } 1}{\text {Volume of cylinder } 2}=\left(\frac{\mathrm{r}}{R}\right)^{2} \times\left(\frac{\mathrm{h}}{\mathrm{H}}\right)$

Substituting we get,

$\frac{\text {Volume of cylinder } 1}{\text {Volume of cylinder } 2}=\left(\frac{2}{3}\right)^{2} \times\left(\frac{2}{3}\right)$

$\frac{\text {Volume of cylinder } 1}{\text {Volume of cylinder } 2}=\frac{2 \times 2 \times 2}{3 \times 3 \times 3}$

$\frac{\text {Volume of cylinder } 1}{\text {Volume of cylinder } 2}=\frac{8}{27}$

Hence, the ratio of volume of two cylinders is 8 : 27

7 0

3 years ago