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2 years ago

14

PLZ HURRY IT'S URGENT!!! Find the volume of the prism.

1 answer:

Mashutka [201]2 years ago

7 0

Answer:

V = 210 cm³

Step-by-step explanation:

The volume (V) of a cube is calculated as

V = Ah ( A is the area of the base and h is the height )

Here A = 5 × 6 = 30 cm² and h = 7 cm , then

V = 30 × 7 = 210 cm³

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raj and his sister zia are both at secondary school. Raj is three years older than zia. the sum of the squares of their ages is

steposvetlana [31]

Let's assume

Raj's age is x years

Zia's age is y years

we are given

Raj is three years older than zia

so, we get

$x=y+3$

now, we can solve for y

$y=x-3$

the sum of the squares of their ages is 369

so, we get

$x^2+y^2=369$

now, we can plug back y

$x^2+(x-3)^2=369$

now, we can solve for x

$2x^2-6x+9=369$

$2x^2-6x-360=0$

$x^2-3x-180=0$

$(x+12)(x-15)=0$

$x=-12 , x=15$

Since, age can not be negative

so, we will only consider positive

$x=15$

now, we can find y-value

$y=15-3$

$y=12$

So,

Raj's age is 15 years

Zia's age is 12 years...........Answer

8 0

3 years ago

Jerri was using a recipe for a large cake that required one-fourth of a tablespoon of cinnamon. Instead of making one large cake

Sergeeva-Olga [200]

Im pretty sure you would put 1/12 tablespoon of cinnamon in the cakes individually. You would divide 1/4 by 3. It's pretty simple. If that's not the answer I apologize dearly!

8 0

3 years ago

Read 2 more answers

The sum of three consecutive integers is 267. What is the largest integer?

Stella [2.4K]

Answer:

let integers be x,2x, 3x respectively

x+2x+3x= 267

6x= 267

x=267/6= 44.5

2x= 44.5*2= 89

3x= 44.5*3= 133.5

largest integer = 133.5

hope it helps

plz mark as brainliest

5 0

3 years ago

Phil received a prize of xxx dollars from a poker tournament. The tournament cost him 100100100 dollars to enter.

nikitadnepr [17]

Answer:

$(x-100)

Step-by-step explanation:

Phil entered the poker tournament with $100.
He received a price of $x.

His net winnings from the tournament will be the price minus the entry fee.

Therefore:

Phil's net winnings=$(x-100)

8 0

3 years ago

find the centre and radius of the following Cycles 9 x square + 9 y square +27 x + 12 y + 19 equals 0

Citrus2011 [14]

Answer:

Radius: $r =\frac{\sqrt {21}}{6}$

$Center = (-\frac{3}{2}, -\frac{2}{3})$

Step-by-step explanation:

Given

$9x^2 + 9y^2 + 27x + 12y + 19 = 0$

Solving (a): The radius of the circle

First, we express the equation as:

$(x - h)^2 + (y - k)^2 = r^2$

Where

$r = radius$

$(h,k) =center$

So, we have:

$9x^2 + 9y^2 + 27x + 12y + 19 = 0$

Divide through by 9

$x^2 + y^2 + 3x + \frac{12}{9}y + \frac{19}{9} = 0$

Rewrite as:

$x^2 + 3x + y^2+ \frac{12}{9}y =- \frac{19}{9}$

Group the expression into 2

$[x^2 + 3x] + [y^2+ \frac{12}{9}y] =- \frac{19}{9}$

$[x^2 + 3x] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}$

Next, we complete the square on each group.

For $[x^2 + 3x]$

1: Divide the $coefficient\ of\ x\ by\ 2$

2: Take the $square\ of\ the\ division$

3: Add this $square\ to\ both\ sides\ of\ the\ equation.$

So, we have:

$[x^2 + 3x] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}$

$[x^2 + 3x + (\frac{3}{2})^2] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}+ (\frac{3}{2})^2$

Factorize

$[x + \frac{3}{2}]^2+ [y^2+ \frac{4}{3}y] =- \frac{19}{9}+ (\frac{3}{2})^2$

Apply the same to y

$[x + \frac{3}{2}]^2+ [y^2+ \frac{4}{3}y +(\frac{4}{6})^2 ] =- \frac{19}{9}+ (\frac{3}{2})^2 +(\frac{4}{6})^2$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =- \frac{19}{9}+ (\frac{3}{2})^2 +(\frac{4}{6})^2$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =- \frac{19}{9}+ \frac{9}{4} +\frac{16}{36}$

Add the fractions

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{-19 * 4 + 9 * 9 + 16 * 1}{36}$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{21}{36}$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{7}{12}$

$[x + \frac{3}{2}]^2+ [y +\frac{2}{3}]^2 =\frac{7}{12}$

Recall that:

$(x - h)^2 + (y - k)^2 = r^2$

By comparison:

$r^2 =\frac{7}{12}$

Take square roots of both sides

$r =\sqrt{\frac{7}{12}}$

Split

$r =\frac{\sqrt 7}{\sqrt 12}$

Rationalize

$r =\frac{\sqrt 7*\sqrt 12}{\sqrt 12*\sqrt 12}$

$r =\frac{\sqrt {84}}{12}$

$r =\frac{\sqrt {4*21}}{12}$

$r =\frac{2\sqrt {21}}{12}$

$r =\frac{\sqrt {21}}{6}$

Solving (b): The center

Recall that:

$(x - h)^2 + (y - k)^2 = r^2$

Where

$r = radius$

$(h,k) =center$

From:

$[x + \frac{3}{2}]^2+ [y +\frac{2}{3}]^2 =\frac{7}{12}$

$-h = \frac{3}{2}$ and $-k = \frac{2}{3}$

Solve for h and k

$h = -\frac{3}{2}$ and $k = -\frac{2}{3}$

Hence, the center is:

$Center = (-\frac{3}{2}, -\frac{2}{3})$

6 0

2 years ago