Ask question

Ask question

All categories

3 years ago

15

Ron has to color in three shapes (square,circle and triangle) with different colors (red,blue,green). In how many different ways

can the shapes be colored??

1 answer:

Nataly [62]3 years ago

3 0

Square-red,blue,green
circle-red,blue,green
triangle-red,blue,green

add them together which will equal 9

You might be interested in

Evaluate the double integral.

Fynjy0 [20]

Answer:

$\iint_D 8y^2 \ dA = \dfrac{88}{3}$

Step-by-step explanation:

The equation of the line through the point $(x_o,y_o)$ & $(x_1,y_1)$ can be represented by:

$y-y_o = m(x - x_o)$

Making m the subject;

$m = \dfrac{y_1 - y_0}{x_1-x_0}$

∴

we need to carry out the equation of the line through (0,1) and (1,2)

i.e

y - 1 = m(x - 0)

y - 1 = mx

where;

$m= \dfrac{2-1}{1-0}$

m = 1

Thus;

y - 1 = (1)x

y - 1 = x ---- (1)

The equation of the line through (1,2) & (4,1) is:

y -2 = m (x - 1)

where;

$m = \dfrac{1-2}{4-1}$

$m = \dfrac{-1}{3}$

∴

$y-2 = -\dfrac{1}{3}(x-1)$

-3(y-2) = x - 1

-3y + 6 = x - 1

x = -3y + 7

Thus: for equation of two lines

x = y - 1

x = -3y + 7

i.e.

y - 1 = -3y + 7

y + 3y = 1 + 7

4y = 8

y = 2

Now, y ranges from 1 → 2 & x ranges from y - 1 to -3y + 7

∴

$\iint_D 8y^2 \ dA = \int^2_1 \int ^{-3y+7}_{y-1} \ 8y^2 \ dxdy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \int ^{-3y+7}_{y-1} \ y^2 \ dxdy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( \int^{-3y+7}_{y-1} \ dx \bigg) dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( [xy^2]^{-3y+7}_{y-1} \bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( [y^2(-3y+7-y+1)]\bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ([y^2(-4y+8)] \bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( -4y^3+8y^2 \bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \bigg [\dfrac{ -4y^4}{4}+\dfrac{8y^3}{3} \bigg ]^2_1$

$\iint_D 8y^2 \ dA =8 \bigg [ -y^4+\dfrac{8y^3}{3} \bigg ]^2_1$

$\iint_D 8y^2 \ dA =8 \bigg [ -2^4+\dfrac{8(2)^3}{3} + 1^4- \dfrac{8\times (1)^3}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ -16+\dfrac{64}{3} + 1- \dfrac{8}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ -15+ \dfrac{64-8}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ -15+ \dfrac{56}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ \dfrac{-45+56}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ \dfrac{11}{3}\bigg]$

$\iint_D 8y^2 \ dA = \dfrac{88}{3}$

4 0

2 years ago

A bicyclist is finishing his repair of a flat tire when a friend rides by with a constant speed of 3.63 m/s. Four seconds later,

castortr0y [4]

From the moment the friend passes the bicyclist, his friend covers a distance over time t of (3.63 m/s)*t.

The bicyclist covers a distance of 1/2*(2.11 m/s^2)*t^2. They meet when these distances are equal:

3.63 t = 1.055 t^2 ==> 1.055 t^2 - 3.63 t = 0

==> t = 0 s or t = 3.44 s

5 0

3 years ago

Can somebody please help i need the help to help me pass my math class thx so much if you can

natulia [17]

Answer:

Step-by-step explanation:

1. x = 11.2

2. 13

3. 9

4. -7/15

5. 15b

6. -12x + 16

7. 4x + 4

8. 11x-10

9. 10a + 5

10. -x, 15, and 2b

Hope that helps, and good luck!

4 0

3 years ago

Ms. Nina has 10 lbs of 25% sugar syrup. How much water does she need to add to make 10% sugar syrup?

ryzh [129]

She has 10lbs of 25% syrup... so, in the 10lbs, 25% of that is syrup, the rest, namely the 75% remaining is water or other substances.

let's say she adds "x" lbs of water, to get "y" lbs for the 10% mixture.

how much is 25% of 10lbs? well, (25/100) * 10, or 2.5.

the water has no sugar syrup in it, so is just pure water, so the amoun of syrup in it is 0%, how much is 0% of "x" lbs? well, (0/100) * x, or 0.00x, which is just 0.

how much is 10% of "y" lbs? well (10/100) * y, or 0.10y.

whatever "x" and "y" are, we know that 10 + x = y.

we also know that the syrup amount in that is also 2.5 + 0.00x = 0.10y, thus

$\bf \begin{array}{lccclll}
&\stackrel{lbs}{syrup}&\stackrel{concentration~\%}{syrup}&\stackrel{concentration}{amount}\\
&------&------&------\\
\textit{25\% syrup}&10&0.25&2.5\\
\textit{pur water}&x&0.00&0.00x\\
------&------&------&------\\
\textit{10\% mixture}&y&0.10&0.10y
\end{array}
\\\\\\
\begin{cases}
10+x=\boxed{y}\\
2.5+0.00x=0.10y\\
----------\\
2.5 = 0.10\left( \boxed{10+x} \right)
\end{cases}
\\\\\\
2.5=1 + 0.10x\implies 1.5=0.10x
\\\\\\
\cfrac{1.5}{0.10}=x\implies 15=x$

8 0

3 years ago

Read 2 more answers

Can someone Please help me ???<br> If you can help me ill give you 20 points and brainliest :)

Phoenix [80]

First, classify each line segments of triangle that are the same in both triangles.

RS = XU

RT = XW

ST = WU

Second, divide to find the scale ratio.

7.5/3 = 2.5

16/6.4 = 2.5

15/6 = 2.5

Since the scale ratios are identical, the triangles are similar.

Therefore, the answer is [ Yes, the sides are in the ratio 2:5 ]

Best of Luck!

4 0

3 years ago