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3 years ago

Which division is shown by the model

Mathematics

2 answers:

cupoosta [38]3 years ago

7 0

Answer:

Im prettuy sure its the last one since theyre are 3 parts and thats the only one thats equal

Step-by-step explanation:

RUDIKE [14]3 years ago

4 0

There is 2 blocks, so that is your whole number. There are 2 colors that match in each pair (2 purple, 2 orange, 2 green). And since the blocks are broken up in thirds, you can assume that that means that you are dividing 2 and 2/3. The answer is 3 because there are three different colors.

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**How do you factor -4x²+14x-6?

MAXImum [283]

2 goes into all the terms so;

2(-2x² + 7x - 3) now you could factor it:

2(-2x - 1)(x + 3)

5 0

2 years ago

What percent of 16 is 0.12?

valentinak56 [21]

Answer:

0.75%

Step-by-step explanation:

Convert the fractions and then multiply by 100.

5 0

3 years ago

Read 2 more answers

What is the area of the trapezoid shown?

Rainbow [258]

BC is 6.08, use distance formula to calculate
height is BE and also 6.08
AD is 18.25
A= h((a+b)/2)
=6.08(6.08+18.25)/2)
= 73.96 units^2 is the area

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3 years ago

Read 2 more answers

Solve 3/2^x-4=16 Two step equations

Lerok [7]

the answer is

x = 40/3

4 0

3 years ago

Evaluate the line integral, where c is the given curve. (x + 9y) dx + x2 dy, c c consists of line segments from (0, 0) to (9, 1)

viktelen [127]

$\displaystyle\int_C(x+9y)\,\mathrm dx+x^2\,\mathrm dy=\int_C\langle x+9y,x^2\rangle\cdot\underbrace{\langle\mathrm dx,\mathrm dy\rangle}_{\mathrm d\mathbf r}$

The first line segment can be parameterized by $\mathbf r_1(t)=\langle0,0\rangle(1-t)+\langle9,1\rangle t=\langle9t,t\rangle$ with $0\le t\le1$ . Denote this first segment by $C_1$ . Then

$\displaystyle\int_{C_1}\langle x+9y,x^2\rangle\cdot\mathbf dr_1=\int_{t=0}^{t=1}\langle9t+9t,81t^2\rangle\cdot\langle9,1\rangle\,\mathrm dt$
$=\displaystyle\int_0^1(162t+81t^2)\,\mathrm dt$
$=108$

The second line segment ( $C_2$ ) can be described by $\mathbf r_2(t)=\langle9,1\rangle(1-t)+\langle10,0\rangle t=\langle9+t,1-t\rangle$ , again with $0\le t\le1$ . Then

$\displaystyle\int_{C_2}\langle x+9y,x^2\rangle\cdot\mathrm d\mathbf r_2=\int_{t=0}^{t=1}\langle9+t+9-9t,(9+t)^2\rangle\cdot\langle1,-1\rangle\,\mathrm dt$
$=\displaystyle\int_0^1(18-8t-(9+t)^2)\,\mathrm dt$
$=-\dfrac{229}3$

Finally,

$\displaystyle\int_C(x+9y)\,\mathrm dx+x^2\,\mathrm dy=108-\dfrac{229}3=\dfrac{95}3$

5 0

3 years ago