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3 years ago

In the equation -10-3b=8 what does b stand for. i need halp soon soo pleaseeeeeeee

Mathematics

2 answers:

Annette [7]3 years ago

5 0

Answer: b = -6

Step-by-step explanation: First add 10 to both sides to get -3b=18, then divide by -3 to get b = -6.

Hope this helped!

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aleksandrvk [35]3 years ago

5 0

-10-3b=8
-10-8=3b
-18=3b
B= -6

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2 years ago

How to simplify this?

cricket20 [7]

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2 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

2x^4–5x^3+x^2+3x+2=?<br><br> X=5

DanielleElmas [232]

Answer:

$2x^4–5x^3+x^2+3x+2=$

$2 \times 5^4–5 \times 5^3+5^2+3 \times 5+2$

= 2 × 625 – 5 × 125 + 25 + 3 × 5 + 2

=1250 – 625 + 25 + 15 +2

= 1292 – 625

= 667

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2 years ago

How do i do this question

oksano4ka [1.4K]

Answer:

A) 4.18

B)7.65

Step-by-step explanation:

You have to use a calculator

7 0

2 years ago