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levacccp [35]
3 years ago
5

Hi... can you tell me is this formula is right or wrong?? ​

Mathematics
1 answer:
Llana [10]3 years ago
6 0

Answer:

this formula is right

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I need to show .y work for 132 × 3​
Lelechka [254]

Answer:

132 × 3 = 396

Step-by-step explanation:

132 3 × 2 = 6

<u>×</u><u> </u><u> </u><u>3</u> 3 × 3 = 9

396 3 × 1 = 3

7 0
3 years ago
Read 2 more answers
Vince has a rectangular rug in his room with an area of 10 ft the length of the rug is 18 inches longer than the width what coul
aev [14]

The length of the rug is 4 ft.

The width of the rug is 2.5 ft.

Explanation:

The area of the rug is 10 ft.

The length of the rug be l.

Let us convert the inches to feet.

Thus, 18 inches =  1.5 ft

Thus, the length of the rug is l=1.5+w

Let the width of the rug be w.

Substituting these values in the formula of area of the rectangle, we get,

A=length\times width

10=(1.5+w)(w)\\10=1.5w+w^2\\w^2+1.5w-10=0

Solving the expression using the quadratic formula,

$w=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Substituting the values, we have,

$w=\frac{-15 \pm \sqrt{15^{2}-4 \cdot 10(-100)}}{2 \cdot 10}\\

$w=\frac{-15 \pm \sqrt{4225}}{20}$

$w=\frac{-15 \pm 65}{2 0}$

Thus,

w=\frac{-15 + 65}{2 0}\\w=\frac{50}{20} \\w=2.5  and   w=\frac{-15 - 65}{2 0}\\w=\frac{-80}{20} \\w=-4

Since, the value of w cannot be negative, the value of w is 2.5ft

Thus, the width of the rug is 2.5ft

Substituting w=2.5 in l=1.5+w, we get,

l=1.5+2.5\\l=4

Thus, the length of the rug is 4 ft.

4 0
3 years ago
On Saturday, Nicole earns no more than $55 at her job. On Sunday, she earns at least $40. Which of these inequalities represent
makkiz [27]

Answer:

B

Step-by-step explanation:

use your head

7 0
3 years ago
Read 2 more answers
F(x)= -x^2+19 find f(3)
Whitepunk [10]
I’m not sure but this could help

4 0
3 years ago
Read 2 more answers
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
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