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3 years ago

Pls help guys ur my only hope pls i need a good grade pls help plssssssss

Mathematics

2 answers:

juin [17]3 years ago

7 0

Answer: The answer is B, 6 1/2

Step-by-step explanation:

6+5= 11

1/2+1/2= 1

11+1= 12

12-3=9

maksim [4K]3 years ago

7 0

6/5. Simple. Including the -3 as well

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Lucy had 3 2/3 gal of paint. After painting a room she had 1 1/4 gal left how many gallons of paint did Lucy use to paint a room

garik1379 [7]

Answer:

$2 \frac{5}{12}$

gallons of paint was used by Lucy.

Step-by-step explanation:

$3\frac{2}{3} - 1\frac{1}{4} = \frac{44}{12} - \frac{15}{12} = \frac{29}{12}$

29/12 = 2 5/12

7 0

3 years ago

Suppose that \nabla f(x,y,z) = 2xyze^{x^2}\mathbf{i} + ze^{x^2}\mathbf{j} + ye^{x^2}\mathbf{k}. if f(0,0,0) = 2, find f(1,1,1).

lesya [120]

The simplest path from (0, 0, 0) to (1, 1, 1) is a straight line, denoted $C$ , which we can parameterize by the vector-valued function,

$\mathbf r(t)=(1-t)(\mathbf i+\mathbf j+\mathbf k)$

for $0\le t\le1$ , which has differential

$\mathrm d\mathbf r=-(\mathbf i+\mathbf j+\mathbf k)\,\mathrm dt$

Then with $x(t)=y(t)=z(t)=1-t$ , we have

$\displaystyle\int_{\mathcal C}\nabla f(x,y,z)\cdot\mathrm d\mathbf r=\int_{t=0}^{t=1}\nabla f(x(t),y(t),z(t))\cdot\mathrm d\mathbf r$

$=\displaystyle\int_{t=0}^{t=1}\left(2(1-t)^3e^{(1-t)^2}\,\mathbf i+(1-t)e^{(1-t)^2}\,\mathbf j+(1-t)e^{(1-t)^2}\,\mathbf k\right)\cdot-(\mathbf i+\mathbf j+\mathbf k)\,\mathrm dt$

$\displaystyle=-2\int_{t=0}^{t=1}e^{(1-t)^2}(1-t)(t^2-2t+2)\,\mathrm dt$

Complete the square in the quadratic term of the integrand: $t^2-2t+2=(t-1)^2+1=(1-t)^2+1$ , then in the integral we substitute $u=1-t$ :

$\displaystyle=-2\int_{t=0}^{t=1}e^{(1-t)^2}(1-t)((1-t)^2+1)\,\mathrm dt$

$\displaystyle=-2\int_{u=0}^{u=1}e^{u^2}u(u^2+1)\,\mathrm du$

Make another substitution of $v=u^2$ :

$\displaystyle=-\int_{v=0}^{v=1}e^v(v+1)\,\mathrm dv$

Integrate by parts, taking

$r=v+1\implies\mathrm dr=\mathrm dv$

$\mathrm ds=e^v\,\mathrm dv\implies s=e^v$

$\displaystyle=-e^v(v+1)\bigg|_{v=0}^{v=1}+\int_{v=0}^{v=1}e^v\,\mathrm dv$

$\displaystyle=-(2e-1)+(e-1)=-e$

So, we have by the fundamental theorem of calculus that

$\displaystyle\int_C\nabla f(x,y,z)\cdot\mathrm d\mathbf r=f(1,1,1)-f(0,0,0)$

$\implies-e=f(1,1,1)-2$

$\implies f(1,1,1)=2-e$

3 0

3 years ago

Solve for x: x - 10 = -12 A. 2 B. -2 C. 22 D. -22

Sindrei [870]

Add 10 to both sides. X is 2. A.

7 0

3 years ago

Read 2 more answers

Yvonne bought a car for $10,300. After 3 years, the value of the car was $6,250. The value of this car decreases exponentially o

Likurg_2 [28]

Answer:

A = $44,778.84

A = P + I where

P (principal) = $10,300.00

I (interest) = $34,478.84 or 12 months

Step-by-step explanation:

Given: Yvonne bought a car for $10,300. After 3 years, the value of the car was $6,250. 50%

To find: In approximately what number of additional months will the value of Yvonne’s car be 50% of the price she originally paid?

Formula: $(\frac{interest rate}{number of payments})$ × $loan principal = interest$

Solution: Divide your interest rate by the number of payments you’ll make in the year (interest rates are expressed annually). So, for example, if you’re making monthly payments, divide by 12. Multiply it by the balance of your loan, which for the first payment, will be your whole principal amount.

(This gives you the amount of interest you pay the first month.)

First, convert R as a percent to r as a decimal

r = R/100

r = 50/100

r = 0.5 rate per year,

Then solve the equation for A

A = P(1 + r/n)nt

A = 10,300.00(1 + 0.5/12)(12)(3)

A = 10,300.00(1 + 0.041666667)(36)

A = $44,778.84

Henceforth:

The total amount accrued, principal plus interest, with compound interest on a principal of $10,300.00 at a rate of 50% per year compounded 12 times per year over 3 years is $44,778.84.

6 0

3 years ago

Simplify 15 to the 18th power over 15 to the 3rd power.

worty [1.4K]

The rule for quotients of similar bases with different exponents is:

(a^c)/(a^b)=a^(c-b) in this case:

15^18/(15^3)=15^15

3 0

3 years ago