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3 years ago

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Mathematics

1 answer:

ss7ja [257]3 years ago

8 0

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Tracy received a score of 85% on her quiz. If Tracey answered 9 questions incorrectly, how many questions were on her quiz?

Vikki [24]

Answer:

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

For which of the following compound inequalities is there no solution?

Bess [88]

Answer:

(a) $6m \le - 36$ and $m + 24 > 20$

Step-by-step explanation:

Required

Which of a to d has no solution

(a) $6m \le - 36$ and $m + 24 > 20$

We have: $m + 24 > 20$

Sole for m

$m >20 - 24$

$m >-4$

$6m \le - 36$

Divide by 6

$m \le -6$

So, we have:

$m \le -6$ and $m >-4$

$m \le -6$ implies that: m = -6,-7,-8,-9.....

$m >-4$ implies that m = -3,-2,-1,0.....

<em>Hence, there is no solution</em>

8 0

3 years ago

Mary Sue and Betty have 603 necklaces altogether. Mary Sue has 115 more necklaces than Betty. How many necklaces does Betty have

Blizzard [7]

Let Mary Sue have m necklaces and Betty have b necklaces.

And they have 603 necklaces altogether, that is

m + b =603

And Mary Sue has 115 more necklaces than Betty, that is

m=b +115

Substituting this value in first equation we will get

b+115+b=603

Subtracting 115

2b= 488

b= 244

So Betty have 244 necklaces .

6 0

3 years ago

If c is the curve given by \mathbf{r} \left( t \right = \left( 1 5 \sin t \right \mathbf{i} \left( 1 3 \sin^{2} t \right \mathbf

jonny [76]

With the curve $C$ parameterized by

$C:\mathbf r(t)=\underbrace{15\sin t}_{x(t)}\,\mathbf i+\underbrace{13\sin^2t}_{y(t)}\,\mathbf j+\underbrace{12\sin^3t}_{z(t)}\,\mathbf k$

with $0\le t\le\dfrac\pi2$ , and given the vector field

$\mathbf f(x,y,z)=x\,\mathbf i+y\,\mathbf j+z\,\mathbf k$

the work done by $\mathbf f$ on a particle moving on along $C$ is given by the line integral

$\displaystyle\int_C\mathbf f\cdot\mathrm d\mathbf r=\int\limits_{t=0}^{t=\pi/2}\mathbf f(x(t),y(t),z(t))\cdot\frac{\mathrm d\mathbf r(t)}{\mathrm dt}\,\mathrm dt$

where

$\mathrm d\mathbf r=(15\cos t\,\mathbf i+26\sin t\cos t\,\mathbf j+36\sin^2t\cos t\,\mathbf k)\,\mathrm dt$

The integral is then

$\displaystyle\int_0^{\pi/2}(15\sin t\,\mathbf i+13\sin^2t\,\mathbf j+12\sin^3t\,\mathbf k)\cdot(15\cos t\,\mathbf i+13\sin2t\,\mathbf j+18\sin t\sin2t\,\mathbf k)\,\mathrm dt$
$=\displaystyle\int_0^{\pi/2}(432\sin^5t\cos t+338\sin^3t\cos t+225\sin t\cos t$
$=269$

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3 years ago

9. If f(x)=x². g(x)=2x-1. and h(x) =find the following.

postnew [5]

Answer:

Please see the attached picture ! :D

-------- HappY LearninG <3 ----------

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2 years ago