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4 years ago

Help me please!!..,,

Mathematics

1 answer:

kari74 [83]4 years ago

5 0

Answer:

Step-by-step explanation:

1.5 x 0.1 = 0.15

0.15 inches

Dont know if this is for sure right. Hope it is.

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Find the rectangular coordinates of the point with the polar coordinates (3, 3 divided by 2 pi ). (1 point)

kifflom [539]

Answer:

The correct choice is (0,-3)

Step-by-step explanation:

To convert from polar coordinates to rectangular coordinates, we use the parametric equations:

$x=r\cos \theta$ and $y=r\sin \theta$

The point given to us in polar coordinates is $(3,\frac{3}{2}\pi)$ .

where $r=3,\theta=\frac{3\pi}{2}$

$x=3\cos (\frac{3\pi}{2})$ and $y=3\sin( \frac{3\pi}{2})$

$x=3(0)$ and $y=3(-1)$

$x=0$ and $y=-3$

The correct choice is (0,-3)

5 0

3 years ago

Solve this inequality and show the result on a number line.

MariettaO [177]

$\\ \bull\tt\longmapsto 5(x-2)\geqslant 4(x-2)$

$\\ \bull\tt\longmapsto 5x-10\geqslant 4x-8$

$\\ \bull\tt\longmapsto 5x-4x\geqslant -8+10$

$\\ \bull\tt\longmapsto x\geqslant 2$

x is either 2 or greater than 2.

8 0

3 years ago

Read 2 more answers

Find the solutions to the equation. Select ALL that apply.

alisha [4.7K]

Answer:

Step-by-step explanation:

-3y(2y+5)=0

-6y+15y=0

9y=0

y=0÷9

y=0

5 0

4 years ago

If the sum of the even integers between 1 and k, inclusive, is equal to 2k, what is the value of k?

Alisiya [41]

If $k$ is odd, then

$\displaystyle\sum_{n=1}^{\lfloor k/2\rfloor}2n=2\dfrac{\left\lfloor\frac k2\right\rfloor\left(\left\lfloor\frac k2\right\rfloor+1\right)}2=\left\lfloor\dfrac k2\right\rfloor^2+\left\lfloor\dfrac k2\right\rfloor$

while if $k$ is even, then the sum would be

$\displaystyle\sum_{n=1}^{k/2}2n=2\dfrac{\frac k2\left(\frac k2+1\right)}2=\dfrac{k^2+2k}4$

The latter case is easier to solve:

$\dfrac{k^2+2k}4=2k\implies k^2-6k=k(k-6)=0$

which means $k=6$ .

In the odd case, instead of considering the above equation we can consider the partial sums. If $k$ is odd, then the sum of the even integers between 1 and $k$ would be

$S=2+4+6+\cdots+(k-5)+(k-3)+(k-1)$

Now consider the partial sum up to the second-to-last term,

$S^*=2+4+6+\cdots+(k-5)+(k-3)$

Subtracting this from the previous partial sum, we have

$S-S^*=k-1$

We're given that the sums must add to $2k$ , which means

$S=2k$
$S^*=2(k-2)$

But taking the differences now yields

$S-S^*=2k-2(k-2)=4$

and there is only one $k$ for which $k-1=4$ ; namely, $k=5$ . However, the sum of the even integers between 1 and 5 is $2+4=6$ , whereas $2k=10\neq6$ . So there are no solutions to this over the odd integers.