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Lesechka [4]
3 years ago
5

What is the value of x? Drawing not to scale​

Mathematics
1 answer:
horsena [70]3 years ago
6 0
Jjdkkfkfkkskskslslslassldccncjffk
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2/3 + 1/2 in simple form fractions
nata0808 [166]
Try and find the common denominator which is 6 since 2×3 is 6 then once u get ur denominator u have to find the numerator which would be 4/6+3/6 And that would equal 7/6 and if you would u would keep it like that but if u would have to change it into a mixed fraction it would be 1 and 1/6

Hoped this helped
3 0
3 years ago
In class, 5/9 of the students have blue Eyes. if 10 students have blue eyes, how many students are in the class in total?
KiRa [710]
Let us cross multiply.

5/9=10/x

9 x 10= 5x

90=5x

90 <u />÷ 5= 5x <span>÷ 5

18=x

So, the answer is 18. I hope this helped you, since I just learned cross multiplication last week.</span>
8 0
3 years ago
Read 2 more answers
Do not pass go. proposition or not?
kolbaska11 [484]
I believe it is proposition
8 0
3 years ago
True or false, 4x-3+x+7 would equal 5x-10​
dimulka [17.4K]

Answer:

false

Step-by-step explanation:

if you simplify you get 5x+4=5x-10, which is false

3 0
3 years ago
Read 2 more answers
Hi guys, can anyone help me with this triple integral? Many thanks:)
Crank

Another triple integral.  We're integrating over the interior of the sphere

x^2+y^2+z^2=2^2

Let's do the outer integral over z.   z stays within the sphere so it goes from -2 to 2.

For the middle integral we have

y^2=4-x^2-z^2

x is the inner integral so at this point we conservatively say its zero.  That means y goes from -\sqrt{4-z^2} and +\sqrt{4-z^2}

Similarly the inner integral x goes between \pm-\sqrt{4-y^2-z^2}

So we rewrite the integral

\displaystyle \int_{-2}^{2} \int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}} \int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dx \; dy \; dz

Let's work on the inner one,

\displaystyle\int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dz

There's no z in the integrand, so we treat it as a constant.

=(x^2+xy+y^2)z \bigg|_{z=-\sqrt{4-y^2-z^2}}^{z=\sqrt{4-y^2-z^2}}

So the middle integral is

\displaystyle\int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}}2(x^2+xy+y^2)\sqrt{4-y^2-z^2} \ dy  

I gotta go so I'll stop here, sorry.

7 0
3 years ago
Read 2 more answers
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