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tensa zangetsu [6.8K]
3 years ago
10

Evaluate the expression 4y/Yfor x=8 and y=0

Mathematics
1 answer:
jeyben [28]3 years ago
8 0
To evaluate 4y/x for x=8 and y=0 we shall have:
4y/x=(4×0)/8
=0/8
=0
Answer; 0
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Nine more than the product of -8 and a number is -7
Yanka [14]

Answer:

-9

Step-by-step explanation:

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3 years ago
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2605 rounded to the nearest hundred
Fittoniya [83]
It would round to 2600
8 0
2 years ago
the weight of an elephant is 10 to the 3rd times the weight of a cat if the elephant weighs 14000 pounds how much does the cat w
Andreyy89
Wouldn't you just divide 14000 by 10^3
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3 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.
5 0
3 years ago
A semicircle with diameter PQ sits on an isosceles triangle PQR to form a region shaped like a two-dimensional ice- cream cone,
Ksivusya [100]

The limit of a(x)/b(x) as x approaches 0 is gotten as;

lim a(x)/b(x); x→0 = 0

  • The image showing the semi circle and isosceles triangle is missing and so i have attached it.

  • Formula for <em>area</em> of a semi circle is;

a(x) = A = ¹/₂πr²

  • <em>Area</em> of triangle is;

b(x) = A =  ¹/₂ × base × height

  • From the image, ∠PQR = θ

Thus; a(x) = ¹/₂π(10 sin (θ/2))²

a(x) =  ¹/₂π(100 sin² (θ/2))

Similarly;

b(x) =  ¹/₂(20 sin (θ/2)) × (10 cos (θ/2))

b(x) = 100 sin (θ/2) cos (θ/2)

  • Thus;

a(x)/b(x) = ¹/₂π(100 sin² (θ/2)) ÷ 100 sin (θ/2) cos (θ/2)

100 will cancel out and the sin (θ/2) at the denominator will be eradicated

upon further simplification to give;

a(x)/b(x) = ¹/₂π(sin (θ/2)) ÷ cos (θ/2)

In trigonometry, we know that; tan θ = sin θ/cos θ

Thus;

a(x)/b(x) = ¹/₂π(sin (θ/2)) ÷ cos (θ/2) =  ¹/₂π(tan (θ/2))

We want to find the limit as θ approaches <em>zero.</em>

Thus;

lim θ → 0 =  ¹/₂π(tan (0/2))

⇒ ¹/₂π tan 0

⇒ 0

Read more at; https://brainly.in/question/9387458

3 0
1 year ago
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