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3 years ago

6

Help me pleaseeeeeeeeeeee...

2 answers:

Aneli [31]3 years ago

8 0

I’m sorry I really need to answer questions

mel-nik [20]3 years ago

3 0

50.24 bc a=3.14(radius to second power) and the radius would be 4, 4•4=16 and 16•3.14=50.24

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A positive integer is 47 more than 9 times another. Their product is 1860. Find the two integers.

svlad2 [7]

X = 47 + 9y
xy = 1860

y(47 + 9y) = 1860
47y + 9y² = 1860
9y² + 47y - 1860 = 0

> using a quadratic equation solver on a calculator but you can also use the quadratic equation = [-b+/- √(b²-4ac)]/(2a)
> only integer solution is x = 12

12y = 1860
y = 155

integers are 12 and 155

8 0

3 years ago

17.55 what is this rounded to nearest pound

vesna_86 [32]

Answer:18

Step-by-step explanation:

The 17.55 rounded to nearest pound is 18.

8 0

3 years ago

There are 200 students in the 8th grade class at Alleghany Middle School. Of these students, 8% play basketball and 14% play soc

sergeinik [125]

Answer:

The probability that a student plays either basketball or soccer is 19% or 0.19.

Step-by-step explanation:

Let A be the event that student play basketball and B be the event that student play soccer.

$P(A)=8\%$

$P(B)=14\%$

It is given that 6 students play on both teams.

$P(A\cap B)=\frac{6}{200}\times 100=3\%$

We have to find the probability that a student plays either basketball or soccer.

$P(A\cup B)=P(A)+P(B)-(A\cap B)$

$P(A\cup B)=8\%+14\%-3\%$

$P(A\cup B)=19\%$

Therefore the probability that a student plays either basketball or soccer is 19% or 0.19.

6 0

4 years ago

The line L is a tangent to the curve with equation y= 4x^2 +1 . The line L cuts the y axis at (0,8) and has a positive gradient.

sergeinik [125]

A generic point on the graph of the curve has coordinates

$(x, 4x^2+1)$

The derivative gives us the slope of the tangent line at a given point:

$f(x) = 4x^2+1 \implies f'(x) = 8x$

Let k be a generic x-coordinate. The tangent line to the curve at this point will pass through $(k, 4k^2+1)$ and have slope $8k$

So, we can write its equation using the point-slope formula: a line with slope m passing through $(x_0, y_0)$ has equation

$y-y_0 = m(x-x_0)$

In this case, $(x_0, y_0)=(k, 4k^2+1)$ and $m=8k$ , so the equation becomes

$y-4k^2-1 = 8k(x-k)$

We can rewrite the equation as follows:

$y-4k^2-1 = 8k(x-k) \iff y = 8kx - 8k^2+4k^2+1 \iff y = 8kx-4k^2+1$

We know that this function must give 0 when evaluated at x=0:

$f(x) = 8kx-4k^2+1 \implies f(0) = -4k^2+1 = 8 \iff -4k^2 = 7 \iff k^2 = -\dfrac{7}{4}$

This equation has no real solution, so the problem looks impossible.

5 0

3 years ago

Suppose that J and K are points on the number line.

rodikova [14]

Answer:

-7, -27

Step-by-step explanation:

J is at -17. K is 10 units from J.

Start at point J, -17:

a) If you go 10 units to the right, you end up at -17 + 10 = -7

b) If you go 10 units to the left, you end up at -17 - 10 = -27

J can be at -7 or at -27

Answer: -7, -27

8 0

4 years ago