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Oksanka [162]
3 years ago
12

2

Mathematics
1 answer:
Slav-nsk [51]3 years ago
7 0

Answer:

area of P=πr^2

area of Q=πr^2/4=π×400/4=100π

acc to ques

A(q)=4× A(p)

100π=4×πr^2.

100=4r^2

r^2=25

radius of p= √25=5cm

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Write an equation of a line that passes through the point (4, 3) and is perpendicular to the graph of the equation y=-1/3 +4.​
likoan [24]

Answer:

y=3x-9

Step-by-step explanation:

Hi there!

we are given a point (4,3), and a line (y=-1/3x+4)

Perpendicular lines have slopes that are negative and reciprocal; if they are multiplied together, the result is -1

first, let's find the slope of the new line.

we can find it with this formula: (m=slope)

-1/3m=-1

multiply both sides by -3

m=3

here's our equation so far, written in y=mx+b form, where m is the slope and b is the y intercept

y=3x+b

we need to find b

Because the line passes through the point (4,3), we can use it to find b

substitute x as 4 and y as 3

3=3(4)+b

multiply

3=12+b

-9=b

substitute into the equation

<u>y=3x-9</u>

Hope this helps!

7 0
3 years ago
Please help me with this problem !!!!!!
iren [92.7K]

Answer:

see explanation

Step-by-step explanation:

Given

s(t) = 10t²

To find s(2) and s(5) substitute t = 2 and t = 5 into f(t)

(a)

s(2) = 10 × 2² = 10 × 4 = 40

s(5) = 10 × 5² = 10 × 25 = 250

(b)

s(5) - s(2) = 250 - 40 = 210

This represents the distance travelled in 3 hours

(c)

The average rate of change is measured as

\frac{s(5)-s(2)}{5-2}

= \frac{250-40}{3}

= \frac{210}{3}

= 70 miles per hour

7 0
3 years ago
What is the smallest value of n such that an algorithm whose running time is 100n 2 runs faster than an algorithm whose running
Semmy [17]

Answer:

n = 15

Step-by-step explanation:

For inputs of the value of n, the running time for the algorithm A is 100n^2 and that of B is 2^n.

If A is to run faster than B, 100n^2 must be smaller than 2^n.

Let's check from n = 1 to know the value of n that fits

n = 1

100(1)^2 > 2^1

100 > 2

n = 2

100(2)^2 > 2^2

400 > 4

n = 4

100(4)^2 > 2^4

1600 > 16

n = 8

100(8)^2 > 2^8

6400 > 2^8

n = 16

100(16)^2 < 2^16

25600 < 2^16

This implies that between n = 8 and 16, A starts to run faster than B

n = (8+16)/2 = 12

100(12)^2 > 2^12

14400 > 2^12

n = (12+16)/2 = 14

100(14)^2 > 2^14

19600 > 2^14

n = (14+16)/2

n = 15

100(15)^2 < 2^15

22500 < 2^15

At n= 15, A starts running faster than B

8 0
4 years ago
MATH LOVERS I NEED YOU!! Please do both pages,
serious [3.7K]
Hopefully that’s right
3 0
3 years ago
Pls help it's due tonight!!!<br> :((((((
statuscvo [17]
I think AB is 3!!

i hope i’m right good luck ,:D
4 0
3 years ago
Read 2 more answers
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