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Sati [7]
3 years ago
14

No links help

Mathematics
1 answer:
Vadim26 [7]3 years ago
8 0
B) 15cm long and 9cm wide
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Can someone please answer this?
rosijanka [135]

Answer:

13 foot rounded to the nearest foot

Step-by-step explanation:

the ratio is

nose to head: 2:9

Rushmore  mountain Roosevelt :

nose to head : x/60

2/9=x/60

120=9x

x=120/9= 13 1/3 or 13.33

x=13 foot rounded to the nearest foot

3 0
3 years ago
Write the number 2,900 in scientific notation.
vitfil [10]
29^100
That’s the answer To the question
8 0
3 years ago
Read 2 more answers
What is the simplified form of squared root 140?
kompoz [17]
Its 2 square root of 35
6 0
3 years ago
Read 2 more answers
my brother and i walk the same route every day. my beother takes 40minutes to get to school and i take 30minutes to get to schoo
Goryan [66]

In 1 minute, I walk = 1/30

In 1 minute, My brother walk = 1/40

In T minutes, I walk = T*1/30

My brother walk 8 minute before = T + 8

In 1 minute, My brother walk = (T + 8)*1/40

When these two distances are same, then I will catch him.

T*(1/30) = (T + 8)*(1/40)

30*(T + 8) = 40*T

30*T + 240 = 40*T

240 = 10*T

T = 24 minutes

hence, I will catch him in 24 minutes.

Take 5 instead of 8.

My brother walk 5 minute before = T + 5

In 1 minute, My brother walk = (T + 5)*1/40

When these two distances are same, then I will catch him.

T*(1/30) = (T + 5)*(1/40)

30*(T + 5) = 40*T

30*T + 150 = 40*T

150 = 10*T

T = 15 minutes

Hence, I will catch him in 15 minutes.


6 0
3 years ago
Explain how to write a quadratic equation given the following three points on the graph (5,31) (3,11) (0,11)
masha68 [24]

Given:

The graph of a quadratic function passes through the points (5,31) (3,11) (0,11).

To find:

The equation of the quadratic function.

Solution:

A quadratic function is defined as:

y=ax^2+bx+c            ...(i)

It is passes through the point (0,11). So, substitute x=0,y=11 in (i).

11=a(0)^2+b(0)+c

11=c

Putting c=11 in (i), we get

y=ax^2+bx+11               ...(ii)

The quadratic function passes through the point (5,31). So, substitute x=5,y=31 in (ii).

31=a(5)^2+b(5)+11

31-11=a(25)+5b

20=25a+5b

Divide both sides by 5.

4=5a+b                  ...(iii)

The quadratic function passes through the point (3,11). So, substitute x=3,y=11 in (ii).

11=a(3)^2+b(3)+11

11-11=a(9)+3b

0=9a+3b

Divide both sides by 3.

0=3a+b                 ...(iv)

Subtracting (iv) from (iii), we get

4-0=5a+b-3a-b

4=2a

\dfrac{4}{2}=a

2=a

Putting a=2 in (iv), we get

0=3(2)+b

0=6+b

-6=b

Putting a=2,b=-6 in (ii), we get

y=(2)x^2+(-6)x+11

y=2x^2-6x+11

Therefore, the required quadratic equation is y=2x^2-6x+11.

7 0
3 years ago
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