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astra-53 [7]
3 years ago
11

Write 625 using exponents as many as you can

Mathematics
1 answer:
dezoksy [38]3 years ago
8 0

Idk if this will help, but her :)

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Solve the system of equations shown. Write your answer as an order pair. (x,y) *
Anon25 [30]

Answer:

The correct answer is (5,3)

Step-by-step explanation:

I just did that one:)

4 0
3 years ago
the surface area of a sphere is 804 .25 m2,to find the radius of the sphere to the nearest world number​
Lady_Fox [76]

Answer:

256

Step-by-step explanation:

804.25 divided by 3.14

4 0
3 years ago
Read 2 more answers
Carry out three steps of the Bisection Method for f(x)=3x−x4 as follows: (a) Show that f(x) has a zero in [1,2]. (b) Determine w
ELEN [110]

Answer:

a) There's a zero between [1,2]

b) There's a zero between [1.5,2]

c) There's a zero between  [1.5,1.75].

Step-by-step explanation:

We have f(x)=3^x-x^4

A)We need to show that f(x) has a zero in the interval [1, 2]. We have to see if the function f is continuous with f(1) and f(2).

f(x)=3^x-x^4\\\\f(1)=3^1-(1)^4=3-1=2\\\\f(2)=3^2-(2)^4=9-16=(-7)

We can see that f(1) and f(2) have opposite signs. And f(1)>f(2) and the function is continuous, this means that exists a real number c between the interval [1,2] where f(c)=0.

B)We have to repeat the same steps of A)

For the subinterval [1,1.5]:

f(x)=3^x-x^4\\\\f(1)=3^1-(1)^4=3-1=2\\\\f(1.5)=3^1^.^5-(1.5)^4=5.19-5.06=0.13

f(1) and f(1.5) have the same signs, this means there's no zero in the subinterval [1,1.5].

For the subinterval [1.5,2]:

f(x)=3^x-x^4\\\\f(1.5)=3^1^.^5-(1.5)^4=5.19-5.06=0.13\\\\f(2)=3^2-(2)^4=9-16=(-7)

f(1.5) and f(2) have opposite signs, this means there's a zero between the subinterval [1.5,2].

C)We have to repeat the same steps of A)

For the subinterval [1,1.25]:

f(x)=3^x-x^4\\\\f(1)=3^1-(1)^4=3-1=2\\\\f(1.25)=3^1^.^2^5-(1.25)^4=3.94-2.44=1.5

f(1) and f(1.25) have the same signs, this means there's no zero in the subinterval [1,1.25].

For the subinterval [1.25,1.5]:

f(x)=3^x-x^4\\\\f(1.25)=3^1^.^2^5-(1.25)^4=3.94-2.44=1.5\\\\f(1.5)=3^1^.^5-(1.5)^4=5.19-5.06=0.13

f(1.25) and f(1.5) have the same signs, this means there's no zero in the subinterval [1.25,1.5].

For the subinterval [1.5,1.75]:

f(x)=3^x-x^4\\\\f(1.5)=3^1^.^5-(1.5)^4=5.19-5.06=0.13\\\\f(1.75)=3^1^.^7^5-(1.75)^4=6.83-9.37=(-2.54)

f(1.5) and f(1.75) have opposite signs, this means there's a zero between the subinterval [1.5,1.75].

For the subinterval [1.75,2]:

f(x)=3^x-x^4\\\\f(1.75)=3^1^.^7^5-(1.75)^4=6.83-9.37=(-2.54)\\\\f(2)=3^2-(2)^4=9-16=(-7)

f(1.75) and f(2) have the same signs, this means there isn't a zero between the subinterval [1.75,2].

The graph of the function shows that the answers are correct.

6 0
3 years ago
What is the length of AB please don’t do it for fun just try to do it for me hurry?
Lerok [7]

Answer:

y6%cf4yergby65thrbgtrvbh

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
I need help on this loll
Serhud [2]

Answer: The translation rule that maps point D ( 7 , − 3 ) onto point D ' ( 2 , 5 )

is (x , y) → (x - 5 , y + 8)

Step-by-step explanation:

Let us revise the translation

If the point (x , y) translated horizontally to the right by h units  then its image is (x + h , y)

If the point (x , y) translated horizontally to the left by h units  then its image is (x - h , y)

If the point (x , y) translated vertically up by k units  then its image is (x , y + k)

If the point (x , y) translated vertically down by k units  then its image is (x , y - k)

(x , y) → (x ± h , y ± k) the right arrow symbol used to show the

translation from a point to its image

Example:

∵ P (0 , 0) → P' (1 , 2)

∴ The rule is (x , y) → (x + 1 , y + 2)

Let us find the translation rule that maps point D ( 7 , − 3 ) onto

point D' (2 , 5)

∵ Point (x , y) = (7 , -3)

∵ Its image after translation (x + h , y + k) = (2 , 5)

∴ x + h = 2

∵ x = 7

∴ 7 + h = 2

- Subtract 7 from both sides

∴ h = -5

∵ y + k = 5

∵ y = -3

∴ -3 + k = 5

- Add 3 to both sides

∴ k = 8

∴ The rule of translation is (x , y) → (x - 5 , y + 8)

The translation rule that maps point D ( 7 , − 3 ) onto point D ' ( 2 , 5 )

is (x , y) → (x - 5 , y + 8)

3 0
3 years ago
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