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2 months ago

Guest #2381

2 months ago

The range is the difference between the highest and lowest values.

The highest value is 61.

The lowest value is 21.

range = highest value - lowest value = 61 - 21 = 40

The highest value is 61.

The lowest value is 21.

range = highest value - lowest value = 61 - 21 = 40

Question

The square root of 136 is between which two whole numbers?

Solution 1

The square root is between 11 and 12 but closer to 12.Â

Hope this helps!Â

Hope this helps!Â

Solution 2

The square root of 136 is between 11 and 12. Hope I helped!

Question

Use the remainder theorem to determine the remainder when
A. 107
B. 57
C. 43
D. 93

Solution 1

That would be 3(5)^2 + 5(5) - 7 =Â 75 + 25 - 7Â = 93 answer

Question

Calculate the equation 6^2/2(3)+4=x

Solution 1

(6^2)/2(3)+4=x

so do the exponent

6^2=36

36/2(3)+4

divide

36/2=18

18(3)+4

54+4

58

x=58

so do the exponent

6^2=36

36/2(3)+4

divide

36/2=18

18(3)+4

54+4

58

x=58

Question

The dot product of u with itself is 12. what is the magnitude of u?

Solution 1

**Answer: Â The magnitude of the vector u is âˆš12 units.**

**Step-by-step explanation: Â **Given that the dot product of a vector u with itself is 12.

We are to **find the magnitude of the vector u.**

Let <a, b> represents the vector u.

That is, **u = <a, b>**

**Then, according to the given information, we have**

**Thus, the magnitude of the vector u is âˆš12 units.**

Solution 2

Let's say u = <a,b>.

we know it's dot product is 12, thus

we know it's dot product is 12, thus

Question

If i = sqrt -1 what is the value of i3
A. -1
B. i
C. 1
D. -i

Solution 1

**Answer:**

The value of i3 is equal to D. -i

**Step-by-step explanation:**

We can find the answer by taking the definition of i :

i = sqrt(-1) Â Â Â Â (1)

i ^2 = (sqrt(-1)) ^2 = -1 Â Â Â Â Â Â (2) because square root and squaring are inverse operations

so from (2)

i ^3= i ^2* i = -1*i = -i

So -i is the final answer or Â â€“ (sqrt(-1)) Â Â Â Â which is the same

Solution 2

I^3Â = i * i * iÂ = i^2 * iÂ = -1 * i = -i Answer

Its D

Its D

Question

The angle of elevation from point A to the top of a cliff is 38 degrees. If point A is 80 feet from the base of the cliff, how high is the cliff? A.
60.50 ft.
B.
62.50 ft.
C.
63.50 ft.
D.
61.50 ft.

Solution 1

Tan 38 = h/80

h = 80 tan 38Â =Â 62.50 ftÂ

Its B

h = 80 tan 38Â =Â 62.50 ftÂ

Its B

Question

What's the area of a square with a radius of 16 root 2

Solution 1

That's impossible to say. The radius of a square is meaningless.

Question

Share out Â£80 between Tom and jerry in the ratio 3:2

Solution 1

Tom gets- 48

Jerry gets- 32

Hope I helped! tell if I missed something! :)

Jerry gets- 32

Hope I helped! tell if I missed something! :)

Solution 2

So let's add variables according to the ratio:

3n + 2n =Â Â£80

5n =Â Â£80

n = 16

3(16):2(16)

48:32

**Tom getsÂ ****Â£48, and Jerry getsÂ ****Â£32**

3n + 2n =Â Â£80

5n =Â Â£80

n = 16

3(16):2(16)

48:32

Question

Will give brainliest-H(t)=-16t^2+28t T is time in seconds.
How long will it take for the dolphin to make one jump, show your work.

Solution 1

For this case we have the following equation:

Â H (t) = - 16t ^ 2 + 28t

Â When the dolphin finishes a jump, then the height is equal to zero.

Â We have then:

Â -16t ^ 2 + 28t = 0

Â We look for the roots of the polynomial:

Â t1 = 0

Â t2 = 1.75 s

Â Therefore, the dolphin takes 1.75 seconds to make a jump.

Â**Answer:**

Â**it will take for the dolphin to make one jump about:**

Â**t = 1.75 s**

Â H (t) = - 16t ^ 2 + 28t

Â When the dolphin finishes a jump, then the height is equal to zero.

Â We have then:

Â -16t ^ 2 + 28t = 0

Â We look for the roots of the polynomial:

Â t1 = 0

Â t2 = 1.75 s

Â Therefore, the dolphin takes 1.75 seconds to make a jump.

Â

Â

Â

Question

X 0 2 4 6 8 f(x) 15 15.5 17 19.5 23 Write the quadratic function that corresponds to the given table of values in standard form, f(x) = ax2 + bx + c, where a, b and c are constants. f(x)=ax^2+bx+c

Solution 1

**Answer with Step-by-step explanation:**

We are given that:

f(x)=axÂ²+bx+c

also at x=0 f(x)=15

Putting x=0 in f(x)

aÃ—0+bÃ—0+c=15

â‡’ **c=15**

now, f(2)=15.5

â‡’ aÃ—2Â²+bÃ—2+c=15.5

â‡’ 4a+2b+15=15.5

â‡’ 4a+2b=0.5 Â -------(1)

f(4)=17

â‡’ aÃ—4Â²+bÃ—4+c=17

â‡’ 16a+4b+15=17

â‡’ 16a+4b=2 Â Â --------(2)

equation (2)- 2Ã—equation(1)

â‡’ 16a+4b-8a-4b=2-2Ã—0.5

â‡’ 8a=2-1

â‡’ 8a=1

â‡’ a=1/8

â‡’** a=0.125**

Putting value of a and c in equation (1)

â‡’ 4Ã—0.125+2b=0.5

â‡’ 0.5+2b=0.5

â‡’ 2b=0

â‡’ **b=0**

**Hence **

**a=0.125**

**b=0**

**c=15**

**and f(x)=0.125xÂ²+15**

Solution 2

F(x)=ax^2=bx=c is the answer

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