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3 years ago

5

If peter has 27 apples and gives Tim 7, how much does peter have?

2 answers:

Salsk061 [2.6K]3 years ago

5 0

Answer: 20

Step-by-step explanation:

27 apples - 7 apples = 20 apples

pentagon [3]3 years ago

3 0

Answer:

Peter has 20 apples

Step-by-step explanation:

27 - 7 = 20

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All answers please(ALL)

astraxan [27]

A.13

B. 35

C. -6

D. -3

E. -1

F. -.75

Solve for the equation or problem.

A. 8

B. 4

C. 1

Add together al variable terms with the same variables.

1. -4x+10

2. 21t-10s+5r-10

Hope this helps!

6 0

4 years ago

An ice sculpture in the form of a sphere melts in such a way that it maintains its spherical shape. The volume of the sphere is

olga nikolaevna [1]

The volume of the sphere is:
V = 4 / 3πr ^ 3
Deriving we have:
V '= (3) (4/3) (π) (r ^ 2) (r')
From here, we clear the value of r ':
r '= (V') / ((4) (π) (r ^ 2))
Substituting values:
r '= (2π) / ((4) (π) ((5) ^ 2))
r '= (2π) / ((4) (π) (25))
r '= (2π) / (100π)
r '= 1/50
Then, the surface area of the sphere is:
A = 4πr ^ 2
Deriving we have:
A '= 8πrr'
Substituting values:
A '= 8π (5) (1/50)
Rewriting:
A '= (40/50) π
A '= (8/10) π
A '= (4/5) π
Answer:
The surface area of the sphere is decreasing at:
(A) 4pi / 5

8 0

4 years ago

10x + 11 - x= 5x - 13

Veseljchak [2.6K]

Answer:

= − 6

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

I need help with finding the answer to a) and b). Thank you!

shtirl [24]

Answer:

$\displaystyle \sin\Big(\frac{x}{2}\Big) = \frac{7\sqrt{58} }{ 58 }$

$\displaystyle \cos\Big(\frac{x}{2}\Big)=-\frac{3 \sqrt{58}}{58}$

$\displaystyle \tan\Big(\frac{x}{2}\Big)=-\frac{7}{3}$

Step-by-step explanation:

We are given that:

$\displaystyle \sin(x)=-\frac{21}{29}$

Where x is in QIII.

First, recall that sine is the ratio of the opposite side to the hypotenuse. Therefore, the adjacent side is:

$a=\sqrt{29^2-21^2}=20$

So, with respect to x, the opposite side is 21, the adjacent side is 20, and the hypotenuse is 29.

Since x is in QIII, sine is negative, cosine is negative, and tangent is also negative.

And if x is in QIII, this means that:

$180$

So:

$\displaystyle 90 < \frac{x}{2} < 135$

Thus, x/2 will be in QII, where sine is positive, cosine is negative, and tangent is negative.

1)

Recall that:

$\displaystyle \sin\Big(\frac{x}{2}\Big)=\pm\sqrt{\frac{1 - \cos(x)}{2}}$

Since x/2 is in QII, this will be positive.

Using the above information, cos(x) is -20/29. Therefore:

$\displaystyle \sin\Big(\frac{x}{2}\Big)=\sqrt{\frac{1 + 20/29}{2}$

Simplify:

$\displaystyle \sin\Big(\frac{x}{2}\Big)=\sqrt{\frac{49/29}{2}}=\sqrt{\frac{49}{58}}=\frac{7}{\sqrt{58}}=\frac{7\sqrt{58}}{58}$

2)

Likewise:

$\displaystyle \cos \Big( \frac{x}{2} \Big) =\pm \sqrt{ \frac{1+\cos(x)}{2} }$

Since x/2 is in QII, this will be negative.

Using the above information, cos(x) is -20/29. Therefore:

$\displaystyle \cos \Big( \frac{x}{2} \Big) =-\sqrt{ \frac{1- 20/29}{2} }$

Simplify:

$\displaystyle \cos\Big(\frac{x}{2}\Big)=-\sqrt{\frac{9/29}{2}}=-\sqrt{\frac{9}{58}}=-\frac{3}{\sqrt{58}}=-\frac{3\sqrt{58}}{58}$

3)

Finally:

$\displaystyle \tan\Big(\frac{x}{2}\Big) = \frac{\sin(x/2)}{\cos(x/2)}$

Therefore:

$\displaystyle \tan\Big(\frac{x}{2}\Big)=\frac{7\sqrt{58}/58}{-3\sqrt{58}/58}=-\frac{7}{3}$

5 0

3 years ago

Please help me thank you and number them

Ipatiy [6.2K]

The answer would b be d 8 I think

7 0

3 years ago