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goblinko [34]
3 years ago
15

What is 38.95 rounded to the nearest tenth

Mathematics
1 answer:
Degger [83]3 years ago
4 0
39. because its close to it lol. 
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PLEASE HELP MEEEEEEEE ILL MARK YOU BRAINLIEST
Andre45 [30]

Answer:

16.8

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
A cylinder and a cone have the same volume. The cylinder has radius x and height y. The cone has radius 2x. Find the height of t
weqwewe [10]
V(cylinder)=πR²H
Radius of the cylinder R=x, height of the cylinder H=y.
We can write for the cylinder
V(cylinder)=πx²y

V(cone) =(1/3)πr²h
Radius of the cone r=2x.
We can write for the cone
V(cone)= (1/3)π(2x)²h=(1/3)π *4*x²h

V(cylinder) =V(cone)
πx²y=(1/3)π *4*x²h
y=(4/3)*h
h=(3/4)*y


7 0
3 years ago
Brody bought 17 chicken wings for $37.40. What's the unit cost of one wing?
lana [24]

<u>2.2</u>

Question:

Brody bought 17 chicken wings for $37.40. What's the unit cost of one wing?

Solution given:

17 chicken wings = $37.40

1chicken wings =$37.40/17*1=$2.2

<u>2</u><u>.</u><u>2</u>

7 0
3 years ago
Whats the answer for number 13?
Mumz [18]

Answer: option 3

Step-by-step explanation:

so you would do

-2=3(1)-5

-2=3-5

-2=-2

3 0
2 years ago
Use the diagram to complete the statement. Triangle J K L is shown. Angle K J L is a right angle. Angle J K L is 52 degrees and
zzz [600]

Answer:

\bold{sin(38^\circ)=cos(52^\circ)}

Step-by-step explanation:

Given that \triangle KJL is a right angled triangle.

\angle JKL = 52^\circ\\\angle KLJ = 38^\circ

and

\angle KJL = 90^\circ

Kindly refer to the attached image of \triangle KJL in which all the given angles are shown.

To find:

sin(38°) = ?

a) cos(38°)

b) cos(52°)

c) tan(38°)

d) tan(52°)

Solution:

Let us use the trigonometric identities in the given \triangle KJL.

We have to find the value of sin(38°).

We know that sine trigonometric identity is given as:

sin\theta =\dfrac{Perpendicular}{Hypotenuse}

sin(\angle JLK) = \dfrac{JK}{KL}\\OR\\sin(38^\circ) = \dfrac{JK}{KL}....... (1)

Now, let us find out the values of trigonometric functions given in options one by one:

cos\theta =\dfrac{Base}{Hypotenuse}

cos(\angle JLK) = \dfrac{JL}{KL}\\OR\\cos(38^\circ) = \dfrac{JL}{KL}....... (2)

By (1) and (2):

sin(38°) \neq cos(38°).

cos(\angle JKL) = \dfrac{JK}{KL}\\OR\\cos(52^\circ) = \dfrac{JK}{KL} ...... (3)

Comparing equations (1) and (3):

we get the both are same.

\therefore \bold{sin(38^\circ)=cos(52^\circ)}

6 0
3 years ago
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