MATHEMATICS HIGH SCHOOL

URGENT!! PLEASE help and don’t link files!!! Find Surface Area of triangular shown.
(If you can please explain, thank you!!)
URGENT!! PLEASE help and don’t link files!!! Find Surface Area - 1

Answers

Answer 1
Answer:

Answer:336cm cube

Step-by-step explanation:

12x4x7 i think im  justlearning this to

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Someone tell me What 2x2 is

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Answer:

y=x+2

Step-by-step explanation:

1cm x ¼ ??????!????????????

Answers

Answer:

2.5 millimeteers

Step-by-step explanation:

Pls brainlest

Two sides of a right triangle have the lengths 4 and 5. What is the product of the possible lengths of the third side? Express the product as a decimal rounded to the nearest tenth.

Answers

Answer:

19.2

Step-by-step explanation:

1st Case:

4 and 5 are legs of the right triangle.

Using the pythagorean therom: a^2+b^2=c^2

We can say that 4^2+5^2=x^2

16+25=x^2

41=x^2

x=√41

√41 is about 6.4

x=6.4

2nd Case

5 is the hypotenuse of the right triangle and 4 is the legs.

Using the pythagorean therom: a^2+b^2=c^2

We can say that 4^2+x^2=5^2

16+x^2=25

x^2=9

x=3

Final Step

We need to multiply the two possible lengths for x. So for case 1 the length of x was 6.4 and for case two the length was 3. 6.4*3=19.2

Anwser: 19.2

Given cos alpha = 8/17, alpha in quadrant IV, and sin beta = -24/25, beta in quadrant III, find sin(alpha-beta)

Answers

Given \cos\alpha=(8)/(17), \alpha is in Quadrant IV,  \sin\beta=-(24)/(25), and \beta is in Quadrant III, find \sin(\alpha-\beta)

We can use the angle subtraction formula of sine to answer this question.

\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta

We already know that \cos\alpha=(8)/(17).

We can use the Pythagorean identity \sin^2\theta+\cos^2\theta=1 to find \sin\alpha.

\sin^2\alpha+((8)/(17))^2=1 \n \sin^2\alpha+(64)/(289)=1 \n \sin^2\alpha=(225)/(289) \n \n\sin\alpha=\pm(15)/(17)

Since \alpha is in Quadrant IV, and sine is represented as y value on the unit circle, we must assume the negative value \sin\alpha=-(15)/(17).

As similar process is then done with  \sin\beta=-(24)/(25).

(-(24)/(25))^2+\cos^2\beta=1 \n (576)/(625)+\cos^2\beta=1 \n \cos^2\beta=(49)/(625) \n \n\cos\beta=\pm(7)/(25)

And since \beta is in Quadrant III, and cosine in represented as x value on the unit cercle, we must assume the negative value \cos\beta=-(7)/(25).

Now we can fill in our angle subtraction formula!

\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta \n\n \sin(\alpha-\beta)=(-(15)/(17)*-(7)/(25))-((8)/(17)*-(24)/(25)) \n\n\sin(\alpha-\beta)=(105)/(425)-(-(192)/(425)) \n\n \boxed{\sin(\alpha-\beta)=(297)/(425)}

What is the perimeter of a triangle that has two sides measuring 7 centimeters and a third side measuring 9 centimeters?

Answers

The answer is 23cm just add the sides

The perimeter is the sum of all of the lengths of the sides. To find the perimeter, add together the length of each side.

For this triangle, our side lengths are 7, 7, and  9.

7 + 7 = 14

14 + 9 = 23

The perimeter of a triangle that has two sides measuring 7 centimeters and a third side measuring 9 centimeters is 23 centimeters.

Hope this helps!! :)

Evaluate the expression for x = 4, y = 1:

x + 12
y + 3

Answers

16 on the top and 4 on the bottom
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