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Mademuasel [1]
3 years ago
6

The hypotenuse is NOT located across from the right angle True or false

Mathematics
2 answers:
azamat3 years ago
6 0

The statement, The hypotenuse is NOT located across from the right angle is false.

<h3>What is a right angled triangle? </h3>

A right-angled triangle is a type of triangle that has an angle that measures 90 degrees. The triangle has three sides. The hypotenuse is the longest side of a right-angled triangle. Other sides are the base and the length. The sum of sides in a a right-angled triangle is 180 degrees.

Please find attached an image of a right-angled triangle. To learn more about  a triangle, please check: brainly.com/question/9329354

Bingel [31]3 years ago
6 0

In a right triangle, the hypotenuse is the longest side, an "opposite" side is the one across from a given angle, and an "adjacent" side is next to a given angle. We use special words to describe the sides of right triangles. The hypotenuse of a right triangle is always the side opposite the right angle.

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There are 527 pencils,646 erasers and 748 sharpeners. These are to be put in separate packets containing the same number of item
sp2606 [1]

Answer:

  • 31 pencils
  • 38 erasers
  • 44 sharpeners

Step-by-step explanation:

The number of packets is the greatest common divisor of the given numbers of pencils, erasers, and sharpeners.

It can be helpful to look at the differences between these numbers:

  748 -646 = 102

  646 -527 = 119

The difference of these differences is 17, suggesting that will be the number of packets possible.

  527 = 17 × 31

  646 = 17 × 38

  748 = 17 × 44

The numbers 31, 38, and 44 are relatively prime (31 is actually prime), so there can be no greater number of packets than 17.

There will be 31 pencils, 38 erasers, and 44 sharpeners in each of the 17 packets.

_____

We may have worked the wrong problem. The way it is worded, the <em>maximum</em> number of items in each packet will be 527 pencils, 646 erasers, and 748 sharpeners in one (1) packet. The <em>minimum</em> number of items in each packet will be the number that corresponds to the maximum number of packets. Since 17 is the maximum number of packets, each packet's contents are as described above.

17 is the only common factor of the given numbers, so will be the number of groups (plural) into which the items can be arranged.

6 0
3 years ago
The product of 2 numbers is -24 and their sum is -5. What are the two numbers and why are they the two numbers?
lana [24]

The numbers are (-8, 3) or (3, -8).

Step-by-step explanation:

  • Step 1: Given the product of the numbers are -24 and their sum are -5. Let the numbers be a and b. Form equations out of these details.

⇒ a + b = -5 ⇒ b = -5 - a

⇒ a × b = -24 -------- (1)

  • Step 2: Substitute the value of b in eq(1)

⇒ a × (-5 - a) = - 24

⇒ -5a - a² = -24

⇒ a² + 5a - 24 = 0

  • Step 3: Solve the quadratic equation for a.

a = (-5 ± √25 - 4 × 1 × -24)/2

  = (-5 ± √121)/2

  = - 16/2, 6/2 = -8 or 3

  • Step 4: For each value of a, find b.

When a = -8, b = -5 +8 = 3

When a = 3, b = -5 - 3 = -8.

6 0
3 years ago
Math question. image shown below
marin [14]

Answer:

The third one is the only one that is not a function.

Step-by-step explanation:


3 0
3 years ago
What's greater then 3/8 or 2/4
iVinArrow [24]
Convert 2/4 to 4/8 and then look at the numerators. since 4 is greater than 3, 2/4 is greater than 3/8
7 0
3 years ago
Read 2 more answers
David found and factored out the GCF of the polynomial 80b4 – 32b2c3 + 48b4c. His work is below. GFC of 80, 32, and 48: 16 GCF o
BlackZzzverrR [31]

Answer: The correct statements are

The GCF of the coefficients is correct.

The variable c is not common to all terms, so a power of c should not have been factored out.

David applied the distributive property.

Step-by-step explanation:

GCF = Greatest common factor

1) GCF of coefficients : (80,32,48)

80 = 2 × 2 × 2 × 2 × 5

32 = 2 × 2 × 2 × 2 × 2

48 = 2 × 2 × 2 × 2 × 3

GCF of coefficients : (80,32,48) is 16.

2) GCF of variables :(b^4,b^2,b^4)

b^4= b × b × b × b

b^2 = b × b

b^4 =b × b × b × b

GCF of variables :(b^4,b^2,b^4) is b^2

3) GCF of c^3 and c: c is not the GCF of the polynomial. The variable c is not common to all terms, so a power of c should not have been factored out.

4) 80b^4-32b^2c^3+48b^4c

=16b^2(5b^2-2c^3+3b^2c)

David applied the distributive property.

7 0
3 years ago
Read 3 more answers
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