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vichka [17]
3 years ago
5

What is the greatest common factor of 55 , 99​

Mathematics
1 answer:
coldgirl [10]3 years ago
5 0
The greatest common factor is 11
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X2-y2-x-y anyone help to solve this matter <br>​
Olegator [25]

Answer:

(x+y)(x-y-1)

Step-by-step explanation:

x²-y²-x-y= (x+y)(x-y)-(x+y)= (x+y)(x-y-1)

4 0
3 years ago
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What is this number written in standard form 3×1,000,00+3×100,00+2×100
pav-90 [236]
3(1,000,000)=3,000,000
3(100,000)=300,000
2(100)=200
3,300,200
6 0
3 years ago
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To convert Celsius temperatures to Fahrenheit, you can use the expression (C + 32). Tom says that another way to write this expr
Alexxandr [17]

Answer:

C) Tom did not distribute to both terms in parentheses.

Step-by-step explanation:

Addition within a paranthesis has a distributive property to the multiplier outside the paranthesis. Ignoring this will lead to a wrong value for the operation.

5 0
3 years ago
In ΔRST, t = 4.1 inches, r = 7.1 inches and ∠S=45°. Find the length of s, to the nearest 10th of an inch.
dem82 [27]

Answer:

The length of s is 5.1 inches to the nearest tenth of an inch

Step-by-step explanation:

In Δ RST

∵ t is the opposite side to ∠T

∵ r is the opposite side to ∠R

∵ s is the opposite side to ∠S

→ To find s let us use the cosine rule

∴ s² = t² + r² - 2 × t × r × cos∠S

∵ t = 4.1 inches, r = 7.1 inches, and m∠S = 45°

→ Substitute them in the rule above

∴ s² = (4.1)² + (7.1)² - 2 × 4.1 × 7.1 × cos(45°)

∴ s² = 16.81 + 50.41 - 41.1677568

∴ s² = 26.0522432

→ Take √ for both sides

∴ s = 5.10413981

→ Round it to the nearest tenth

∴ s = 5.1 inches

∴ The length of s is 5.1 inches to the nearest tenth of an inch

3 0
3 years ago
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Find the angle between the given vectors to the nearest tenth of a degree. u = , v = (2 points)
Liono4ka [1.6K]

Answer:

<h2>3.6°</h2>

Step-by-step explanation:

The question is incomplete. Here is the complete question.

Find the angle between the given vectors to the nearest tenth of a degree.

u = <8, 7>, v = <9, 7>

we will be using the formula below to calculate the angle between the two vectors;

u*v = |u||v| cos \theta

\theta is the angle between the two vectors.

u = 8i + 7j and v = 9i+7j

u*v = (8i + 7j )*(9i + 7j )

u*v = 8(9) + 7(7)

u*v = 72+49

u*v = 121

|u| = √8²+7²

|u| = √64+49

|u| = √113

|v| = √9²+7²

|v| = √81+49

|v| = √130

Substituting the values into the formula;

121= √113*√130 cos θ

cos θ = 121/121.20

cos θ = 0.998

θ = cos⁻¹0.998

θ = 3.6° (to nearest tenth)

Hence, the angle between the given vectors is 3.6°

5 0
3 years ago
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