MATHEMATICS COLLEGE

Is 1.3 greater than negative 5

Answers

Answer 1

Answer: 1.3 would be considered greater than negative 5. Although 1.3 is a decimal it is still a positive number and positives are worth more than negatives regardless

Answer 2

Answer:

Yes it is jndnscjoixglcogkg oyvyvh

Related Questions

1 3/7+2 2/6 the answer
I need help question 1
7 is what percent of 8
Prove the trigonometric identity(tan x + cot x)/(csc x * cos x) = sec^2 x
CE is an angle bisector of angle ACB.If m angle ACE = 2x - 15 and m angle ECB = x +5,determine the value of x.

Which number produces an irrational number when added to? 3\4

Answers

Answer:

π

Step-by-step explanation:

Any other irrational number, actually, so √2 would also do the trick.

Answer:

Pi, since pi is irrational

Step-by-step explanation:

Any irrational number, such as √2

Which expression represents "6 more than x"?
x - 6
6x
x + 6
6 - x

Answers

x+6
this is because + means more, so 6 more

An animal gained 8 kilograms over 32 years.what is the unit rate per each year

Answers

Answer:

0.25 kilograms per 1 year

Step-by-step explanation:

8/32 since ur trying to find the unit rate per each year

PLZZZ HELP FAST What is the product of Three-fourths and Negative StartFraction 6 over 7 EndFraction? Negative StartFraction 7 over 8 EndFraction Negative StartFraction 9 over 14 EndFraction StartFraction 9 over 14 EndFraction StartFraction 7 over 8 EndFraction

Answers

Answer:

-9/14 so the answer is B

Step-by-step explanation:

Hope this helps

Hey can you please help me posted picture of question

Answers

it would be false i believe so and if it is wrong i am sorry

Find the perimeter of WXYZ. Round to the nearest tenth if necessary.

Answers

Answer:

C. 15.6

Step-by-step explanation:

Perimeter of WXYZ = WX + XY + YZ + ZW

Use the distance formula, $d = √((x_2 - x_1)^2 + (y_2 - y_1)^2)$ to calculate the length of each segment.

✔️Distance between W(-1, 1) and X(1, 2):

Let,

$W(-1, 1) = (x_1, y_1)$

$X(1, 2) = (x_2, y_2)$

Plug in the values

$WX = √((1 - (-1))^2 + (2 - 1)^2)$

$WX = √((2)^2 + (1)^2)$

$WX = √(4 + 1)$

$WX = √(5)$

$WX = 2.24$

✔️Distance between X(1, 2) and Y(2, -4)

Let,

$X(1, 2) = (x_1, y_1)$

$Y(2, -4) = (x_2, y_2)$

Plug in the values

$XY = √((2 - 1)^2 + (-4 - 2)^2)$

$XY = √((1)^2 + (-6)^2)$

$XY = √(1 + 36)$

$XY = √(37)$

$XY = 6.08$

✔️Distance between Y(2, -4) and Z(-2, -1)

Let,

$Y(2, -4) = (x_1, y_1)$

$Z(-2, -1) = (x_2, y_2)$

Plug in the values

$YZ = √((-2 - 2)^2 + (-1 -(-4))^2)$

$YZ = √((-4)^2 + (3)^2)$

$YZ = √(16 + 9)$

$YZ = √(25)$

$YZ = 5$

✔️Distance between Z(-2, -1) and W(-1, 1)

Let,

$Z(-2, -1) = (x_1, y_1)$

$W(-1, 1) = (x_2, y_2)$

Plug in the values

$ZW = √((-1 -(-2))^2 + (1 - (-1))^2)$

$ZW = √((1)^2 + (2)^2)$

$ZW = √(1 + 4)$

$ZW = √(5)$

$ZW = 2.24$

✅Perimeter = 2.24 + 6.08 + 5 + 2.24 = 15.56

≈ 15.6

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Step-by-step explanation:

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