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3 years ago

Is 1.65 greather than 16.5

Mathematics

2 answers:

Bas_tet [7]3 years ago

5 0

Answer:

Step-by-step explanation:

nadezda [96]3 years ago

5 0

Answer: No , 16.5 is greater though than 1.65 .

Step-by-step explanation:

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The given lengths are two sides of a right triangle. All three side lengths of the triangle are integers and together form a pyt

Leni [432]

If all three side lengths are INTEGERS and form PYTHAGOREAN TRIPLE then we know decimals are not allowed

if 24 and 51 are both legs then
$24^{2} + 51^{2} = x^{2}$
x = 3177
and squre root of that gives us 56.364.... NOT an integer

So that means one of them is not a leg and since hypotenuse is longer than both legs, 51 must be the hypotenuse so to get the third side

$51^{2} - 24^{2} = x^{2}[tex] [tex] x^{2}$ =2025
x=45, leg B

7 0

3 years ago

Please help me on this problem

Anna71 [15]

Answer:

The pairs of integer having two real solution for $ax^{2} -6x+c = 0$ are

$a = -4, c = 5$
$a = 1, c = 6$
$a = 2, c = 3$
$a = 3, c = 3$

Step-by-step explanation:

Given

$ax^{2} -6x+c = 0$

Now we will solve the equation by putting all the 6 pairs so we get the following

$-3x^{2} -6x-5 = 0$ for $a = -3 , c=-5$

$-4x^{2} -6x+5 = 0$ for $a = -4 , c=5$

$1x^{2} -6x+6 = 0$ for $a = 1 , c=6$

$2x^{2} -6x+3 = 0$ for $a = 2 , c=3$

$3x^{2} -6x+3 = 0$ for $a = 3 , c=3$

$5x^{2} -6x+4 = 0$ for $a = 5 , c=4$

The above all are Quadratic equations inn general form $ax^{2} +bx+c=0$

where we have a,b and c constant values

So for a real Solution we must have

$Disciminant , b^{2} -4\timesa\timesc \geq 0$

for $a = -3 , c=-5$ we have

$Discriminant =-24$ which is less than 0 ∴ not a real solution.

for $a = -4 , c=5$ we have

$Discriminant = 116$ which is greater than 0 ∴ a real solution.

for $a = 1 , c=6$ we have

$Discriminant =12$ which is greater than 0 ∴ a real solution.

for $a = 2 , c=3$ we have

$Discriminant =12$ which is greater than 0 ∴ a real solution.

for $a = 3 , c=3$ we have

$Discriminant =0$ which is equal to 0 ∴ a real solution.

for $a = 5 , c=4$ we have

$Discriminant =-44$ which is less than 0 ∴ not a real solution.

7 0

3 years ago

HELP QUICKLY what is the equation of the line through (-10,-7) and (-5,-9)

hichkok12 [17]

This is your answer

7 0

3 years ago

Read 2 more answers

1. Trapezoid BEAR has a height of 8.5 centimeters and parallel bases that measure 6.5 centimeters and 11.5 centimeters. To the n

e-lub [12.9K]

<h3>Given</h3>

1) Trapezoid BEAR with bases 11.5 and 6.5 and height 8.5, all in cm.

2) Regular pentagon PENTA with side lengths 9 m

The area of each figure, rounded to the nearest integer

<h3>Solution</h3>

1) The area of a trapezoid is given by

... A = (1/2)(b1 +b2)h

... A = (1/2)(11.5 +6.5)·(8.5) = 76.5 ≈ 77

The area of BEAR is about 77 cm².

2) The conventional formula for the area of a regular polygon makes use of its perimeter and the length of the apothem. For an n-sided polygon with side length s, the perimeter is p = n·s. The length of the apothem is found using trigonometry to be a = (s/2)/tan(180°/n). Then the area is ...

... A = (1/2)ap

... A = (1/2)(s/(2tan(180°/n)))(ns)

... A = (n/4)s²/tan(180°/n)

We have a polygon with s=9 and n=5, so its area is

... A = (5/4)·9²/tan(36°) ≈ 139.36

The area of PENTA is about 139 m².

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3 years ago

A teacher interested in determining the effect of a new computer program on learning to read conducted a study. One hundred stud

djyliett [7]

Answer:2

Step-by-step explanation:

4 0

2 years ago