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2 years ago

A^2 = 4 and B^2 = 3. What is C^2 × 15?

Mathematics

2 answers:

adoni [48]2 years ago

7 0

Answer:

105

Step-by-step explanation:

Assuming A^2 + B^2 = C^2:

C^2 = 4 + 3 = 7

So,

C^2 * 15 = 49(15) = 105

Note: Since the relationship between A, B, and C isn't known, it cannot be said for certain that 105 is the answer.

LenaWriter [7]2 years ago

5 0

Step-by-step explanation:

there is no C in the first 2 equations. There is no A or B in the last expression. I do not think you can solve this.

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1.Which is an equivalent form of the equation 3x-3y=15?

defon

1) The given equation is 3x-3y=15

To solve for y we subtract 3x both sides:

3x-3x-3y=-3x+15

-3y=-3x+15

With y term -3 is multiplied .We perform the opposite operation both sides that is divide by -3 both sides we have:

y=x-5

Option A :y=x-5 is the right answer.

2) |2x-1|=3

We can remove the absolute sign by forming equations taking both the positive and negative signs:

2x-1=3 or 2x-1=-3

Solving the two equations for x:

2x=3+1 or 2x=-3+1

2x=4 ,x=2. 2x=-2 , x=-1 .

Option a)x=2 or x=-1 is the right answer.

3) 2< 3x-1 ≤ 5

Adding 1 to all the sides we have:

3<3x≤ 6

Dividing by 3 :

1<x≤ 2.

8 0

3 years ago

Read 2 more answers

The perimeter of a flower garden with three sides is 46 feet. One side of the garden is 11 feet long. Another side of the garden

saul85 [17]

Answer:

Step-by-step explanation:

Step one:

given that the orientation of the garden is a triangular shape(it has three sides)

the perimeter is given as=46ft

one side y=11ft

the second side z= 13ft

let the third side be xft

Step two:

the expression for the perimeter = z+y+x

46=11+13+x

46=24+x

solve for x

x=46-24

x=22ft

Therefore the third side is 22ft long

7 0

2 years ago

Trigonometric ratios in right triangles pls help

Ket [755]

The answer is 4/3 if I am not mistaken

8 0

3 years ago

How many solutions does the equation have, x1 + x2 + x3 = 10 , where x1 , x2, and x3 are non-negative integers?

MAVERICK [17]

Non-negative integers are positive integers or zero.

1. When $x_1=0,$ then there are such possible cases for $x_2$ and $x_3:$

$x_2=0,\ x_3=10;$
$x_2=1,\ x_3=9;$
$x_2=2,\ x_3=8;$
$x_2=3,\ x_3=7;$
$x_2=4,\ x_3=6;$
$x_2=5,\ x_3=5;$
$x_2=6,\ x_3=4;$
$x_2=7,\ x_3=3;$
$x_2=8,\ x_3=2;$
$x_2=9,\ x_3=1;$
$x_2=10,\ x_3=0.$

In total 11 solutions for $x_1=0.$

2. For $x_1=1,$ there are such possible cases for $x_2$ and $x_3:$

$x_2=0,\ x_3=9;$
$x_2=1,\ x_3=8;$
$x_2=2,\ x_3=7;$
$x_2=3,\ x_3=6;$
$x_2=4,\ x_3=5;$
$x_2=5,\ x_3=4;$
$x_2=6,\ x_3=3;$
$x_2=7,\ x_3=2;$
$x_2=8,\ x_3=1;$
$x_2=9,\ x_3=0.$

In total 10 solutions for $x_1=1.$

3. This process gives you

for $x_1=2$ - 9 solutions;
for $x_1=3$ - 8 solutions;
for $x_1=4$ - 7 solutions;
for $x_1=5$ - 6 solutions;
for $x_1=6$ - 5 solutions;
for $x_1=7$ - 4 solutions;
for $x_1=8$ - 3 solutions;
for $x_1=9$ - 2 solutions;
for $x_1=10$ - 1 solution.

4. Add all numbers of solutions:

$11+10+9+8+7+6+5+4+3+2+1=66.$

Answer: there are 66 possible solutions (with non-negative integer variables)

8 0

3 years ago

Read 2 more answers

in a family of cats,there are 2 uncool cats for every 3 coolcats. if these cats have 120 paws together,how many cool cats are th

rjkz [21]

One cat has four paws, so there are 120/4=30 cats. If a "set" of cats is two uncool cats and three cool cats, there are five cats in a set, so there are 30/5=6 sets. Since each of the six sets contains three cool cats, there are 18 cool cats in the family.

7 0

3 years ago