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tekilochka [14]

3 years ago

9

Nicola is multiplying 1/6 by 3/2. Use the drop down menu to complete the statement

2 answers:

cestrela7 [59]3 years ago

8 0

Answer:

Greater than

Step-by-step explanation:

sertanlavr [38]3 years ago

6 0

Answer:

Step-by-step explanation:

1/6 x 3/2 = 1/4

Hope it helps

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Line a is parallel to line b, m 6=85 and m 3=2x. Find m 1.

FromTheMoon [43]

The sum of the angles <6 and <3 must be 180. Since <3 and <1 are opposed to the top, their measure are equal. We need then to calculate the measure of angle >3 = 180 - <6 = 180-85=95

m(<1) = 95

8 0

3 years ago

Find the value of x , if log₂ [log₅ (log₃x)] = 0

Lina20 [59]

$\log_2 (\log_5( \log_3x)) = 0\\
D:x>0 \wedge \log_3x>0 \wedge \log_5(\log_3x)>0\\
D:x>0\wedge x>1 \wedge\log_3x>1\\
D:x>1 \wedge x>3\\
D:x>3\\
\log_5(\log_3x)=1\\
\log_3x=5\\
x=3^5\\
x=243$

3 0

3 years ago

One side of triangle is 4cm longer than the shortest side and 2cm shorter than the longest side the perimeter is 38cm what are t

o-na [289]

9.3, 13.3, 15.3
Hope it helps!

7 0

2 years ago

Read 2 more answers

can someone please tell me how to do 2x=3y and 4x-3y=12 using substitution to solve the system of equations.

maw [93]

Answer:

Step-by-step explanation:

you divide the two to the y, which is x=3/2y

then, you simplify the second equation to y, which is -3y= -4x+12

then, you substitute the x=3/2y into the second equation,

-3y = -6+12

-3y = 6

y = -2

Substitute the y into the first equation, x= 3/2y,

x = -3

5 0

3 years ago

Substitute the values for a, b, and c into b2 – 4ac to determine the discriminant. Which quadratic equations will have two real

hoa [83]

Answer: $0=2x^2-7x-9\\0=4x^2-3x-1\\0=x^2-2x-8$

Step-by-step explanation:

We know that the standard quadratic equation is ax^2+bx+c=0

Let's compare all the given equation to it and , find discriminant.

1. a=2, b= -7, c=-9

$b^2-4ac=(-7)^2-4(2)(-9)=49+72>0$

So it has 2 real number solutions.

2. a=1, b=-4, c=4

$b^2-4ac=(-4)^2-4(1)(4)=16-16=0$

So it has only 1 real number solution.

3. a=4, b=-3, c=-1

$b^2-4ac=(-3)^2-4(4)(-1)=9+16=25>0$

So it has 2 real number solutions.

4. a=1, b=-2, c=-8

$b^2-4ac=(-2)^2-4(1)(-8)=4+32=36>0$

So it has 2 real number solutions.

5. a=3, b=5, c=3

$b^2-4ac=(5)^2-4(3)(3)=25-36=-9$

Thus it does not has real solutions.

5 0

3 years ago

Read 2 more answers