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julia-pushkina [17]
3 years ago
14

Solve the equation for -2/3x= -12 A. -18 B. -8 C. 8 D. 18

Mathematics
2 answers:
stira [4]3 years ago
6 0
D.18 is the answer for that
Olegator [25]3 years ago
5 0

Answer:

D, 18.

Step-by-step explanation:

1. We need to separate any value from x, so we must remove (-2/3) from the left side of the equation and place it on the right by dividing by (-2/3).

-2/3(x) = -12

2. We recognize that the negatives cancel each other out, so it cannot be A or B. Also, remember that dividing by a fraction or decimal will always increase the value of your answer. Therefore, the answer must be 18.

x = -12 times -3/2

x = 36/2

x = 18

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the half-life of uranium -238 is 4.5 × 10 to the power of 9 years. The half-life of uranium-238 than that of uranium-234
Anika [276]

Answer:

Uranium-234 is an isotope of uranium. In natural uranium and in uranium ore, U-234 occurs as an indirect decay product of uranium-238, but it makes up only 0.0055% (55 parts per million) of the raw uranium because its half-life of just 245,500 years is only about 1/18,000 as long as that of U-238.

Step-by-step explanation:


5 0
3 years ago
The area of the triangle below is 8.91 square inches. What is the length of the base?<br> 2.2 in
Inessa05 [86]

Answer:

8.1

Step-by-step explanation:

Area of a triangle = ½ × base × height = Area

Rearranged =

area/1/2xheight

The base is the length.

8.91/1/2x2.2

=8.1 inches

4 0
3 years ago
Cindy's truck get 16 miles per gallon plans to buy a hybrid car that gets 48 miles per gallon and cost 20,599 how many miles wha
mina [271]
So the questioon is asking for how many gass refils she would have saved so we want to find how many gas refils equals to  the cost of the car so
3.49x=20,599
divide both sides by 3.49 and get x=miles=5902 gas refils
each gass refil fo r the truck is every 16 mile so 16 times 5902 or 94432 miles
5 0
3 years ago
Prove that the roots of x2+(1-k)x+k-3=0 are real for all real values of k​
masha68 [24]

Answer:

Roots are not real

Step-by-step explanation:

To prove : The roots of x^2 +(1-k)x+k-3=0x

2

+(1−k)x+k−3=0 are real for all real values of k ?

Solution :

The roots are real when discriminant is greater than equal to zero.

i.e. b^2-4ac\geq 0b

2

−4ac≥0

The quadratic equation x^2 +(1-k)x+k-3=0x

2

+(1−k)x+k−3=0

Here, a=1, b=1-k and c=k-3

Substitute the values,

We find the discriminant,

D=(1-k)^2-4(1)(k-3)D=(1−k)

2

−4(1)(k−3)

D=1+k^2-2k-4k+12D=1+k

2

−2k−4k+12

D=k^2-6k+13D=k

2

−6k+13

D=(k-(3+2i))(k+(3+2i))D=(k−(3+2i))(k+(3+2i))

For roots to be real, D ≥ 0

But the roots are imaginary therefore the roots of the given equation are not real for any value of k.

6 0
3 years ago
Please help it is math related
soldi70 [24.7K]

Answer:

The slope is 3/8!

Explanation:

You put your equation in slope-intercept form, that is how you find the slope.

3x-8y=-20

-8y=-3x-20

y=3/8x-20

the fraction is now positive because the numerator and the denominator are both negative so they are changed into positive.

Hope This Helps!

4 0
3 years ago
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