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3 years ago

Please help ASAP work and answer please please

Mathematics

1 answer:

tiny-mole [99]3 years ago

5 0

Answers:

8. 12x+3= 27

move 3 to the other side , sign changes from +3 to -3

12x+3-3=27-3

12x= 24

divide by 12 for both sides

12x/12=24/12

x= 2

9. 6(y-10)=42

mutiply the bracket by 6

(6)(y)(6)(-10)= 6y-60

6y-60= 42

move -60 to the other side, sign changes from -60 to +60

6y-60+60=42+60

6y=102

divide by 6

6y/6=102/6

y=17

10. 9x-2=4x+13

move +4x to the other side

9x-4x-2= 4x-4x+13

5x-2= 13

5x-2+2= 13+2

5x= 15

divide by 5 for both sides

5x/5= 15/5

x= 3

11. 4/3y+5/2y= 1

find the common denominator for both of the fractions which is 6.

Mutiply by 2 for 4/3y .

4(2)/3(2)y=8/6y

Mutiply by 3 for 5/2y

5(3)/2(3y)= 15/6y

8/6y+15/6y= 23/6y

23/6y= 1

Mutiply both sides by 6/23

23/6y(6/23)= 1 (6/23)

Cross out 6 and 6 , divide by 6 and then becomes 1.

Cross out 23 and 23, divide by 23 and then becomes 1.

1*1*y=y

y= 6/23

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Is relation t a function? Is the inverse of relation t a function?

bonufazy [111]

Answer:

Hence, Relation t is a function. The inverse of relation t is a function.

Step-by-step explanation:

We are given the relation as:

x: 0 , 2 , 4 , 6

y: -10 , -1 , 4 , 8

Clearly from the y-values corresponding to the x-values we could see that each x has a single image (single y-value).

Hence, the corresponding relation is a function.

Now we have to find whether the inverse of this relation is a function or not.

When we take the inverse of this function that is the y-values will behave as a pre-image and x-values as its image.

Hence we will see that corresponding to each y-value there is a unique image hence the inverse relation is also a function.

Hence, Relation t is a function. The inverse of relation t is a function.

3 0

3 years ago

Read 2 more answers

Ben said the graph of the inequality

Ksivusya [100]

Answer:

Ben is incorrect

The solutions are x>-3 and x < 3

Step-by-step explanation:

we have

$-x^{2} +9 > 0$

Multiply by -1 both sides

$x^{2} -9 < 0$

Adds 9 both sides

$x^{2} < 9$

The solutions are

$x < 3$

and

$-x < 3$ -----> Multiply by -1 both sides ----> $x > -3$

therefore

The solution is the interval (-3,3)

Ben is incorrect

7 0

3 years ago

Geometry ! No links please

DENIUS [597]

The Answer Is CD=5

AB/BC=2

2=BE/CD

2/1=10/X

2/1=10/5

2=2

6 0

3 years ago

Use the Venn diagram to calculate probabilities. Circles A and B overlap. Circle A contains 15, circle B contains 10, and the in

grin007 [14]

Answer:

$(B) P(B)=\dfrac{16}{31}$

Step-by-step explanation:

From the Venn diagram, n(AUB)=31

The given probabilities in the option are calculated below:

$P(A)=\dfrac{21}{31}\\\\P(B)=\dfrac{16}{31}\\\\P(A|B)=\dfrac{P(A\cap B)}{P(B)}= \dfrac{6/31}{16/31} =\dfrac{6}{16}=\dfrac{3}{8} \\\\P(B|A)=\dfrac{P(B\cap A)}{P(A)}= \dfrac{6/31}{21/31} =\dfrac{6}{21}=\dfrac{3}{7}$

The only correct option is the probability of B which is $\frac{16}{31}$

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3 years ago

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avanturin [10]

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3 years ago