All categories

3 years ago

What is the answer to x=3x-20 and y=-6+30

Mathematics

1 answer:

NISA [10]3 years ago

5 0

Answer:y=24 and x=10

Step-by-step explanation:

To find the value of x and y in x=3x-20 and y=-6+30

We gets

y=24

To get x

x=3x-20

Collect like terms

x-3x=-20

Divide through by -2

-2x=-20

x=10

You might be interested in

a golf ball is hit down a fairway. the golfer relates the time passed to the height of the ball. is this a function?

andre [41]

Yes, considering that the ball cannot have two different heights at the same exact time.

7 0

3 years ago

If FGH QRS, find the measure of Q A. 9 B.28 C.11 D.34

Ne4ueva [31]

Answer:

28°

Step-by-step explanation:

Since ΔFHG ≅ QRC, the corresponding angles are congruent:

∠F = ∠Q
(2x + 10)° = (3x + 1)°
3x - 2x = (10 - 1)°
x = 9°

Then, the angle Q measures:

∠Q = (3*9 + 1)° = 28°

6 0

3 years ago

Th inequality 6-2/3x9 B. x<9 C. x>-3/5 D. x<-3/5

Delvig [45]

Is there an a choice ? Or are those all the choices, just making sure

5 0

3 years ago

Robert was using a bucket that holds 243.5 ounces (oz) of water when full. The bucket loses 0.3 oz of water per second. After 20

Assoli18 [71]

Answer:

237.5

Step-by-step explanation:

Since the bucket loses 0.3 oz of water per second, and 20 secs passed, you would have to multiply 0.3 and 20 to find out how much water leaked. And 0.3x20=6, so then, we would have to subtract 234.5 and 6 to find out how much water there is left in the bucket. 234.5-6=237.5

So 237.5 is your answer

4 0

3 years ago

1.Arsenic-74 is used to locate brain tumors. It has a half-life of 17.5 days. 90 mg wereused in a procedure. Write an equation t

Sophie [7]

$1)\text{ }N_t\text{ = 90\lparen}\frac{1}{2})^{\frac{t}{17.5}}$

2) 5.625 mg will be left

Explanation:

1) Half-life = 17.5 days

initial amount of Arsenic-74 = 90 mg

To get the equation, we will use the equation of half-life:

$\begin{gathered} N_t\text{ = N}_0(\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}} \\ where\text{ N}_t\text{ = amount remaining} \\ N_0\text{ = initial amount} \\ t_{\frac{1}{2}\text{ }}\text{ = half-life} \end{gathered}$ $N_t\text{ = 90\lparen}\frac{1}{2})^{\frac{t}{17.5}}$

2) we need to find the remaining amount of Arsenic-74 after 70 days

t = 70

$\begin{gathered} N_t=\text{ 90\lparen}\frac{1}{2})^{\frac{70}{17.5}} \\ N_t\text{ = 5.625 mg} \end{gathered}$

So after 70 days, 5.625 mg will be left

4 0

1 year ago