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2 years ago

Solve for y -2x + 5y - 6 = -11 HELP ASAPPP

Mathematics

2 answers:

Leona [35]2 years ago

5 0

Answer:

It would be

y=2x/5 - 1

7nadin3 [17]2 years ago

4 0

Answer:

y= 2 /5 x − 1

Step-by-step explanation:

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The scale from a playground to a scale drawing of the playground is 4 meters per centimeter.The length of the drawing of the pla

Scilla [17]

Answer:

322.24 meters

Step-by-step explanation:

The scale is 4 meter per centimeter. Multiply 5.3 by 4 to get 21.2 meters. Multiply 5.3 by 4 to get 15.2 meters. Multiply 21.2 by 15.2 to get 322.24

6 0

2 years ago

-3/4x>12 solve for x

Ad libitum [116K]

Answer:

$\Huge\boxed{\mathsf{\rightarrow X$

Step-by-step explanation:

$\Large\boxed{\mathsf{SUBJECT: MATH}}$

Isolate x on one side of the equation.

Find the value of x.

First, multiply -1 from both sides.

(-3/4x)(-1)<12(-1)

Solve.

12(-1)=-12

3/4x<-12

Next, multiply 4 from both sides.

4*3/4x<4(-12)

Solve.

Multiply the numbers from left to right.

4(-12)=-48

3x<-48

Then, divide by 3 from both sides.

3x/3<-48/3

Solve.

Divide the numbers from left to right.

-48/3=-16

X<-16

$\Large\boxed{\Rightarrow \mathsf{X$

The correct answer is x<-16.

6 0

2 years ago

Q5 Q3.) How far, to the nearest foot, is the ship from the statue's base?

dusya [7]

Distance from base = 305 / tan(29.5deg)
= 539 ft

7 0

2 years ago

On a school bus, 22 of the 40 students are in window seats. What fraction

Gnoma [55]

Answer:

11/20

Step-by-step explanation:

simplify 22/40

8 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