Simplify 3a x 4, thank youuuu

If the point (a,3) lies on the graph of the equation 5x + y = 8 then a =?

Contact [7]

Answer:

a = 1

Step-by-step explanation:

Given the point (a, 3) is on the line: 5x + y = 8.

Any point on the line satisfies the equation.

So, (a, 3) should satisfy the equation 5x + y = 8

Substituting x = a and y = 3, we get:

5(a) + 3 = 8

$\implies 5a = 5$

⇒ a = 1

Therefore, (1, 3) is the point on the line.

3 0

3 years ago

Please help me with this!!!!!

KengaRu [80]

-1 is the first answer and for #2 is the 4th answer.

Please let me know if you need additional help in private messages
Thanks,

-GeVontae The BRAINLY Team.

4 0

3 years ago

Read 2 more answers

Instructions: Write the explicit rule. Remember to simplify and do not put spaces in your answer. Sequence: 30,0,−30,−60,… 30 ,

kodGreya [7K]

Answer:

a(n)=60-30n

Step-by-step explanation:

Equation to use:

a(n) = a(1) + d(n-1)

where a(1) is the first term, d is the common difference between terms, and n is the nth term you are trying to find.

In this case, a(1) is 30, d is -30, and n is just n.

If you plug that into the equation and use the distributive property, you would get: a(n) = 30 -30n + 30.

Adding the two 30's gets you: a(n) = 60 - 30n, which is the solution.

7 0

4 years ago

Can I have help on this one

GaryK [48]

Answer:

4. 40°

5. 15

6.

I hope this helps. I don't know the last one.

4 0

3 years ago

Chlorofluorocarbons (CFCs) were used as propellants in spray cans until their buildup in the atmosphere started destroying the o

melamori03 [73]

Answer:

a) C(2000)=1915

C(2014)=1915

b) $C'(2000)=-\frac{275}{14}$

$C'(2014)=-\frac{275}{14}$

c) $C(t)=-\frac{275}{14}t+\frac{288405}{7}$

d) t=2021

e) too early

Step-by-step explanation:

a)

Since C(t) is the concentration of CFCs in ppt in year t, all we need to do to solve this part is determine what the concentration of CFCs is in years 2000 and 2014. Luckily for us, the problem already gives us those values, so:

C(2000)=1915 concentration in year 2000

C(2014)=1915 concentration in year 2014

b)

By definition, the derivative of a function at a given point is interpreted as the slope of the tangent line to the point of interest, so in order to find this answer, we need to find the slope of the line. Since the problem specifies that the behavior is linear, this means that the slope will always be the same no matter the year, so we get:

$m=C'(t)=\frac{C'(t_{2})-C'(t_{1})}{t_2-t_1}$

so:

$C'(t)=\frac{1640-1915}{2014-2000}=-\frac{275}{14}\approx -19.64$

therefore:

$C'(2000)=-\frac{275}{14}$

$C'(2014)=-\frac{275}{14}$

c)

Since the behavior is linear, we can calculate it with the point-slope form of the line which is:

$y-y_{1}=m(x-x_{1})$

in this case:

$C(t)-C(t_1)=C'(t)(t-t_{1})$

so we get:

$C(t)-1915=-\frac{275}{14}(t-2000)$

and we can now solve for C(t) so we get:

$C(t)=-\frac{275}{14}t+\frac{275000}{t}+1915$

for a final answer of:

$C(t)=-\frac{275}{14}t+\frac{288405}{7}$

d)

So next we solve the equation for C(t)=1500 so we get:

$1500=-\frac{275}{14}t+\frac{288405}{7}$

$1500-\frac{288405}{7}=-\frac{275}{14}t$

$-\frac{277905}{7}=-\frac{275}{14}t$

$t=2021 \frac{7}{55}$

so we will reach that concentration level at the year 2021 approximately.

e)

Since the second derivative of the concentration function is greater than zero, this means that the original function might be a function in the form: .

This means that the decrease of the concentration levels is slower than that of a linear equation. So the projected date will be too early than the real date.

8 0

3 years ago