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3 years ago

Solve this root2x-7 =7

Mathematics

2 answers:

torisob [31]3 years ago

7 0

Answer:

28.

Step-by-step explanation:

1) the range of x is: 2x-7>0; ⇔ x>3.5.

2) the solution is:

$\sqrt{2x-7}=7;$

2x-7=49;

2x=56; ⇔ x=28 (28>3.5, belongs to the range of x above).

Karo-lina-s [1.5K]3 years ago

5 0

-14??
Idek tbh it’s kinda hard

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12 less than the quotient of a number and seven is -2 find the number

kiruha [24]

A quotient is the result of a division.
Let the unknown number be x.

We translate the problem into an equation, piece by piece.

"quotient of a number and seven" ----> x/7

"12 less than the quotient of a number and seven" ----> x/7 - 12

The word "is" means "="
x/7 - 12 =

All the above is -2, so it equals -2.

x/7 - 12 = -2

$\dfrac{x}{7} - 12 = -2$

First, add 12 to both sides and simplify.

 $\dfrac{x}{7} - 12 + 12 = -2 + 12$


 $\dfrac{x}{7} = 10$ 

Next, multiply both sides by 7.

$\dfrac{x}{7} \times 7 = 10 \times 7$

$x = 70$

Answer: The number is 70.

5 0

3 years ago

Read 2 more answers

X + x * 7 * 14= 200 what is the x's

xxMikexx [17]

X= 200/99

Have a great day.

3 0

3 years ago

Read 2 more answers

Please Help me pleaseeeeee

lions [1.4K]

I think the answer is the second one

3 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:

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

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:

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

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

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:

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

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.

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

Info is on the picture

natali 33 [55]

Answer: C

Step-by-step explanation:

1 ml = 0.001 l

1 l = 1000 kl

6 0

4 years ago

Read 2 more answers