All categories

3 years ago

Find the x intercept of the line graphed by the equation 1/2x-7y=7

Mathematics

1 answer:

Rina8888 [55]3 years ago

3 0

Answer:

1/2*x - 7y = 7

1/2*x - 7 = 7y

1/14*x - 1 = y

y = 1/14*x - 1

You might be interested in

Sammi has $125.75 in her savings account. She deposits $25.50 into the account each month for the next 6 months. How much is in

AnnZ [28]

Hi,

Sammy has $125.75 in her savings account. She deposits $25.50 into the account each month for the next 6 months.

How much is in Sammy's account at the end of the 6 months?

25.50 × 6 = $153

In 6 months she deposits $153 into the account.

So :

125.75 + 153 = $278.75

At the end of the 6 months Sammy has $278.75 in her savings account.

7 0

3 years ago

Find the slope of the line with the given equation. Show your work. 6x – 2y = 18

Vera_Pavlovna [14]

Currently the equation 6x - 2y = 18 is in standard form. Convert the standard form equation into a slope-intercept form and we can find the slope easily.

Solve for y.

6x - 2y = 18
-2y = 18 - 6x <-- Subtract 6x from each side. This is to isolate the 2y term
2y / -2 = $\frac{18 - 6x}{2}$ <-- Divide each side by 2. This it to
get rid of the 2 coefficient.
y = -9 + 3x

Rearrange the right-hand side a bit.

y = -9 + 3x becomes y = 3x - 9

Now it is in slope-intercept form.
The slope is the coefficient of the x variable.

So, 3 is the slope.

7 0

3 years ago

Max is taking 38 children for rides in

dem82 [27]

Answer: 8 turns

Step-by-step explanation: 5 children at a time, so 5*8=40, which is the closest to 38

7 0

3 years ago

For a square with a side length x + 1 , what is the area if x=9

Svetach [21]

Answer:

100 units squared

Step-by-step explanation:

The area of a square is A=s^2, where s is the side length. So, in your problem, the area would be A=(x+1)^2. If x = 9, then the formula would be A = (9+1)^2 = 100.

6 0

2 years ago

Read 2 more answers

The graph of the continuous function g, the derivative of the function f, is shown above. The function g is piecewise linear for

4vir4ik [10]

A) $g=f'$ is continuous, so $f$ is also continuous. This means if we were to integrate $g$ , the same constant of integration would apply across its entire domain. Over $0$ , we have $g(x)=2x$ . This means that

$f_{0$

For $f$ to be continuous, we need the limit as $x\to1^-$ to match $f(1)=3$ . This means we must have

$\displaystyle\lim_{x\to1}x^2+C=1+C=3\implies C=2$

Now, over $x$ , we have $g(x)=-3$ , so $f_{x$ , which means $f(-5)=17$ .

b) Integrating over [1, 3] is easy; it's just the area of a 2x2 square. So,

$\displaystyle\int_1^6g(x)=4+\int_3^62(x-4)^2\,\mathrm dx=4+6=10$

c) $f$ is increasing when $f'=g>0$ , and concave upward when $f''=g'>0$ , i.e. when $g$ is also increasing.

We have $g>0$ over the intervals $0$ and $x>4$ . We can additionally see that $g'>0$ only on $0$ and $x>4$ .

d) Inflection points occur when $f''=g'=0$ , and at such a point, to either side the sign of the second derivative $f''=g'$ changes. We see this happening at $x=4$ , for which $g'=0$ , and to the left of $x=4$ we have $g$ decreasing, then increasing along the other side.

We also have $g'=0$ along the interval $-1$ , but even if we were to allow an entire interval as a "site of inflection", we can see that $g'>0$ to either side, so concavity would not change.

5 0

3 years ago