A trapezoid with area of 195cmhas bases 7 cm long and 32cm long. What is

I need help pleaseeee

stealth61 [152]

Answer:

its 65

Step-by-step explanation:

welcome girl

3 0

3 years ago

7x – 2y = -5 which is the correct answer help a homie out pls

Elis [28]

Neither of those work for that equation

7 0

3 years ago

Find the circumference of the circle. Round your answer to the nearest hundredth. Use 3.14 or pi circle is 3 ft in radius

densk [106]

Answer:

18.84 feet

Step-by-step explanation:

The circumference of a circle can be found using:

c=π*d

The radius is given, but we have to find diameter. The diameter is twice the radius, or

d=2r

The radius is 3 feet. Substitute 3 in for r.

d=2*3

d=6

The diameter is 6 feet.

Now we have the diameter. We can substitute 6 in for d in the circumference formula. We can also substitute 3.14 in for pi.

c=π*d

c=3.14*6

c=18.84

The circumference is 18.84 feet

8 0

4 years ago

Read 2 more answers

Solve <img src="https://tex.z-dn.net/?f=%5Csf%20%5Cdfrac%7B1%7D%7Bp%7D%20%2B%20%5Cdfrac%7B1%7D%7Bq%7D%20%2B%20%5Cdfrac%7B1%7D

Nostrana [21]

Answer:

$\displaystyle \begin{cases} \displaystyle {x} _{1} = - p \\ \displaystyle x _{2} = - q \end{cases}$

Step-by-step explanation:

we would like to solve the following equation for x:

$\displaystyle \frac{1}{p} + \frac{1}{q} + \frac{1}{x} = \frac{1}{p + q + x}$

to do so isolate $\frac{1}{x}$ to right hand side and change its sign which yields:

$\displaystyle \frac{1}{p} + \frac{1}{q} = \frac{1}{p + q + x} - \frac{1}{x}$

simplify Substraction:

$\displaystyle \frac{1}{p} + \frac{1}{q} = \frac{x - (q + p + x)}{x(p + q + x)}$

get rid of only x:

$\displaystyle \frac{1}{p} + \frac{1}{q} = \frac{ - (q + p )}{x(p + q + x)}$

simplify addition of the left hand side:

$\displaystyle \frac{q + p}{pq} = \frac{ - (q + p )}{x(p + q + x)}$

divide both sides by q+p Which yields:

$\displaystyle \frac{1}{pq} = \frac{ -1}{x(p + q + x)}$

cross multiplication:

$\displaystyle x(p + q + x) = - pq$

distribute:

$\displaystyle xp + xq + {x}^{2} = - pq$

isolate -pq to the left hand side and change its sign:

$\displaystyle xp + xq + {x}^{2} + pq = 0$

rearrange it to standard form:

$\displaystyle {x}^{2} + xp + xq + pq = 0$

now notice we end up with a quadratic equation therefore to solve so we can consider factoring method to use so

factor out x:

$\displaystyle x( {x}^{} + p ) + xq + pq = 0$

factor out q:

$\displaystyle x( {x}^{} + p ) +q (x + p)= 0$

group:

$\displaystyle ( {x}^{} + p ) (x + q)= 0$

by Zero product property we obtain:

$\displaystyle \begin{cases} \displaystyle {x}^{} + p = 0 \\ \displaystyle x + q= 0 \end{cases}$

cancel out p from the first equation and q from the second equation which yields:

$\displaystyle \begin{cases} \displaystyle {x}^{} = - p \\ \displaystyle x = - q \end{cases}$

and we are done!

3 0

3 years ago

If you plotted the points in the table and connected

castortr0y [4]

Answer:

U shaped.

Step-by-step explanation:

When x = 0 , f(x) = 6

when x =1 yf(x) = 0

when x = 2 f(x) = -2

x = 3 f(x) = 0

x = 4 f(x) = 6.

So the graphs falls from the left and rises to the right in the form of a U.

You can draw a rough graph to confirm this.

3 0

3 years ago