Select the diagram showing a pair of alternate interior angles:

Mei does 6 problems in 18 minutes. How many can she complete in 12 minutes

Dmitriy789 [7]

Do cross multiplication

6 = x
18 12

Now cross multiply

(6)(12) = (18)(x)

72 = 18x

Divide both sides by 18

72 = 18x
18 18

4 = x

So tMei can complete 4 problems in 12 minutes.

5 0

3 years ago

Find all the critical points of h(x) = x^3 − 3x^4 and categorize them as local

neonofarm [45]

Answer:

0 is an inflection point

1/4 is a local maximum.

Step-by-step explanation:

To begin with you find the first derivative of the function and get that

$h'(x) = 3x^2 - 12x^3$

to find the critical points you equal the first derivative to 0 and get that

$3x^2 - 12x^3 = 0, x = 0,1/4$

To find if they are maximums or local minimums you use the second derivative.

$h''(x) = 6x-36x^2$

since $h''(0) = 0$ is neither an inflection point, and since $h''(1/4) = -3/4$ then 1/4 is a maximum.

6 0

3 years ago

While walking for exercise, Mia is able to walk only 3\4 as far on Saturday she walks 1 1/2 miles. How many miles did she walk F

Ket [755]

$1\frac{1}{2} - \frac{3}{4}$
$\frac{3}{2} - \frac{3}{4}$
$\frac{6}{4} - \frac{3}{4}$
= $\frac{3}{4}$

3 0

3 years ago

How does the slope of g(x) compare to the slope of f(x)?

Maru [420]

Answer:

The answer is 'The slope of g(x) is less than the slope of f(x)'

Step-by-step explanation:

Given the graphs of f(x) and g(x). we have to compare the slops of these two.

The graph of f(x) passes through the points (1,0) and (2,2)

∴ $\text{The slope of f(x)=}\frac{y_2-y_1}{x_2-x_1}=\frac{2-0}{2-1}=2$

The graph of g(x) passes through the points (0,2) and (2,3)

∴ $\text{The slope of g(x)=}\frac{y_2-y_1}{x_2-x_1}=\frac{3-2}{2-0}=\frac{1}{2}$

As $\frac{1}{2}$

This shows that the

The slope of g(x) is less than the slope of f(x).

8 0

3 years ago

Read 2 more answers

The sum of first three terms of a finite geometric series is -7/10 and their product is -1/125. [Hint: Use a/r, a, and ar to rep

Alchen [17]

Ooh, fun

geometric sequences can be represented as
$a_n=a(r)^{n-1}$
so the first 3 terms are
$a_1=a$
$a_2=a(r)$
$a_2=a(r)^2$

the sum is -7/10
$\frac{-7}{10}=a+ar+ar^2$
and their product is -1/125
$\frac{-1}{125}=(a)(ar)(ar^2)=a^3r^3=(ar)^3$

from the 2nd equation we can take the cube root of both sides to get
$\frac{-1}{5}=ar$
note that a=ar/r and ar²=(ar)r
so now rewrite 1st equation as
$\frac{-7}{10}=\frac{ar}{r}+ar+(ar)r$
subsituting -1/5 for ar
$\frac{-7}{10}=\frac{\frac{-1}{5}}{r}+\frac{-1}{5}+(\frac{-1}{5})r$
which simplifies to
$\frac{-7}{10}=\frac{-1}{5r}+\frac{-1}{5}+\frac{-r}{5}$
multiply both sides by 10r
-7r=-2-2r-2r²
add (2r²+2r+2) to both sides
2r²-5r+2=0
solve using quadratic formula
for $ax^2+bx+c=0$
$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$
so
for 2r²-5r+2=0
a=2
b=-5
c=2

$r=\frac{-(-5) \pm \sqrt{(-5)^2-4(2)(2)}}{2(2)}$
$r=\frac{5 \pm \sqrt{25-16}}{4}$
$r=\frac{5 \pm \sqrt{9}}{4}$
$r=\frac{5 \pm 3}{4}$
so
$r=\frac{5+3}{4}=\frac{8}{4}=2$ or $r=\frac{5-3}{4}=\frac{2}{4}=\frac{1}{2}$

use them to solve for the value of a
$\frac{-1}{5}=ar$
$\frac{-1}{5r}=a$
try for r=2 and 1/2
$a=\frac{-1}{10}$ or $a=\frac{-2}{5}$

test each
for a=-1/10 and r=2
a+ar+ar²= $\frac{-1}{10}+\frac{-2}{10}+\frac{-4}{10}=\frac{-7}{10}$
it works

for a=-2/5 and r=1/2
a+ar+ar²= $\frac{-2}{5}+\frac{-1}{5}+\frac{-1}{10}=\frac{-7}{10}$
it works

both have the same terms but one is simplified

the 3 numbers are $\frac{-2}{5}$ , $\frac{-1}{5}$ , and $\frac{-1}{10}$

6 0

3 years ago