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3 years ago

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There are currently 3 students signed up for a trip. The van can transport only 7 students. Which graph shows all the possible v

alues for the extra number of students that need to sign up so that more than one van is needed to transport them?

1 answer:

oksano4ka [1.4K]3 years ago

5 0

You did not enter the graph but if only 7 kids can ride one bus then more then 8 kids must join

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78 students are separated into groups of 8 for a field trip how many groups are there there are many students from a smaller gro

mixas84 [53]

Nine groups of 8 students and one group of 6 students. 9 x 8 = 72. 78 - 72 = 6.

5 0

3 years ago

51.83 to the nearest whole numbe

Westkost [7]

Answer:

52

Step-by-step explanation:

it's 52 I think, yeah .

4 0

3 years ago

If each interior angle of a regular polygon is 162 how many sides does it have

Yuri [45]

Answer:

20 sides

Step-by-step explanation:

The formula that relates interior angles with number of sides is:

$Each \ Interior = \frac{(n-2)*180}{n}$

Where

n is the number of sides.

We know each interior is 162 degrees, so we substitute and cross multiply and solve for n:

$Each \ Interior = \frac{(n-2)*180}{n}\\162=\frac{(n-2)180}{n}\\162n=(n-2)180\\162n=180n-360\\18n=360\\n=20$

THus, the polygon has 20 sides

4 0

3 years ago

What is the solution of the equation<br> 3.4 = 2p

liq [111]

We have that

3.4=2p----------> divide by 2 both members
3.4/2=2p/2------------> 1.7=p
p=1.7

the answer is p=1.7

7 0

3 years ago

Read 2 more answers

For which of the following compound inequalities is there no solution? Select one: a. 8m+5>=5 and -4m+6<=-10 b. 6m<=-36

enyata [817]

<span> $a.\\8m+5\geq5\
\ \ |-5\\8m\geq0\ \ \ |:8\\m\geq0\to m\in\left$

$b.\\6m\leq-36\ \ \ |:6\\m\leq-6\to m\in\left(-\infty;\
-6\right>\\\\m+24 > 20\ \ \ |-24\\m > -4\to m\in(-4;\
\infty)\\\\solution:\ m\in\left(-\infty;\ -6\right>\ \cap\ (-4;\ \infty)\to
m\in\O\to NO\ SOLUTION$

$c.\\-2m < 6\ \ \ |:(-2)\\m > -3\to m\in(-3;\ \infty)\\\\3m > 24\
\ \ |:3\\m > 8\to m\in(8;\ \infty)\\\\solution:\ m\in(-3;\ \infty)\ \cap\
(8;\ \infty)\to m\in(8;\ \infty)$

$d.\\m+9 < 8\ \ \ |-9\\m < -1\to m\in(-\infty;\ -1)\\\\-2m \geq10\ \
\ |:(-2)\\m\leq-5\to m\in\left(-\infty;-5\right>\\\\solution:\
m\in(-\infty;\ -1)\ \cap\ \left(-\infty;-5\right>\to
m\in\left(-\infty;-5\right>$ </span>

6 0

4 years ago