All categories

3 years ago

The fifth grade classes at Brookfield school used five identical buses to go on a field trip

1 answer:

7 0

Total no. of seats:

40×5=200

4 buses were filled- 40 × 4= 160

fifth bus was 4/5 filled- 32 seats

1/8 of total passengers is adults.

192 ÷ 8 = 24

You might be interested in

Jamie wants to run 2 3/4 miles today. He has already run 1 1/4 miles. How much farther must he run to reach his goal?

GaryK [48]

Working out:
2 3/4 ー 1 1/4 = 1 2/4 = 1 1/2

Answer:
He needs to run 1 1/2 more miles.

4 0

3 years ago

Read 2 more answers

4x + 8 = 32 pleas eexplain

Sedbober [7]

Answer:

x = 6

Step-by-step explanation:

Subtract 8 from both sides of the equation
Simplify
Subtract the numbers
Divide both sides of the equation by the same term
Simplify
Cancel terms that are in both the numerator and denominator
Divide the numbers

5 0

3 years ago

Read 2 more answers

ALG II NEED HELP QUICK! Will mark the brainliest answer

lbvjy [14]

I think it’s the second one because The formula would be
x=
2a
−b±
b
2
−4ac

4 0

3 years ago

The population of a local species of dragonfly can be found using an infinite geometric series where a1=42 and the common ratio

san4es73 [151]

$\bf \textit{sum of an infinite geometric serie}\\\\
\stackrel{for~~|r|\ \textless \ 1}{S=\sum\limits_{i=0}^{\infty}~a_1r^i\implies \cfrac{a_1}{1-r}}\qquad 
\begin{cases}
a_1=\textit{first term's value}\\
r=\textit{common ratio}\\
----------\\
a_1=42\\
r=\frac{3}{4}
\end{cases}
\\\\\\
S=\cfrac{42}{1-\frac{3}{4}}\implies S=\cfrac{42}{\frac{1}{4}}\implies S=164$

bearing in mind that, the geometric sequence is "convergent" only when |r|<1, or namely "r" is a fraction between 0 and 1.

3 0

3 years ago

What is the 23rd term of the arithmetic sequence where a1 = 8 and a9 = 48?

joja [24]

Answer:

23rd term of the arithmetic sequence is 118.

Step-by-step explanation:

In this question we have been given first term a1 = 8 and 9th term a9 = 48

we have to find the 23rd term of this arithmetic sequence.

Since in an arithmetic sequence

$T_{n}=a+(n-1)d$

here a = first term

n = number of term

d = common difference

since 9th term a9 = 48

48 = 8 + (9-1)d

8d = 48 - 8 = 40

d = 40/8 = 5

Now $T_{23}= a + (n-1)d$

= 8 + (23 -1)5 = 8 + 22×5 = 8 + 110 = 118

Therefore 23rd term of the sequence is 118.

4 0

3 years ago