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3 years ago

9

I need help on these 3 questions. Please help. “Find the area of each figure. Round your answer to the nearest tenth, show your

work.

1 answer:

My name is Ann [436]3 years ago

6 0

The first one is 6x10+6x7/2=81

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A train is going at 200/KM per hour

olya-2409 [2.1K]

Answer:

<u>It</u><u> </u><u>moved</u><u> </u><u>for</u><u> </u><u>1</u><u>5</u><u>2</u><u>.</u><u>7</u><u> </u><u>seconds</u>

Step-by-step explanation:

Time taken/ period, T :

${ \sf{T = \frac{distance}{speed} }}$

substitute the variables:

${ \sf{T = \frac{8482}{(200 \times 1000)} }} \\ \\ { \sf{T =0.042 \: hours }}$

In seconds:

${ \sf{T = (0.042 \times 3600) \: s}} \\ { \sf{T = 152.7 \: seconds}}$

6 0

2 years ago

Read 2 more answers

20 is a factor of 24 true or false

igor_vitrenko [27]

False, 20 is not a factor of 24

3 0

2 years ago

Read 2 more answers

The value 5 is an upper bound for the zeros of the function shown below.

Mice21 [21]

Answer:

The given statement that value 5 is an upper bound for the zeros of the function f(x) = x⁴ + x³ - 11x² - 9x + 18 will be true.

Step-by-step explanation:

Given

$f\left(x\right)\:=\:x^2\:+\:x^3\:-\:11x^2\:-\:9x\:+\:18$

We know the rational zeros theorem such as:

if $x=c$ is a zero of the function $f(x)$ ,

then $f(c) = 0$ .

As the $f\left(x\right)\:=\:x^2\:+\:x^3\:-\:11x^2\:-\:9x\:+\:18$ is a polynomial of degree $4$ , hence it can not have more than $4$ real zeros.

Let us put certain values in the function,

$f(5) = 448$ , $f(4) = 126$ , $f(3) = 0$ , $f(2) = -20$ ,

$f(1) = 0$ , $f(0) = 18$ , $f(-1) = 16$ , $f(-2) = 0$ , $f(-3) = 0$

From the above calculation results, we determined that $4$ zeros as

$x = -3, -2, 1,$ and $3$ .

Hence, we can check that

$f(x) = (x+3)(x+2)(x-1)(x-3)$

Observe that,

$for x > 3$ , $f(x)$ increases rapidly, so there will be no zeros for $x>3$ .

Therefore, the given statement that value 5 is an upper bound for the zeros of the function f(x) = x⁴ + x³ - 11x² - 9x + 18 will be true.

5 0

3 years ago

If 8 identical blackboards are to be divided among 4 schools, how many divisions are possible? how many if each school must rece

DaniilM [7]

Alright, so we'd use the combinations with repetition formula, so we choose from 4 schools to distribute to and distribute 8 blackboards. It's then

( 8+4-1)!/8!(4-1)!=11!/(3!*8!)=165

For at least one blackboard, we first distribute 1 to each school and then have 4 blackboards left, getting (4+4-1)!/4!(4-1)!=7!/(4!*3!)=35

8 0

3 years ago

A school has 400 students. They all come to school by bus, and each bus carries the same number of students. How many students m

rodikova [14]

Answer:

100 students on each bus

Step-by-step explanation:

100*4 is 400 so there might be just 4

5 0

3 years ago

Read 2 more answers