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3 years ago

(05.07 MC)

Mathematics

2 answers:

Pavlova-9 [17]3 years ago

6 0

Answer:

0.65 square inch

Step-by-step explanation:

Each triangle has an area given by ...

A = 1/2bh = 1/2·(0.5 in)(0.4 in) = 0.10 in²

The square base has an area given by ...

A = s² = (0.5 in)² = 0.25 in²

The total area is that of the square base and 4 triangles:

S = 0.25 in² + 4×0.10 in² = 0.65 in²

vesna_86 [32]3 years ago

3 0

Answer:

0.65 square inch

Step-by-step explanation:

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HELP PLEASE!! this is 8th grade math btw

natima [27]

Answer:

the 2nd one

Step-by-step explanation:

5 0

3 years ago

How many pairs of consecutive natural numbers have a product of less than 40000? I am in 5th grade. This is supposed to be easy

Ugo [173]

Answer:

There are 199 pairs of consecutive natural numbers whose product is less than 40000.

Step-by-step explanation:

We notice that such statement can be translated into this inequation:

$n \cdot (n+1) < 40000$

Now we solve this inequation to the highest value of $n$ that satisfy the inequation:

$n^{2}+n < 40000$

$n^{2}+n -40000$

The Quadratic Formula shows that roots are:

$n_{1,2} = \frac{-1\pm\sqrt{1^{2}-4\cdot (1)\cdot (-40000)}}{2\cdot (1)}$

$n_{1,2} = -\frac{1}{2}\pm \frac{1}{2} \cdot \sqrt{160001}$

$n_{1} = -\frac{1}{2}+\frac{1}{2}\cdot \sqrt{160001}$

$n_{1} \approx 199.501$

$n_{2} = -\frac{1}{2}-\frac{1}{2}\cdot \sqrt{160001}$

$n_{2} \approx -200.501$

Only the first root is valid source to determine the highest possible value of $n$ , which is $n_{max} = 199$ . Each natural number represents an element itself and each pair represents an element as a function of the lowest consecutive natural number. Hence, there are 199 pairs of consecutive natural numbers whose product is less than 40000.