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3 years ago

Will enyone help me answer this question please

Mathematics

1 answer:

SSSSS [86.1K]3 years ago

6 0

You could use 4(division) sign 25 + 3 (division sign) 15 * 10, ok?

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Estimate the value of √70 to the nearest integer. Describe your process.

Anastaziya [24]

Answer:

$\large\boxed{\sqrt{70}\approx8}$

Step-by-step explanation:

$\text{We know:}\\\\\sqrt{a}=b\iff a^2=b,\ \text{for}\ a\geq0\ \wedge\ b\geq0\\\\5^2=25$

5 0

3 years ago

Cole is studying employment trends.

natali 33 [55]

Answer:

225

Step-by-step explanation:

So we know that 20% of the social workers he surveyed was 45 people. The easiest way to find all 100% of the social workers is to do 45 x 5 because 20% x 5 = 100%. So 45 x 5 = 225.

Hope this helped! :)

3 0

3 years ago

What Is The scale factor of the dilation?<br> A)-1/2<br> B)1/2<br> C)-2<br> D)2

Vadim26 [7]

Answer:

B) 1/2

Step-by-step explanation:

5 0

3 years ago

If a tank holds 5000 gallons of water, which drains from the bottom of the tank in 40 minutes, then Torricelli's Law gives the v

pochemuha

Answer:

V'(t) = $-250(1 - \frac{1}{40}t)$

If we know the time, we can plug in the value for "t" in the above derivative and find how much water drained for the given point of t.

Step-by-step explanation:

Given:

V = $5000(1 - \frac{1}{40}t )^2$ , where 0≤t≤40.

Here we have to find the derivative with respect to "t"

We have to use the chain rule to find the derivative.

V'(t) = $2(5000)(1 - \frac{1}{40} t)d/dt (1 - \frac{1}{40}t )$

V'(t) = $2(5000)(1 - \frac{1}{40} t)(-\frac{1}{40} )$

When we simplify the above, we get

V'(t) = $-250(1 - \frac{1}{40}t)$

If we know the time, we can plug in the value for "t" and find how much water drained for the given point of t.

4 0

3 years ago

Lowest common multiple of 12,18,56

Ira Lisetskai [31]

Step-by-step explanation:

12 = 2 × 2 × 3

18 = 2 × 3 × 3

56 = 2 ×2 × 2 × 7

Now

Common factor = 2

Remaining factor = 2 × 3 × 2 × 3 × 7

LCM = RF × CF

= 504

hence the lCM of 12 , 18 and 56 is 504...

$...$

8 0

2 years ago

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