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3 years ago

7

What the difference of -5-(-9)?

2 answers:

Svet_ta [14]3 years ago

5 0

Answer:

4

Step-by-step explanation:

-5-(-9)

when subtracting by a negative like : -(-9) it will cancel into it being + that number so -5 + 9 is 4

Dafna11 [192]3 years ago

3 0

Answer:

4

Step-by-step explanation:

Subtracting a negative number from a negative number – a minus sign followed by a negative sign, turns the two signs into a plus sign. So, instead of subtracting a negative, you are adding a positive. Basically, - (-4) becomes +4, and then you add the numbers. For example, say we have the problem -2 - –4.

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Evaluate the function.<br><br> g(x)=10x+3<br><br> g(19.6)=

Zolol [24]

Answer:

199

Step-by-step explanation:

plug in x=19.6, then

g(19.6) = 10 * 19.6 +3= 199

7 0

3 years ago

How to solve a system of two linear equations:

antiseptic1488 [7]

Answer:

Step-by-step explanation:

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3 years ago

In a classroom of 28 students: 19 of the students have brown eyes 13 of the students wear glasses 7 of the students do not have

V125BC [204]

Total students = 28

Students with brown eyes = 19

students wear glasses = 13

Not brown eyes or no glasses = 7

Not brown eyes = 28 - 19 = 9

Not wearing glasses = 28 - 13 = 15

Not brown eyes and wearing glasses = 15 - 9 = 6

Brown eyes and wearing glasses = 13 - 6 = 7 <==== answer

7 0

3 years ago

Read 2 more answers

Suppose that a spherical droplet of liquid evaporates at a rate that is proportional to its surface area: where V = volume (mm3

Alex

Answer:

V = 20.2969 mm^3 @ t = 10

r = 1.692 mm @ t = 10

Step-by-step explanation:

The solution to the first order ordinary differential equation:

$\frac{dV}{dt} = -kA$

Using Euler's method

$\frac{dVi}{dt} = -k *4pi*r^2_{i} = -k *4pi*(\frac {3 V_{i} }{4pi})^(2/3)\\ V_{i+1} = V'_{i} *h + V_{i} \\$

Where initial droplet volume is:

$V(0) = \frac{4pi}{3} * r(0)^3 = \frac{4pi}{3} * 2.5^3 = 65.45 mm^3$

Hence, the iterative solution will be as next:

i = 1, ti = 0, Vi = 65.45

$V'_{i} = -k *4pi*(\frac{3*65.45}{4pi})^(2/3) = -6.283\\V_{i+1} = 65.45-6.283*0.25 = 63.88$

i = 2, ti = 0.5, Vi = 63.88

$V'_{i} = -k *4pi*(\frac{3*63.88}{4pi})^(2/3) = -6.182\\V_{i+1} = 63.88-6.182*0.25 = 62.33$

i = 3, ti = 1, Vi = 62.33

$V'_{i} = -k *4pi*(\frac{3*62.33}{4pi})^(2/3) = -6.082\\V_{i+1} = 62.33-6.082*0.25 = 60.813$

We compute the next iterations in MATLAB (see attachment)

Volume @ t = 10 is = 20.2969

The droplet radius at t=10 mins

$r(10) = (\frac{3*20.2969}{4pi})^(2/3) = 1.692 mm\\$

The average change of droplet radius with time is:

Δr/Δt = $\frac{r(10) - r(0)}{10-0} = \frac{1.692 - 2.5}{10} = -0.0808 mm/min$

The value of the evaporation rate is close the value of k = 0.08 mm/min

Hence, the results are accurate and consistent!

5 0

3 years ago

Venita is factoring the expression 32ab-ab. She determines the GCF and writes the factored expression as 8b(4a-0) Which best des

Ymorist [56]

A) she incorrectly determined the GCF.

The GCF of 32ab - ab would be ab, not 8b.

7 0

3 years ago

Read 2 more answers