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2 years ago

Give one example of when you would use the friendly numbers strategy to subtract. Explain why.

Mathematics

2 answers:

Harman [31]2 years ago

8 0

U can also use agarothim strategy

ziro4ka [17]2 years ago

6 0

In the <span>addition process</span> this usually involves:

<span><span>rounding numbers up to the nearest multiple of 10 or 100 </span>subtracting the extras that were added at the end.</span>

In the subtraction process this usually involves subtracting too much and completing the calculation by putting some back.

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I don't understand how to do this. Can anyone help. Preferably with steps. Thank you in advance!

Andrew [12]

You put the x=6
6³-4(6)²-12(6)+15=15

3 0

2 years ago

Julie will build a rectangular pen for her dog against a barn. A wall from the barn will form one side of the pen. She has 32 m

MAXImum [283]

Answer:

<h3> The width (side perpedicular to the barn): <u>x = 8 m</u></h3><h3> The lenght (side parallel to the barn): <u>y = 16 m</u> </h3>

Step-by-step explanation:

x - the width of the barn

She has 32 m of fencing so for the lenght remain (32-2x) m of fencing:

y = 32 - 2x

Area of the fencing: A = x•y

A(x) = x•(32 - 2x)

A(x) = -2x² + 32x ← quadratic function

The maximum value of quadratic function occurs at: $x=-\frac b{2a}$

a = -2, b = 32

$x=-\frac b{2a}=-\frac{32}{2\cdot(-2)}=-(-8)=8$

32-2x = 32 - 2•8 = 16

4 0

2 years ago

Hi guys, can anyone help me with this triple integral? Many thanks:)

Crank

Another triple integral. We're integrating over the interior of the sphere

$x^2+y^2+z^2=2^2$

Let's do the outer integral over z. z stays within the sphere so it goes from -2 to 2.

For the middle integral we have

$y^2=4-x^2-z^2$

x is the inner integral so at this point we conservatively say its zero. That means y goes from $-\sqrt{4-z^2}$ and $+\sqrt{4-z^2}$

Similarly the inner integral x goes between $\pm-\sqrt{4-y^2-z^2}$

So we rewrite the integral

$\displaystyle \int_{-2}^{2} \int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}} \int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dx \; dy \; dz$

Let's work on the inner one,

$\displaystyle\int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dz$

There's no z in the integrand, so we treat it as a constant.

$=(x^2+xy+y^2)z \bigg|_{z=-\sqrt{4-y^2-z^2}}^{z=\sqrt{4-y^2-z^2}}$

So the middle integral is

$\displaystyle\int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}}2(x^2+xy+y^2)\sqrt{4-y^2-z^2} \ dy$

I gotta go so I'll stop here, sorry.

7 0

2 years ago

Read 2 more answers

What is the quotient of 6 and 2?<br> ?<br> -12.<br> 1<br> 12.<br> 0<br><br> 12.<br> 0 12.

Julli [10]

To find the quotient you do 6/ (-1/2). You keep, change, flip when dealing with fractions.

7 0

3 years ago

I have another question, please explain don't just give me an answer. I need help on 18 part a, b, and c

morpeh [17]

Fo part a: You would have the answer of 6x +12 since both side is x (2x together) then the longer side would be 2x+6 but then u would have to multiply by 2 since there are 2 sides (4x+12) Then if u combined it you would get 6x+12 for part a

7 0

2 years ago

Read 2 more answers