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3 years ago

I need help please?

Mathematics

1 answer:

ludmilkaskok [199]3 years ago

7 0

Answer:

Step-by-step explanation:

the blue parabola shows (1/4x)^2, the red shoes x^2, you can see the horizontal stretch, hope this helps :)

Download pdf

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How do you do a ratio tape diagram

sp2606 [1]

15 degrees hope this helps mark as brainliest

7 0

3 years ago

How to find factors of any numbers?

JulijaS [17]

Answer:Find all the numbers less than or equal to the given number.

Divide the given number by each of the numbers.

The divisors that give the remainder to be 0 are the factors of the number.

Step-by-step explanation:Find all the numbers less than or equal to the given number.

Divide the given number by each of the numbers.

The divisors that give the remainder to be 0 are the factors of the number.

3 0

3 years ago

On the distant plant, Mathology, a sports area covers 7400 yodels2. How many square deckles would this be if 1 deckle = 75.9 yod

STatiana [176]

Answer:

It would be <u>97.50</u> square deckles.

Step-by-step explanation:

Given:

On the distant plant, Mathology, a sports area covers 7400 yodels².

1 deckle = 75.9 yodels.

Now, to get the square deckles.

As given, 1 deckle = 75.9 yodels.

So, to get the square deckles by using conversion factor:

<em>75.9 yodels = 1 deckle.</em>

7400 yodels² = $7400\div 75.9$

= $97.50\ deckels ^2.$

Therefore, it would be 97.50 square deckles.

8 0

4 years ago

Consider f and c below. f(x, y, z) = (y2z + 2xz2)i + 2xyzj + (xy2 + 2x2z)k,

qaws [65]

$\dfrac{\partial f}{\partial x}=y^2z+2xz^2$
$f(x,y,z)=xy^2z+x^2z^2+g(y,z)$

$\dfrac{\partial f}{\partial y}=2xyz+\dfrac{\partial g}{\partial y}=2xyz$
$\dfrac{\partial g}{\partial y}=0\implies g(y,z)=h(z)$

$\dfrac{\partial f}{\partial z}=xy^2+2x^2z+\dfrac{\mathrm dh}{\mathrm dz}=xy^2+2x^2z$
$\dfrac{\mathrm dh}{\mathrm dz}=0\implies h(z)=0$

$\implies f(x,y,z)=xy^2z+x^2z^2+C$

8 0

3 years ago

Need help ASAP

valentinak56 [21]

Answer:

Step-by-step explanation:

700x.10=630

630x.05=31.5

630-31.5=598.50

7 0

3 years ago

I need help please?​

I need help please?