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3 years ago

What's 10 times 3 tens

2 answers:

8 0

Hey there!

You can think of these questions in terms of dollar bills. For example, "tens" can also be ten dollar bills. So, three tens would be:

$10 + $10 + $10 = $30

Now, multiply that number of dollars by 10. Remember a really good trick for doing mental math with multiplying by 10: just add a zero onto the end of the other number.

10 × $30 = 300

Your answer will be 300.

Hope this helped you out! :-)

nadya68 [22]3 years ago

6 0

$10+$10+$10 = $30

$30 x $10 = $300

Answer= $300

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Which relation is a function? {(1, 2), (2,4), (−1, 2), (0,3)}

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Step-by-step explanation:

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3 years ago

A triangle with vertices at A(0, 0), B(0, 4), and C(6, 0) is dilated to yield a triangle with vertices at A′(0, 0), B′(0, 10), a

Vesnalui [34]

Answer:

2.5

Step-by-step explanation:

A triangle ABC with vertices A(0, 0), B(0, 4), and C(6, 0) is dilated to form a triangle A'B'C' with vertices A′(0, 0), B′(0, 10), and C′(15, 0). If the center of dilation is point A or A' (the origin), then

$\dfrac{A'B'}{AB}=\dfrac{A'C'}{AC}.$

Since

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then the scale factor of the dilation is

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3 years ago

For fun question

ZanzabumX [31]

Consider the function $f(x)=x^{1/3}$ , which has derivative $f'(x)=\dfrac13x^{-2/3}$ .

The linear approximation of $f(x)$ for some value $x$ within a neighborhood of $x=c$ is given by

$f(x)\approx f'(c)(x-c)+f(c)$

Let $c=64$ . Then $(63.97)^{1/3}$ can be estimated to be

$f(63.97)\approxf'(64)(63.97-64)+f(64)$
$\sqrt[3]{63.97}\approx4-\dfrac{0.03}{48}=3.999375$

Since $f'(x)>0$ for $x>0$ , it follows that $f(x)$ must be strictly increasing over that part of its domain, which means the linear approximation lies strictly above the function $f(x)$ . This means the estimated value is an overestimation.

Indeed, the actual value is closer to the number 3.999374902...

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