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3 years ago

In the following number, which digit is in the hundredths place? 745312.269784

Mathematics

2 answers:

fenix001 [56]3 years ago

8 0

Answer:

6 is in the hundreths place

Step-by-step explanation:

After the decimal point it goes tenths, hundredths, thousandths, ten thousandths, and so on.

Greeley [361]3 years ago

6 0

Answer:

The second number after the decimal point- the 6

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My base is 10. My exponent is an odd number. If you write a multiplication sentence to show my value, it will be have fewer than

bearhunter [10]

You need to write 1×5×2=10

8 0

3 years ago

Sarah used 2.5 cups of cheese in a dish that serves 10 people. Amyn used 1.6 cups of cheese in a dish that serves 8 people.

Drupady [299]

Is D
Explanation : you have to divide the amounts too see how many cheese for one person and then you subtract them and you get you answer that will be like this
2.5 divided by 10 = 0.25
1.6 divided by 8 = 0.2
0.25 - 0.2 = 0.05

5 0

3 years ago

Find the coordinates of P ' if D o, 2.4 (x,y) is applied to the triangle below

guajiro [1.7K]

Answer:

The coordinates of P' are (4.8,-4.8).

Step-by-step explanation:

The rule of dilation $D_{o,2.4}(x,y)$ represent the dilation with scale factor 2.4 and center at origin.

If the scale factor of the dilation is k and the center is (0,0), then

$(x,y)\rightarrow (kx,ky)$

Since the scale factor is 2.4, therefore

$(x,y)\rightarrow (2.4x,2.4y)$

From the given figure it is noticed that the coordinates of P are (2,-2). The coordinates of P' are

$P(2,-2)\rightarrow P'(2.4(2),2.4(-2))$

$P(2,-2)\rightarrow P'(4.8,-4.8)$

Therefore the coordinates of P' are (4.8,-4.8).

3 0

3 years ago

1 pt) If a parametric surface given by r1(u,v)=f(u,v)i+g(u,v)j+h(u,v)k and −4≤u≤4,−4≤v≤4, has surface area equal to 1, what is t

natta225 [31]

The area of the surface given by $\vec r_1(u,v)$ is 1. In terms of a surface integral, we have

$1=\displaystyle\int_{-4}^4\int_{-4}^4\left\|\frac{\partial\vec r_1(u,v)}{\partial u}\times\frac{\partial\vec r_1(u,v)}{\partial v}\right\|\,\mathrm du\,\mathrm dv$

By multiplying each component in $\vec r_1$ by 5, we have

$\dfrac{\partial\vec r_2(u,v)}{\partial u}=5\dfrac{\partial\vec r_1(u,v)}{\partial u}$

and the same goes for the derivative with respect to $v$ . Then the area of the surface given by $\vec r_2(u,v)$ is

$\displaystyle\int_{-4}^4\int_{-4}^425\left\|\frac{\partial\vec r_1(u,v)}{\partial u}\times\frac{\partial\vec r_1(u,v)}{\partial v}\right\|\,\mathrm du\,\mathrm dv=\boxed{25}$